t-Formulae

Topic

Course Support Current PowerPoint version

A2 June 2025 Q5

EdexcelCurrent spec8 markst-Formulae

(a) Use the substitution \(t = \tan\left(\dfrac{x}{2}\right)\) to show that\[\int \frac{1}{1 + \operatorname{cosec} x}\,\mathrm{d}x = \int \frac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2}\,\mathrm{d}t\] (3)

(b) Hence show that\[\int \frac{1}{1 + \operatorname{cosec} x}\,\mathrm{d}x = Ax + \frac{B}{1 + \tan\left(\frac{x}{2}\right)} + k\]where \(A\) and \(B\) are constants to be determined and \(k\) is an arbitrary constant. (5)

Mark scheme (a)

Scheme	Marks	AO
\(1 + \operatorname{cosec} x = 1 + \dfrac{1}{\sin x} = 1 + \dfrac{1}{\left(\frac{2t}{1 + t^2}\right)}\) or \(1 + \dfrac{1 + t^2}{2t}\left(= \dfrac{t^2 + 2t + 1}{2t}\right)\) or \(1 + \operatorname{cosec} x = \dfrac{\sin x + 1}{\sin x} = \dfrac{\frac{2t}{1 + t^2} + 1}{\frac{2t}{1 + t^2}}\)	B1	1.1a
\(\displaystyle\int \frac{1}{1 + \operatorname{cosec} x}\,\mathrm{d}x = \int \frac{2t}{t^2 + 2t + 1} \times \frac{2}{1 + t^2}\,\mathrm{d}t\)	M1	2.1
\(\displaystyle\int \frac{1}{1 + \operatorname{cosec} x}\,\mathrm{d}x = \int \frac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2}\,\mathrm{d}t\ *\)	A1*	1.1b
	(3)

Notes

B1: Selects the correct formulae to express \(1 + \operatorname{cosec} x\) in terms of \(t\). May be seen within the integral

M1: Makes a complete substitution to obtain an integral in terms of \(t\) only, including \(\mathrm{d}x = \dfrac{2}{1 + t^2}\,\mathrm{d}t\)

A1*: Correct proof with no errors or omissions, must see the factorisation if the denominator is expanded first. Long division or equivalent with \(t^2 + 1\) or \((t + 1)^2\) or \(t + 1\)

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2} \equiv \dfrac{At + B}{1 + t^2} + \dfrac{C}{1 + t} + \dfrac{D}{(1 + t)^2}\) or \(\dfrac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2} \equiv \dfrac{At + B}{1 + t^2} + \dfrac{Ct + D}{(1 + t)^2}\) or \(\dfrac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2} \equiv \dfrac{At + B}{1 + t^2} + \dfrac{C}{1 + t} + \dfrac{Dt + E}{(1 + t)^2}\) Leading to finding the value for at least one constant	M1	3.1a
\(\dfrac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2} \equiv \dfrac{2}{1 + t^2} - \dfrac{2}{(1 + t)^2}\)	A1	1.1b
\(\displaystyle\int \frac{2}{1 + t^2} - \frac{2}{(1 + t)^2}\,\mathrm{d}t = 2\arctan t + \frac{2}{1 + t}\)	A1	2.2a
\(= 2\arctan\left(\tan\frac{x}{2}\right) + \dfrac{2}{1 + \tan\frac{x}{2}} + k = x + \dfrac{2}{1 + \tan\frac{x}{2}} + k\) cso	M1 A1	1.1b 2.1
	(5)
	(8 marks)

Notes

M1: Realises the need to express the integrand in terms of partial fractions and attempts the correct form, condone the use of \(x\) instead of \(t\) on their numerator with \(t\)’s on the denominator. Must attempt to find at least one constant term

A1: Correct partial fractions.

A1: Correct integration, must have scored the previous marks, no need for the constant of integration
Note an incorrect form such as \(\dfrac{A}{1 + t^2} + \dfrac{B}{1 + t} + \dfrac{C}{(1 + t)^2}\) can lead to a correct answer but is M0A0A0

M1: Reverses the substitution to express in terms of \(x\)

A1: Correct answer with no errors including “\(+ k\)” must have scored all the previous marks in part (b)

Note Verification method using \(\dfrac{A}{1 + t^2} + \dfrac{B}{(1 + t)^2}\) to find the values for \(A\) and \(B\), please send to review

AS June 2025 Q1

EdexcelAS paperCurrent spec9 markst-Formulae

In this question you must show all stages of your working.

Solutions based entirely on calculator technology are not acceptable.

The surface temperature of the water in a lake during a particular year is modelled by the equation

\[S = 12 - \frac{15}{2}\cos x^\circ - \frac{27}{10}\sin x^\circ \qquad (\text{I})\]

where \(S\) is the temperature in degrees Celsius and \(x\) is the number of days after the start of the year.

(a) Use the model to write down the surface temperature of the water in the lake at the start of the year. (1)

Using the substitution \(t = \tan\left(\dfrac{x}{2}\right)\)

(b) show that equation \((\text{I})\) can be rewritten as\[S = \frac{At^2 + Bt + C}{10(1 + t^2)}\]where \(A\), \(B\) and \(C\) are integers to be determined. (3)

(c) Hence determine, according to the model, the number of days after the start of the year when the surface temperature of the water in the lake is \(10^\circ\mathrm{C}\) for the second time that year. Give your answer to the nearest day. (5)

Scheme	Marks	AO
\(x = 0 \Rightarrow 12 - 7.5 = 4.5^\circ\mathrm{C}\)	B1	3.4
	(1)

Mark scheme (b)

Scheme	Marks	AO
\(S = 12 - \dfrac{15}{2}\left(\dfrac{1 - t^2}{1 + t^2}\right) - \dfrac{27}{10}\left(\dfrac{2t}{1 + t^2}\right)\)	M1	1.1b
\(= \dfrac{120(1 + t^2) - 75(1 - t^2) - 54t}{10(1 + t^2)}\)	M1	1.1b
\(= \dfrac{195t^2 - 54t + 45}{10(1 + t^2)}\)	A1	2.1
	(3)

Notes

M1: Uses the correct t-formulae to obtain \(S\) in terms of \(t\)

M1: Correct method to obtain a common denominator,
\(A + \dfrac{B(1 - t^2)}{1 + t^2} + \dfrac{Ct}{1 + t^2} = \dfrac{A(1 + t^2) + B(1 - t^2) + Ct}{(1 + t^2)}\)

A1: Correct answer in the correct form

(Corrected from the printed mark scheme: the second line is printed as \(\dfrac{120(1 + t^2) - 75(1 - t^2)3 - 54t}{10(1 + t^2)}\), with a stray 3.)

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{195t^2 - 54t + 45}{10(1 + t^2)} = 10 \Rightarrow 195t^2 - 54t + 45 = 100 + 100t^2\)	M1	3.4
\(95t^2 - 54t - 55 = 0\)	A1	1.1b
\(t = 1.09\ldots, \Rightarrow \dfrac{x}{2} = \tan^{-1} 1.09\ldots = \ldots\{47.4\ldots \text{ or } 0.828\ldots\}\) or \(-0.528\ldots \Rightarrow \dfrac{x}{2} = \tan^{-1}(-0.528) = \ldots\{-27.83\ldots \text{ or } -0.485\ldots\}\) Alternative: Leading to a value for x \(t = 1.09\ldots \Rightarrow \sin x = \dfrac{2(1.09)}{1 + (1.09)^2}\) or \(\cos x = \dfrac{1 - (1.09)^2}{1 + (1.09)^2}\) Or \(t = -0.528\ldots \Rightarrow \sin x = \dfrac{2(-0.528)}{1 + (-0.528)^2}\) or \(\cos x = \dfrac{1 - (-0.528)^2}{1 + (-0.528)^2}\)	dM1	3.4
\(\dfrac{x}{2} = \tan^{-1}(-0.528) + 180 = 152.1\ldots \Rightarrow x = \ldots\) \(x = 360 + \sin^{-1}\left(\dfrac{2(-0.528)}{1 + (-0.528)^2}\right)\) \(x = 360 - \cos^{-1}\left(\dfrac{1 - (-0.528)^2}{1 + (-0.528)^2}\right)\)	ddM1	3.1b
304 days	A1	3.2a
	(5)
	(9 marks)

Notes

M1: Uses \(S = 10\) with the model and multiplies up to obtain a quadratic equation in \(t\)

A1: Correct 3TQ

dM1: Solves their 3TQ in \(t\) and proceeds to obtain at least one value of \(\dfrac{x}{2}\) as suggested by the model. Alternative uses \(t\) formulae for \(\sin x\) or \(\cos x\) to find a value for \(x\), degrees or radians

ddM1: A fully correct strategy to find the required value of \(x\) from the negative root of the quadratic equation in \(t\). They must be using degrees for this mark. This mark may be award even if more than one value found e.g \(x = 95.2,\ 304.3\)

A1: Correct number of days, must be clear that this is their answer

A2 June 2024 Q7

EdexcelCurrent spec7 markst-Formulae

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

(a) Use the substitution \(t = \tan\left(\dfrac{\theta}{2}\right)\) to show that\[\int \frac{1}{2\sin\theta + \cos\theta + 2}\,\mathrm{d}\theta = \int \frac{a}{(t + b)^2 + c}\,\mathrm{d}t\]where \(a\), \(b\) and \(c\) are constants to be determined. (3)

(b) Hence show that\[\int_{\frac{\pi}{2}}^{\frac{2\pi}{3}} \frac{1}{2\sin\theta + \cos\theta + 2}\,\mathrm{d}\theta = \ln\left(\frac{2\sqrt{3}}{3}\right)\] (4)

Mark scheme (a)

Scheme	Marks	AO
\(\displaystyle\int \frac{1}{2\sin\theta + \cos\theta + 2}\,\mathrm{d}\theta = \int \frac{1}{2\left(\frac{2t}{1 + t^2}\right) + \left(\frac{1 - t^2}{1 + t^2}\right) + 2} \times \frac{2}{1 + t^2}\,\mathrm{d}t\)	M1	2.1
\(\displaystyle\int \frac{2}{4t + \left(1 - t^2\right) + 2\left(1 + t^2\right)}\,\mathrm{d}t = \int \frac{2}{t^2 + 4t + 3}\,\mathrm{d}t\)	M1	1.1b
\(\displaystyle\int \frac{2}{(t + 2)^2 - 1}\,\mathrm{d}t\)	A1	2.2a
	(3)

Notes

M1: Applies the correct substitutions \(\sin\theta = \dfrac{2t}{1 + t^2}\), \(\cos\theta = \dfrac{1 - t^2}{1 + t^2}\) and \(\mathrm{d}\theta = \dfrac{2}{1 + t^2}\,\mathrm{d}t\) The \(\mathrm{d}t\) may be missing for this and the next M.

M1: Simplifies the integrand to the form \(\dfrac{a}{pt^2 + qt + r}\) allowing slips in coefficients but the “\(1 + t^2\)” must be correctly dealt with.

A1: Correct answer in the form required, including the \(\displaystyle\int \ldots\,\mathrm{d}t\)

Mark scheme (b)

Scheme	Marks	AO
\(\displaystyle\int \frac{a}{(t + b)^2 - c}\,\mathrm{d}t = \frac{a}{2\sqrt{c}}\ln\left\|\frac{(t + b) - \sqrt{c}}{(t + b) + \sqrt{c}}\right\|, c \gt 0\) or \(\displaystyle\int \frac{a}{(t + b)^2 + c}\,\mathrm{d}t = \frac{a}{\sqrt{c}}\arctan\left(\frac{t + b}{\sqrt{c}}\right), c \gt 0\) Note correct answer is \(\displaystyle\int \frac{2}{(t + 2)^2 - 1}\,\mathrm{d}t = \frac{2}{2(1)}\ln\left\|\frac{(t + 2) - 1}{(t + 2) + 1}\right\|\)	M1	3.1a
Correctly uses the limits \(t = 1\) and \(t = \sqrt{3} \to \ln\left\|\dfrac{\sqrt{3} + 1}{\sqrt{3} + 3}\right\| - \ln\left\|\dfrac{2}{4}\right\|\) if correct.	M1	1.1b
\(= \ln\left(\dfrac{\sqrt{3} + 1}{\sqrt{3} + 3} \div \dfrac{1}{2}\right)\)	M1	1.1b
\(= \ln\left(2 \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 3} \times \dfrac{\sqrt{3} - 3}{\sqrt{3} - 3}\right) = \ln\left(\dfrac{2\sqrt{3}}{3}\right)\ *\)	A1*	2.1
	(4)
	(7 marks)

Notes

M1: A correct method for the integration, by use of standard formula. Modulus signs are not necessary. Note that \(-\dfrac{a}{\sqrt{c}}\operatorname{artanh}\left(\dfrac{t + b}{\sqrt{c}}\right) = -\dfrac{a}{2\sqrt{c}}\ln\left|\dfrac{t + b + \sqrt{c}}{t + b - \sqrt{c}}\right|\) is also a correct version. If by error they end up with a positive \(c\) then accept for the arctan integral shown in scheme.

M1: Correctly uses the limits \(t = 1\) and \(t = \sqrt{3}\) in their integral to produce an exact expression. Use of the original limits is M0, must have changed to \(t\) (unless they reverse the substitution).

M1: Correctly combines the ln terms with at least one step showing the result of substituting limits before the final answer, and their final answer must follow this step if no further working is shown. May be seen before or after the rationalisation. (M0 if no ln terms.)

A1*cso: Shows the method to rationalise the denominator and correctly completes to the given answer. Some suitable working must be seen. Note that \(\sqrt{3} + 3 = \sqrt{3}\left(1 + \sqrt{3}\right)\) may be used in the denominator, followed by cancelling.

(b) Alt

Scheme	Marks	AO
\(\displaystyle\int \frac{2}{(t + 2)^2 - 1}\,\mathrm{d}t = \int \frac{2}{(t + 1)(t + 3)}\,\mathrm{d}t\) \(\displaystyle = \int \frac{1}{t + 1} - \frac{1}{t + 3}\,\mathrm{d}t = \ln\|t + 1\| - \ln\|t + 3\|\)	M1	3.1a
Correctly uses the limits \(t = 1\) and \(t = \sqrt{3}\) \(\ln\left(\sqrt{3} + 1\right) - \ln\left(\sqrt{3} + 3\right) - \ln 2 + \ln 4\)	M1	1.1b
\(= \ln\left(2 \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 3}\right)\)	M1	2.1
\(= \ln\left(2 \times \dfrac{\sqrt{3} + 1}{\sqrt{3}\left(1 + \sqrt{3}\right)}\right) = \ln\left(\dfrac{2\sqrt{3}}{3}\right)\ *\)	A1*	1.1b
	(4)

M1: A correct method for the integration, applying partial fractions and integrating to ln terms.

M1: Correctly uses the limits \(t = 1\) and \(t = \sqrt{3}\) (or correct \(\theta\) limits if substitution is reversed) to produce an exact expression.

AS June 2024 Q4

EdexcelAS paperCurrent spec12 markst-Formulae

(a) Given that \(t = \tan\dfrac{x}{2}\) prove that\[\cos x \equiv \frac{1 - t^2}{1 + t^2}\] (3)

(b) Show that the equation\[3\tan x - 10\cos x = 10\]can be written in the form\[(t + 2)(at^2 + bt + c) = 0\]where \(t = \tan\dfrac{x}{2}\) and \(a\), \(b\) and \(c\) are integers to be determined. (4)

Mark scheme (a)

Scheme	Marks	AO
\(\cos x = \cos^2\dfrac{x}{2} - \sin^2\dfrac{x}{2}\) or \(2\cos^2\dfrac{x}{2} - 1\) or \(1 - 2\sin^2\dfrac{x}{2}\) or \(\tan x = \dfrac{\tan\frac{x}{2} + \tan\frac{x}{2}}{1 - \tan^2\frac{x}{2}}\)	B1	1.2
\(\Rightarrow \cos x = \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2 - \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2\) or \(2\left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2 - 1\) or \(1 - 2\left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2\) or \(\tan x = \dfrac{2t}{1 - t^2} \Rightarrow \cos x = \dfrac{1 - t^2}{\sqrt{(1 - t^2)^2 + 4t^2}}\)	M1	1.1b
\(= \dfrac{1}{1 + t^2} - \dfrac{t^2}{1 + t^2} = \dfrac{1 - t^2}{1 + t^2}\ \) or \(= \dfrac{2}{1 + t^2} - 1 = \dfrac{2 - 1 - t^2}{1 + t^2} = \dfrac{1 - t^2}{1 + t^2}\ \) or \(1 - \dfrac{2t^2}{1 + t^2} = \dfrac{1 + t^2 - 2t^2}{1 + t^2} = \dfrac{1 - t^2}{1 + t^2}\ \) or \(\dfrac{1 - t^2}{\sqrt{(1 - t^2)^2 + 4t^2}} = \dfrac{1 - t^2}{\sqrt{t^4 + 2t^2 + 1}} = \dfrac{1 - t^2}{\sqrt{(1 + t^2)^2}} = \dfrac{1 - t^2}{1 + t^2}\ \)	A1*	2.1
	(3)

Notes

B1: Uses any correct appropriate double angle identity for \(\cos x\) or possibly \(\tan x\).

M1: Substitutes correct expressions for \(\cos\dfrac{x}{2}\) and/or \(\sin\dfrac{x}{2}\) in terms of \(t\)

A1*: Completes the proof with sufficient working shown with no errors.

Other alternatives are possible and can be marked in a similar way e.g.

\(\cos x = \dfrac{\sin x}{\tan x} = \dfrac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\left(\dfrac{2\tan\frac{x}{2}}{1 - \tan^2\frac{x}{2}}\right)}\)
Scores B1 for a correct double angle identity for \(\cos x\)

\(= 2\left(\dfrac{1}{\sqrt{1 + t^2}}\right)\left(\dfrac{t}{\sqrt{1 + t^2}}\right) \times \dfrac{1 - t^2}{2t}\)
M1 for substituting correct expressions for \(\cos\dfrac{x}{2}\), \(\sin\dfrac{x}{2}\) and \(\tan\dfrac{x}{2}\) in terms of \(t\)

\(= \dfrac{2t}{1 + t^2} \times \dfrac{1 - t^2}{2t} = \dfrac{1 - t^2}{1 + t^2}\ *\)
A1* for completing the proof with sufficient working shown with no errors.
Allow \(\theta\) for \(x\) for the first 2 marks but must obtain \(\cos x = \ldots\) for A1*

(a) Special Case – candidates who quote results for tan x and/or sin x to verify the result:

e.g.
\(t = \tan\dfrac{x}{2} \Rightarrow \tan x = \dfrac{2t}{1 - t^2},\ \sin x = \dfrac{2t}{1 + t^2}\)
\(\cos x = \dfrac{\sin x}{\tan x} = \dfrac{2t}{1 + t^2} \times \dfrac{1 - t^2}{2t} = \dfrac{1 - t^2}{1 + t^2}\)

or e.g.
\(t = \tan\dfrac{x}{2} \Rightarrow \sin x = \dfrac{2t}{1 + t^2}\)
\(\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \dfrac{4t^2}{(1 + t^2)^2}} = \sqrt{\dfrac{1 + 2t^2 + t^4 - 4t^2}{(1 + t^2)^2}} = \sqrt{\dfrac{(1 - t^2)^2}{(1 + t^2)^2}} = \dfrac{1 - t^2}{1 + t^2}\)

Scores SC B1 only

Mark scheme (b)

Scheme	Marks	AO
\(3\tan x - 10\cos x = 10 \Rightarrow 3 \times \dfrac{2t}{1 - t^2} - 10 \times \dfrac{1 - t^2}{1 + t^2} = 10\)	B1	1.1a
\(\Rightarrow 6t(1 + t^2) - 10(1 - t^2)^2 = 10(1 - t^2)(1 + t^2)\) \(\Rightarrow 6t + 6t^3 - 10 + 20t^2 - 10t^4 = 10 - 10t^4\) \(\Rightarrow 6t^3 + 20t^2 + 6t - 20 = 0\) or e.g. \(3t^3 + 10t^2 + 3t - 10 = 0\)	M1	2.1
\(\Rightarrow (t + 2)(\text{“}6\text{”}t^2 + \ldots t + \ldots) = 0\) or e.g. \(\Rightarrow (t + 2)(\text{“}3\text{”}t^2 + \ldots t + \ldots) = 0\)	M1	1.1b
\(\Rightarrow (t + 2)(6t^2 + 8t - 10) = 0\) or \((t + 2)(3t^2 + 4t - 5) = 0\)	A1	2.2a
	(4)

Notes

B1: Applies part (a) and the half angle formula for tan to give a correct equation in \(t\).

M1: Attempts to multiply by \(1 - t^2\) and \(1 + t^2\), expands and simplifies to a cubic in \(t\).

M1: Attempts to take a factor \((t + 2)\) out of their cubic.
May be done by inspection or long division but must obtain the correct coefficient for \(t^2\) and reach a 3 term quadratic expression.

A1: Correct equation achieved as shown. Accept with any integer multiple of the quadratic.
Must see the equation written down not just values for \(a\), \(b\) and \(c\).

Mark scheme (c)

Scheme	Marks	AO
\(t + 2 = 0 \Rightarrow x = 2 \times \arctan(-2) = \ldots\)	M1	1.1b
\(x = -126.86\ldots^\circ\) (allow awrt \(-127^\circ\))	A1	2.2a
\(3t^2 + 4t - 5 = 0 \Rightarrow t = \dfrac{-4 \pm \sqrt{16 - 4 \times 3 \times -5}}{6} = \dfrac{-2 \pm \sqrt{19}}{3}\) \((= 0.786\ldots, -2.11\ldots)\) \(\Rightarrow x = 2 \times \arctan(\ldots) = \ldots\)	M1	3.1a
\(x = \text{(awrt) } -129^\circ, 76.4^\circ \quad (-129.486\ldots, 76.355\ldots)\)	A1 (one) A1 (both)	1.1b 1.1b
	(5)
	(12 marks)

Notes

M1: Attempts to solve the \(t + 2 = 0\) equation, look for an attempt at \(\arctan(\pm 2)\) and an attempt to double. Allow in radians for the M mark.

A1: awrt \(-127^\circ\)

M1: Solves their quadratic and attempts arctan and doubles.
The usual rules apply for solving the quadratic and may be implied by their values.

A1: Either awrt \(-129^\circ\) or awrt \(76.4^\circ\) (degrees symbol not required) or allow for this mark one of these in radians (awrt \(-2.25\) or awrt 1.33)

A1: Both awrt \(-129^\circ\) (or better e.g. \(-129.5^\circ\)) and awrt \(76.4^\circ\) (degrees symbol not required)
There must be no other values in range for this mark. Values outside the range can be ignored.

A2 June 2023 Q5

EdexcelCurrent spec8 markst-Formulae

(a) Show that the substitution \(t = \tan\left(\dfrac{x}{2}\right)\) transforms the integral\[\int \frac{1}{2\sin x - \cos x + 5}\,\mathrm{d}x\]into the integral\[\int \frac{1}{3t^2 + 2t + 2}\,\mathrm{d}t\] (4)

(b) Hence determine\[\int \frac{1}{2\sin x - \cos x + 5}\,\mathrm{d}x\] (4)

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{1 + t^2}{2}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{2}{1 + t^2}\) or \(\mathrm{d}x = \dfrac{2\,\mathrm{d}t}{1 + t^2}\) or \(\mathrm{d}t = \dfrac{1 + t^2}{2}\mathrm{d}x\) oe	B1	1.1b
\(2\sin x - \cos x + 5 = 2\left(\dfrac{2t}{1 + t^2}\right) - \left(\dfrac{1 - t^2}{1 + t^2}\right) + 5\)	M1	1.1a
\(\displaystyle\int \frac{1}{2\sin x - \cos x + 5}\,\mathrm{d}x = \int \frac{1}{2\left(\dfrac{2t}{1 + t^2}\right) - \left(\dfrac{1 - t^2}{1 + t^2}\right) + 5} \times \frac{2\,\mathrm{d}t}{1 + t^2}\)	M1	2.1
\(\displaystyle = \int \frac{2}{4t - 1 + t^2 + 5 + 5t^2} \times \mathrm{d}t = \int \frac{1}{3t^2 + 2t + 2}\,\mathrm{d}t\ *\)	A1*	2.1
	(4)

Notes

B1: Correct equation in terms of \(\mathrm{d}x\), \(\mathrm{d}t\) and \(t\) – can be implied if seen as part of their substitution.

M1: Express \(2\sin x - \cos x + 5\) in terms of \(t\) using half angle formulae with at least one correct. Allow if e.g. there are slips in sign when substituting or the \(+5\) is missed.

M1: Makes a complete substitution to obtain an integral in terms of \(t\) only. Allow slips with the substitution of “\(\mathrm{d}x\)” but must be \(\mathrm{d}x = \mathrm{f}(t)\,\mathrm{d}t\) where \(\mathrm{f}(t) \ne 1\). This mark is also available if the candidate makes errors when attempting to simplify \(2\left(\dfrac{2t}{1 + t^2}\right) - \left(\dfrac{1 - t^2}{1 + t^2}\right)\) before attempting the substitution. Condone omission of the integral and \(\mathrm{d}t\) if the intent is clear.

A1*: Obtains the printed answer (including \(\mathrm{d}t\)) with at least one intermediate line or aside working showing elimination of the fractions and no errors seen.

Mark scheme (b)

Scheme	Marks	AO
\(\displaystyle\int \frac{1}{3t^2 + 2t + 2}\,\mathrm{d}t = \frac{1}{3}\int \frac{1}{t^2 + \frac{2}{3}t + \frac{2}{3}}\,\mathrm{d}t = \frac{1}{3}\int \frac{1}{\left(t + \frac{1}{3}\right)^2 + \ldots}\,\mathrm{d}t\) or \(\displaystyle\int \frac{1}{\left(\sqrt{3}t + \frac{1}{\sqrt{3}}\right)^2 + \ldots}\,\mathrm{d}t\)	M1	3.1a
\(\displaystyle\frac{1}{3}\int \frac{1}{\left(t + \frac{1}{3}\right)^2 + \frac{5}{9}}\,\mathrm{d}t\) or \(\displaystyle\int \frac{1}{\left(\sqrt{3}t + \frac{\sqrt{3}}{3}\right)^2 + \frac{5}{3}}\,\mathrm{d}t\) (oe)	A1	1.1b
\(= \dfrac{1}{3} \times \dfrac{1}{\frac{\sqrt{5}}{3}}\tan^{-1}\left(\dfrac{t + \frac{1}{3}}{\frac{\sqrt{5}}{3}}\right)\ (+c) = \dfrac{1}{\sqrt{5}}\tan^{-1}(\mathrm{f}(x))\ (+c)\) (oe)	M1	3.1a
\(= \dfrac{1}{\sqrt{5}}\tan^{-1}\left(\dfrac{3\tan\left(\frac{x}{2}\right) + 1}{\sqrt{5}}\right)\ (+c)\) (oe)	A1	1.1b
	(4)
	(8 marks)

Notes

M1: Adopts the correct strategy of completing the square in order to attempt the integration. Must achieve one of the forms shown in scheme. May be seen separately but must be applied to the integral, not just as an attempt to solve the quadratic.

A1: Correct integral with completed square integrand.

M1: For recognising the arctan form for the integration (look for e.g \(k\arctan\left(\dfrac{\text{“}t + \frac{1}{3}\text{”}}{b}\right)\)) and makes further progress by undoing the substitution to obtain an answer in terms of \(x\)

A1: Correct answer. Accept equivalents with simplified (and no nested) fractions. No need for \(c\).

AS June 2023 Q2

EdexcelAS paperCurrent spec7 markst-Formulae

(a) Use the substitution \(t = \tan\left(\dfrac{x}{2}\right)\) to show that the equation\[3\cos x - 2\sin x = 1\]can be written in the form\[2t^2 + 2t - 1 = 0\] (3)

(b) Hence solve, for \(-180^\circ \lt x \lt 180^\circ\), the equation\[3\cos x - 2\sin x = 1\]giving your answers to one decimal place. (4)

Mark scheme (a)

Scheme	Marks	AO
\(3\cos x - 2\sin x \Rightarrow 3\left(\dfrac{1 - t^2}{1 + t^2}\right) - 2\left(\dfrac{2t}{1 + t^2}\right)\)	M1	1.1b
\(3\left(\dfrac{1 - t^2}{1 + t^2}\right) - 2\left(\dfrac{2t}{1 + t^2}\right) = 1 \Rightarrow 3(1 - t^2) - 4t = 1 + t^2\)	M1	1.1b
\(2t^2 + 2t - 1 = 0\ *\)	A1*	2.1
	(3)

Notes

M1: Uses at least one of \(\sin x = \dfrac{2t}{1 + t^2}\) or \(\cos x = \dfrac{1 - t^2}{1 + t^2}\) to express \(3\cos x - 2\sin x\) in terms of \(t\) only.

M1: Uses both correct formulae for \(\sin x = \dfrac{2t}{1 + t^2}\) and \(\cos x = \dfrac{1 - t^2}{1 + t^2}\) in \(3\cos x - 2\sin x\) equates their expression to 1 and multiplies through by \(1 + t^2\) to achieve a quadratic expression in \(t\) (there may be slips in coefficients). Alternatively, gather all terms to a single fraction over denominator \(1 + t^2\).

A1*: Collects terms to one side and simplifies to achieve the printed answer, with no errors seen.

(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)

Mark scheme (b)

Scheme	Marks	AO
\(t = \dfrac{-2 \pm \sqrt{(2)^2 - 4(2)(-1)}}{2(2)}\ \left\{= \dfrac{-1 \pm \sqrt{3}}{2} = -1.366\ldots, 0.366\ldots\right\}\)	M1	1.1b
\(\dfrac{x}{2} = \arctan(-1.366\ldots)\) or \(\dfrac{x}{2} = \arctan(0.366\ldots)\) and \(\Rightarrow x = \ldots\)	dM1	3.1a
\(x = \text{awrt } -107.6^\circ\) or awrt \(40.2^\circ\)	A1	1.1b
\(x = -107.6^\circ,\ 40.2^\circ\) cao	A1	1.1b
	(4)
	(7 marks)

Notes

Note: The question says “hence solve”, so there needs to be evidence of use of part (a). Answers only scores no marks. Minimal requirement would be to see the solutions for \(t\) before stating the final answers.

M1: Selects a correct process to solve \(2t^2 + 2t - 1 = 0\) (calculator, quadratic formula, completing the square) to obtain at least one value for \(t\). Attempts by factorisation are M0.

dM1: Adopts a correct strategy of taking arctan(their value for \(t\)) and multiplies the result by 2 to obtain at least one value for \(x\) within the range \(-180^\circ \lt x \lt 180^\circ\). Allow slips on, or miscopies of their roots if the process is correct.

A1: One correct answer awrt 1 d.p.

A1: Both correct answers to 1 d.p. and no incorrect answers in the range \(-180^\circ \lt x \lt 180^\circ\)

(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)

AS June 2022 Q3

EdexcelAS paperCurrent spec7 markst-Formulae

(a) Use \(t = \tan\dfrac{\theta}{2}\) to show that, where both sides are defined\[\frac{29 - 21\sec\theta}{20 - 21\tan\theta} \equiv \frac{5t + 2}{2t + 5}\] (4)

(b) Hence, again using \(t = \tan\dfrac{\theta}{2}\), prove that, where both sides are defined\[\frac{20 + 21\tan\theta}{29 + 21\sec\theta} \equiv \frac{29 - 21\sec\theta}{20 - 21\tan\theta}\] (3)

Mark scheme (a)

Scheme	Marks	AO
Sight of \(\sec\theta = \dfrac{1 + t^2}{1 - t^2}\) and \(\tan\theta = \dfrac{2t}{1 - t^2}\) at least once each.	B1	1.2
\(\dfrac{29 - 21\sec\theta}{20 - 21\tan\theta} = \dfrac{29 - 21\left(\dfrac{1 + t^2}{1 - t^2}\right)}{20 - 21\left(\dfrac{2t}{1 - t^2}\right)}\)	M1	1.1b
\(= \dfrac{29(1 - t^2) - 21(1 + t^2)}{20(1 - t^2) - 21(2t)} = \dfrac{8 - 50t^2}{-20t^2 - 42t + 20}\) or \(= \dfrac{\left(\dfrac{29(1 - t^2) - 21(1 + t^2)}{(1 - t^2)}\right)}{\left(\dfrac{20(1 - t^2) - 21(2t)}{(1 - t^2)}\right)} = \dfrac{8 - 50t^2}{-20t^2 - 42t + 20}\)	M1	2.1
\(= \dfrac{-2(5t - 2)(5t + 2)}{-2(5t - 2)(2t + 5)} = \dfrac{5t + 2}{2t + 5}\ \) cso \(= \dfrac{2(2 + 5t)(2 - 5t)}{2(2 - 5t)(5 + 2t)} = \dfrac{5t + 2}{2t + 5}\ \) cso	A1*	1.1b
	(4)

Notes

B1: Uses the correct identities at least once each. May be seen in (a) or (b)

M1: Substitutes their identities into the equation, they need not be correct

M1: Multiplies numerator and denominator by \(1 - t^2\) and simplifies to a quadratic with all terms collected
Special case: If they make an error with the identities award M1 if they use the correct method to write the numerator and denominator as a single fraction and then divides to achieve an expression of the form \(\dfrac{a}{b}\)

A1*: Cancels factors to achieve the given expression, with no errors seen
Condone \(\dfrac{50t^2 - 8}{20t^2 + 42t - 20} = \dfrac{(5t + 2)(5t - 2)}{(5t - 2)(2t + 5)} = \dfrac{5t + 2}{2t + 5}\) and \(\dfrac{-8 + 50t^2}{-20t^2 - 42t + 20} = \dfrac{(5t + 2)(5t - 2)}{(5t - 2)(2t + 5)} = \dfrac{5t + 2}{2t + 5}\)

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{20 + 21\tan\theta}{29 + 21\sec\theta} = \dfrac{20 + 21\left(\dfrac{2t}{1 - t^2}\right)}{29 + 21\left(\dfrac{1 + t^2}{1 - t^2}\right)}\) \(= \dfrac{20(1 - t^2) + 21(2t)}{29(1 - t^2) + 21(1 + t^2)}\)	M1	2.1
\(\dfrac{20 + 42t - 20t^2}{50 - 8t^2} = \dfrac{2(5 - 2t)(5t + 2)}{2(5 - 2t)(5 + 2t)} = \ldots\) \(\dfrac{20 + 42t - 20t^2}{50 - 8t^2} = \dfrac{-2(2t - 5)(5t + 2)}{-2(2t + 5)(2t - 5)} = \ldots\)	M1	1.1b
Achieves form correct working \(= \dfrac{5t + 2}{2t + 5}\ *\) Then concludes hence the result is true or = RHS or \(= \dfrac{5t + 2}{2t + 5} = \dfrac{29 - 21\sec\theta}{20 - 21\tan\theta}\) or \(\dfrac{20 + 21\tan\theta}{29 + 21\sec\theta} = \dfrac{29 - 21\sec\theta}{20 - 21\tan\theta}\) LHS = RHS	A1*	2.2a
	(3)
	(7 marks)

Notes

M1: Substitutes into LHS of equation and multiplies numerator and denominator by \(1 - t^2\) to work towards the other side.

M1: Simplifies to a quadratic with all terms collected and factorises then cancels terms in the LHS in an attempt to try and match expressions.

A1*: Achieves correct expressions for both sides and gives a conclusion deducing that the result is true.

M1A0 for \(\dfrac{20t^2 - 42t - 20}{8t^2 - 50} = \dfrac{(5t + 2)(2t - 5)}{(2t - 5)(2t + 5)} = \dfrac{5t + 2}{2t + 5}\) and
\(\dfrac{-20t^2 - 42t + 20}{-8 + 50t^2} = \dfrac{(5t + 2)(2t - 5)}{(2t - 5)(2t + 5)} = \dfrac{5t + 2}{2t + 5}\)

A2 June 2022 Q2

EdexcelCurrent spec7 markst-Formulae

2. During 2029, the number of hours of daylight per day in London, \(H\), is modelled by the equation

\[H = 0.3\sin\left(\frac{x}{60}\right) - 4\cos\left(\frac{x}{60}\right) + 11.5 \qquad 0 \leqslant x \lt 365\]

where \(x\) is the number of days after 1st January 2029 and the angle is in radians.

(a) Show that, according to the model, the number of hours of daylight in London on the 31st January 2029 will be 8.13 to 3 significant figures. (1)

(b) Use the substitution \(t = \tan\left(\dfrac{x}{120}\right)\) to show that \(H\) can be written as\[H = \frac{at^2 + bt + c}{1 + t^2}\]where \(a\), \(b\) and \(c\) are constants to be determined. (2)

(c) Hence determine, according to the model, the date of the first day of 2029 when there will be at least 12 hours of daylight in London. (4)

Mark scheme (a)

Scheme	Marks	AO
\((H =)\,0.3\sin\left(\dfrac{30}{60}\right) - 4\cos\left(\dfrac{30}{60}\right) + 11.5 = 8.13\) {hours}*	B1*	3.4
	(1)

Notes

B1*: Uses \(x = 30\) to show that \(H = 8.13\). Accept \(x = 30\) seen substituted followed by 8.13, or \(x = 30\) identified followed by 8.133… before rounding to 8.13.

Mark scheme (b)

Scheme	Marks	AO
Substitutes \(\sin\left(\dfrac{x}{60}\right) = \dfrac{2t}{1 + t^2}\) and \(\cos\left(\dfrac{x}{60}\right) = \dfrac{1 - t^2}{1 + t^2}\) into \(H\) \((H =)\,0.3\left(\dfrac{2t}{1 + t^2}\right) - 4\left(\dfrac{1 - t^2}{1 + t^2}\right) + 11.5\)	M1	1.1b
\((H =)\,\dfrac{0.6t - 4 + 4t^2 + 11.5\left(1 + t^2\right)}{1 + t^2} = \dfrac{15.5t^2 + 0.6t + 7.5}{1 + t^2}\)	A1	2.1
	(2)

Notes

M1: Uses the correct \(t\)-formulae \(\sin\left(\dfrac{x}{60}\right) = \dfrac{2t}{1 + t^2}\) and \(\cos\left(\dfrac{x}{60}\right) = \dfrac{1 - t^2}{1 + t^2}\), attempts to substitute into \(H\).

A1: Fully correct method, expresses as a single fraction with a denominator of \(1 + t^2\) to achieve \(H = \dfrac{15.5t^2 + 0.6t + 7.5}{1 + t^2}\) (oe with fractions or accept values for \(a\), \(b\) and \(c\) stated).

Mark scheme (c)

Scheme	Marks	AO
\(H = \dfrac{15.5t^2 + 0.6t + 7.5}{1 + t^2} = 12 \Rightarrow 3.5t^2 + 0.6t - 4.5 = 0\)	M1	3.4
\(\Rightarrow t = \dfrac{-0.6 \pm \sqrt{0.6^2 - 4(3.5)(-4.5)}}{7}\) \(= \ldots\ (1.051\ldots, -1.222\ldots) \Rightarrow x = 120\tan^{-1}(\text{“}1.051\ldots\text{”}) = \ldots\ (97.254\ldots)\)	dM1	3.1b
\(x =\) awrt 97	A1	1.1b
8^th or 9^th April	A1	3.2a
	(4)
	(7 marks)

Notes

M1: Sets \(H = 12\) (or any inequality in between) and rearranges to form a quadratic equation for \(t\).

dM1: Dependent on the previous method mark. Solves the quadratic by any means (accept one correct answer for their quadratic if no method shown) and uses this to find a value for \(x\).

A1: Correct value for \(x =\) awrt 97 or accept 98 following a correct value for \(t\).

A1: Correct day of the year. Accept 8^th or 9^th April following awrt 97 from a correct method.

Note: Question says hence, so answers by graphical methods or trial and improvement are not acceptable for full credit. They can score a SC M0dM0A0B1 for achieving a correct date.

A2 October 2021 Q2

EdexcelCurrent spec10 markst-Formulae

(i) Use the substitution \(t = \tan\dfrac{x}{2}\) to prove the identity\[\frac{\sin x - \cos x + 1}{\sin x + \cos x - 1} \equiv \sec x + \tan x \qquad x \neq \frac{n\pi}{2} \quad n \in \mathbb{Z}\] (5)

(ii) Use the substitution \(t = \tan\dfrac{\theta}{2}\) to determine the exact value of\[\int_0^{\frac{\pi}{2}} \frac{5}{4 + 2\cos\theta}\,\mathrm{d}\theta\]giving your answer in simplest form. (5)

Mark scheme (i)

Scheme	Marks	AO
\(\dfrac{\sin x - \cos x + 1}{\sin x + \cos x - 1} = \dfrac{\frac{2t}{1 + t^2} - \frac{1 - t^2}{1 + t^2} + 1}{\frac{2t}{1 + t^2} + \frac{1 - t^2}{1 + t^2} - 1} = \ldots\)	M1	1.1b
\(= \dfrac{2t - \left(1 - t^2\right) + 1 + t^2}{2t + 1 - t^2 - \left(1 + t^2\right)}\) or \(= \dfrac{2t - 1 + t^2 + 1 + t^2}{2t + 1 - t^2 - 1 - t^2}\) numerator \(= \dfrac{2t - \left(1 - t^2\right) + 1 + t^2}{1 + t^2}\) denominator \(= \dfrac{2t + 1 - t^2 - \left(1 + t^2\right)}{1 + t^2}\) and divides	M1	2.1
\(= \dfrac{2t^2 + 2t}{2t - 2t^2}\left(= \dfrac{1 + t}{1 - t}\right)\)	A1	1.1b
\(= \dfrac{t + 1}{1 - t} \times \dfrac{1 + t}{1 + t} = \dfrac{t^2 + 2t + 1}{1 - t^2} = \dfrac{1 + t^2}{1 - t^2} + \dfrac{2t}{1 - t^2}\) Alt: \(\sec x + \tan x = \dfrac{1 + t^2}{1 - t^2} + \dfrac{2t}{1 - t^2} = \dfrac{1 + 2t + t^2}{1 - t^2}\)	M1	3.1a
\(= \dfrac{1}{\cos x} + \tan x = \sec x + \tan x\ \) Alt: \(= \dfrac{(t + 1)^2}{(1 - t)(1 + t)} = \dfrac{1 + t}{1 - t} = LHS\) hence result proved.	A1*	2.1
	(5)

Notes

M1: Applies the \(t\)-formulae to the left-hand side of expression. Allow slips in signs of the terms.

M1: Multiplies numerator and denominator through by \(1 + t^2\) (allow if they forget to multiply the 1’s). Alternative works separately on the numerator and denominator to combine terms and then divides.

A1: Correct \(\dfrac{\text{quadratic}}{\text{quadratic}}\) with terms gathered, award where first seen, need not have cancelled \(2t\) for this mark.

M1: Cancels \(2t\), multiplies numerator and denominator by \(1 + t\) and splits to sum of two terms. If working from both sides, this mark is for substituting the \(t\)-formulae into the right-hand side and combining to single fraction.

A1*: Correct completion to given result. No errors in proof. If working from both sides, a suitable conclusion is needed, e.g “hence proven”.

Mark scheme (ii)

Scheme	Marks	AO
\(\displaystyle\int_{(0)}^{\left(\frac{\pi}{2}\right)} \frac{5}{4 + 2\cos\theta}\,\mathrm{d}\theta = \int_{(0)}^{(1)} \frac{5}{4 + 2\frac{1 - t^2}{1 + t^2}} \times \frac{2}{1 + t^2}\,\mathrm{d}t\) Alternatively \(\dfrac{\mathrm{d}t}{\mathrm{d}\theta} = \dfrac{1}{2}\sec^2\left(\dfrac{\theta}{2}\right) = \dfrac{1}{\cos\theta + 1}\) leading to \(\displaystyle\int_{(0)}^{\left(\frac{\pi}{2}\right)} \frac{5}{4 + 2\cos\theta}\,\mathrm{d}\theta = \int_{(0)}^{(1)} \frac{5\cos\theta + 5}{4 + 2\cos\theta}\,\mathrm{d}t = \int_{(0)}^{(1)} \frac{5\left(\frac{1 - t^2}{1 + t^2}\right) + 5}{4 + 2\left(\frac{1 - t^2}{1 + t^2}\right)}\,\mathrm{d}t\)	M1	2.1
\(\displaystyle = \int_{(0)}^{(1)} \frac{10}{4\left(1 + t^2\right) + 2\left(1 - t^2\right)}\,\mathrm{d}t = \int_{(0)}^{(1)} \frac{5}{3 + t^2}\,\mathrm{d}t\) o.e.	A1	1.1b
\(= \left[5 \times \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{t}{\sqrt{3}}\right)\right]_{(0)}^{(1)}\)	M1	1.1b
\(= \dfrac{5}{\sqrt{3}}\left(\arctan\left(\dfrac{1}{\sqrt{3}}\right) - 0\right)\) or \(\dfrac{5}{\sqrt{3}}\left(\arctan\left(\dfrac{\tan\left(\frac{\pi}{4}\right)}{\sqrt{3}}\right) - \arctan\left(\dfrac{\tan(0)}{\sqrt{3}}\right)\right)\)	M1	2.2a
\(= \dfrac{5\pi\sqrt{3}}{18}\) oe in a surd form e.g. \(\dfrac{5\pi}{6\sqrt{3}}\)	A1	1.1b
	(5)
	(10 marks)

Notes

M1: Applies the substitution including the use of \(\mathrm{d}\theta = \dfrac{2}{1 + t^2}\,\mathrm{d}t\) (Limits not needed for first three marks).

A1: Simplifies correctly to a recognisable integrable form.

M1: Integrates to the form \(K\arctan\left(\dfrac{t}{a}\right)\) where \(a^2\) is their constant term.

M1: Deduces correct limits and applies them the correct way round OR deduces integral in terms of \(\theta\) from their integration and applies original limits the correct way.

A1: \(\dfrac{5\pi\sqrt{3}}{18}\) or equivalent in surd form.

Note use of calculator does not lead to the exact value required in the question \(= 1.51149947\)
This can score M1 A1 M0 M1 A0

A2 October 2020 Q8

EdexcelCurrent spec16 markst-Formulae

\[\mathrm{f}(x) = \frac{3}{13 + 6\sin x - 5\cos x}\]

Using the substitution \(t = \tan\left(\dfrac{x}{2}\right)\)

(a) show that \(\mathrm{f}(x)\) can be written in the form \[\frac{3(1 + t^2)}{2(3t + 1)^2 + 6}\] (3)

(b) Hence solve, for \(0 \lt x \lt 2\pi\), the equation \[\mathrm{f}(x) = \frac{3}{7}\] giving your answers to 2 decimal places where appropriate. (5)

(c) Use the result of part (a) to show that \[\int_{\frac{\pi}{3}}^{\frac{4\pi}{3}}\mathrm{f}(x)\,\mathrm{d}x = K\left(\arctan\left(\frac{\sqrt{3} - 9}{3}\right) - \arctan\left(\frac{\sqrt{3} + 3}{3}\right) + \pi\right)\] where \(K\) is a constant to be determined. (8)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{f}(x) = \dfrac{3}{13 + 6 \times \dfrac{2t}{1 + t^2} - 5 \times \dfrac{1 - t^2}{1 + t^2}}\)	M1	1.1b
\(= \dfrac{3(1 + t^2)}{13(1 + t^2) + 12t - 5(1 - t^2)}\)	M1	1.1b
\(= \dfrac{3(1 + t^2)}{18t^2 + 12t + 8} \Rightarrow\) for example \(\dfrac{3(1 + t^2)}{2(9t^2 + 6t + 1) + 6}\) or \(\dfrac{3(1 + t^2)}{2\left[(3t + 1)^2 - 1\right] + 8}\) \(\Rightarrow \dfrac{3(1 + t^2)}{2(3t + 1)^2 + 6}\ *\)	A1*	2.1
	(3)

Notes

M1: Uses one correct substitution

M1: Both substitutions correct and attempts to multiply through numerator and denominator by \(1 + t^2\).

A1*: Completes to the correct expression with no errors seen. Must see an intermediate step simplifying the denominator – most likely one of the ones seen in the scheme.

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{f}(x) = \dfrac{3}{7} \Rightarrow \dfrac{3(1 + t^2)}{2(3t + 1)^2 + 6} = \dfrac{3}{7} \Rightarrow 21 + 21t^2 = 54t^2 + 36t + 24\) \(\Rightarrow 11t^2 + 12t + 1 = 0\)	M1	1.1b
\(\Rightarrow (11t + 1)(t + 1) = 0 \Rightarrow t = \ldots\)	M1	1.1b
\(t = -1,\ t = -\dfrac{1}{11}\)	A1	1.1b
\(\Rightarrow x = 2\arctan\left(\text{“}\textit{their } t\text{”}\right) + 2\pi\) for a negative \(t\)	dM1	3.1a
\(x = \dfrac{3\pi}{2}\) or awrt \(4.71\) and awrt \(x = 6.10\)	A1	1.1b
	(5)

Notes

M1: Equates the result in (a) to \(\dfrac{3}{7}\) and simplifies to a 3TQ

M1: Solves their equation by any valid means.

A1: Correct values for \(t\)

dM1: Dependent on first method mark. Applies the correct process to find at least one value for \(x\) from a negative value for \(t\). (If two positive values are found in error, this mark cannot be scored.)

A1: Both answers correct and no others in range.

Mark scheme (c)

Scheme	Marks	AO
\(\displaystyle\int \mathrm{f}(x) = \int \dfrac{3(1 + t^2)}{2(3t + 1)^2 + 6} \times \dfrac{2}{1 + t^2}\,\mathrm{d}t = \int \dfrac{3}{(3t + 1)^2 + 3}\,\mathrm{d}t\)	B1	2.1
\(= K\arctan\left(M(3t + 1)\right)\) or \(u = (3t + 1) \Rightarrow K\arctan\left(Mu\right)\)	M1	1.1b
\(= \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{3t + 1}{\sqrt{3}}\right)\) or \(= \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{u}{\sqrt{3}}\right)\)	A1	1.1b
\(\displaystyle\int_{\frac{\pi}{3}}^{\frac{4\pi}{3}}\mathrm{f}(x)\,\mathrm{d}x = \int_{\frac{\pi}{3}}^{\pi}\mathrm{f}(x)\,\mathrm{d}x + \int_{\pi}^{\frac{4\pi}{3}}\mathrm{f}(x)\,\mathrm{d}x\) \(\displaystyle = \int_{\frac{\sqrt{3}}{3}}^{\infty}\ldots\,\mathrm{d}t + \int_{-\infty}^{-\sqrt{3}}\ldots\,\mathrm{d}t\) or \(\displaystyle\int_{\sqrt{3} + 1}^{\infty}\ldots\,\mathrm{d}u + \int_{-\infty}^{1 - 3\sqrt{3}}\ldots\,\mathrm{d}u\)	B1	3.1a
\(= \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{3\left(-\sqrt{3}\right) + 1}{\sqrt{3}}\right) - \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{3\left(\frac{\sqrt{3}}{3}\right) + 1}{\sqrt{3}}\right) + \ldots\)	M1	1.1b
\(= \dfrac{\sqrt{3}}{3}\left(\arctan\left(\dfrac{\sqrt{3} - 9}{3}\right) - \arctan\left(\dfrac{\sqrt{3} + 3}{3}\right)\right) + \ldots\)	A1	1.1b
\(= \ldots + \lim\limits_{t \to \infty}\dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{3t + 1}{\sqrt{3}}\right) - \lim\limits_{t \to -\infty}\dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{3t + 1}{\sqrt{3}}\right)\) \(= \ldots + \dfrac{\pi}{2\sqrt{3}} - \left(-\dfrac{\pi}{2\sqrt{3}}\right)\)	M1	3.1a
\(= \dfrac{\sqrt{3}}{3}\left(\arctan\left(\dfrac{\sqrt{3} - 9}{3}\right) - \arctan\left(\dfrac{\sqrt{3} + 3}{3}\right) + \pi\right)\)	A1	2.1
	(8)
	(16 marks)

Notes

B1: Applies the substitution including the use of \(\mathrm{d}x = \dfrac{2}{1 + t^2}\,\mathrm{d}t\)

M1: Attempts the integration to achieve \(K\arctan\left(M(1 + 3t)\right)\) or \(K\arctan\left(Mu\right)\) if using a substitution of \(u = (3t + 1)\).

May use substitution \(3t + 1 = \sqrt{3}\tan\theta \Rightarrow \dfrac{\mathrm{d}t}{\mathrm{d}\theta} = \dfrac{\sqrt{3}}{3}\sec^2\theta \Rightarrow \displaystyle\int\dfrac{\sqrt{3}}{3}\,\mathrm{d}\theta = \dfrac{\sqrt{3}}{3}\theta = \dfrac{\sqrt{3}}{3}\arctan\left(\dfrac{3t + 1}{\sqrt{3}}\right)\)

A1: Correct integral.

B1: Changes the limits and splits the integral around \(\pi\)

M1: Applies their limits ‘\(\frac{1}{\sqrt{3}}\)’ and ‘\(-\sqrt{3}\)’ to their integrand.

A1: Correct “arctan” expressions.

M1: Correct work to evaluate the \(\pm\infty\) limits

A1: Fully correct solution.

AS October 2020 Q3

EdexcelAS paperCurrent spec11 markst-Formulae

(i) Use the substitution \(t = \tan\left(\dfrac{x}{2}\right)\) to prove that\[\cot x + \tan\left(\frac{x}{2}\right) = \operatorname{cosec} x \qquad x \neq n\pi,\ n \in \mathbb{Z}\] (2)

(ii)

Figure 1: wind turbine on horizontal ground, with the height H metres marked from the ground up to the tip of one blade — Figure 1

An engineer models the vertical height above the ground of the tip of one blade of a wind turbine, shown in Figure 1. The ground is assumed to be horizontal.

The vertical height of the tip of the blade above the ground, \(H\) metres, at time \(x\) seconds after the wind turbine has reached its constant operating speed, is modelled by the equation

\[H = 90 - 30\cos(120x)^\circ - 40\sin(120x)^\circ \qquad (\text{I})\]

(a) Show that \(H = 60\) when \(x = 0\) (1)

Using the substitution \(t = \tan(60x)^\circ\)

(b) show that equation \((\text{I})\) can be rewritten as\[H = \frac{120t^2 - 80t + 60}{1 + t^2}\] (3)

(c) Hence find, according to the model, the value of \(x\) when the tip of the blade is 100 m above the ground for the first time after the wind turbine has reached its constant operating speed. (5)

Mark scheme (i)

Scheme	Marks	AO
\(\text{lhs} = \cot x + \tan\left(\dfrac{x}{2}\right) = \dfrac{1 - t^2}{2t} + t\)	M1	1.1a
\(\dfrac{1 - t^2}{2t} + t = \dfrac{1 + t^2}{2t}\left(= \dfrac{1}{\sin x}\right) = \operatorname{cosec} x\ *\)	A1*	2.1
	(2)

Notes

M1: Selects the correct expression for \(\cot x\) in terms of \(t\) and substitutes this and \(t\) into the lhs

A1*: Fully correct proof. Allow correct work leading to \(\dfrac{1 + t^2}{2t} = \operatorname{cosec} x\)

Scheme	Marks	AO
\(x = 0 \Rightarrow H = 90 - 30\cos(0) - 40\sin(0) = 90 - 30 = 60\)	B1	1.1b
	(1)

Mark scheme (ii)(b)

Scheme	Marks	AO
\(H = 90 - 30\cos 120x - 40\sin 120x = 90 - 30\left(\dfrac{1 - t^2}{1 + t^2}\right) - 40\left(\dfrac{2t}{1 + t^2}\right)\)	M1	1.1b
\(= \dfrac{90 + 90t^2 - 30 + 30t^2 - 80t}{1 + t^2}\)	M1	1.1b
\(= \dfrac{120t^2 - 80t + 60}{1 + t^2}\ *\)	A1*	2.1
	(3)

Notes

M1: Uses the correct formulae to obtain \(H\) in terms of \(t\)

M1: Correct method to obtain a common denominator

A1*: Collects terms and simplifies to obtain the printed answer with no errors

Mark scheme (ii)(c)

Scheme	Marks	AO
\(\dfrac{120t^2 - 80t + 60}{1 + t^2} = 100 \Rightarrow 120t^2 - 80t + 60 = 100 + 100t^2\)	M1	3.4
\(20t^2 - 80t - 40 = 0\)	A1	1.1b
\(t = \dfrac{4 \pm \sqrt{16 + 8}}{2} \Rightarrow 60x = \tan^{-1}\left(2 + \sqrt{6}\right)\) or \(60x = \tan^{-1}\left(2 - \sqrt{6}\right)\)	M1	3.4
\(60x = \tan^{-1}\left(2 + \sqrt{6}\right) = 77.33\ldots \Rightarrow x = \ldots\)	dM1	3.1b
\(x = 1.29\)	A1	3.2a
	(5)
	(11 marks)

Notes

M1: Uses \(H = 100\) with the model and multiplies up to obtain a quadratic equation in \(t\)

A1: Correct 3TQ

M1: Solves their 3TQ in \(t\) and proceeds to obtain values of \(60x\) as suggested by the model

M1: A fully correct strategy to identify the required value of \(x\) from the positive root of the quadratic equation in \(t\)

A1: awrt 1.29

Attempts in radians can score all but the final mark in (c). (Gives \(60x = 1.3\ldots\) etc.)

A2 June 2019 Q5

EdexcelCurrent spec8 markst-Formulae

\[I = \int\frac{1}{4\cos x - 3\sin x}\,\mathrm{d}x \qquad 0 \lt x \lt \frac{\pi}{4}\]

Use the substitution \(t = \tan\left(\dfrac{x}{2}\right)\) to show that

\[I = \frac{1}{5}\ln\left(\frac{2 + \tan\left(\frac{x}{2}\right)}{1 - 2\tan\left(\frac{x}{2}\right)}\right) + k\]

where \(k\) is an arbitrary constant.

(8)

Mark scheme

Scheme	Marks	AO
\(4\cos x - 3\sin x = 4\left(\dfrac{1 - t^2}{1 + t^2}\right) - 3\left(\dfrac{2t}{1 + t^2}\right)\)	B1	1.1a
\(\dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{1 + t^2}{2}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{2}{1 + t^2}\) or \(\mathrm{d}x = \dfrac{2\,\mathrm{d}t}{1 + t^2}\) or \(\mathrm{d}t = \dfrac{1 + t^2}{2}\,\mathrm{d}x\) oe	B1 M1 on ePEN	2.1
\(\displaystyle\int\frac{1}{4\cos x - 3\sin x}\,\mathrm{d}x = \int\frac{1}{4\left(\dfrac{1 - t^2}{1 + t^2}\right) - 3\left(\dfrac{2t}{1 + t^2}\right)} \times \frac{2\,\mathrm{d}t}{1 + t^2}\)	M1	2.1
\(\displaystyle = \int\frac{2}{4 - 4t^2 - 6t}(\mathrm{d}t)\) or \(\displaystyle\int\frac{1}{2 - 2t^2 - 3t}(\mathrm{d}t)\) or \(\displaystyle\int\frac{-1}{2t^2 + 3t - 2}(\mathrm{d}t)\) etc.	A1	1.1b
\(\begin{aligned}&\dfrac{-2}{4t^2 + 6t - 4} = \dfrac{-1}{(t + 2)(2t - 1)} = \dfrac{A}{(t + 2)} + \dfrac{B}{(2t - 1)}\\[6pt] &\dfrac{-1}{(t + 2)(2t - 1)} = \dfrac{1}{5(t + 2)} + \dfrac{2}{5(1 - 2t)}\end{aligned}\)	M1	3.1a
\(\displaystyle\Rightarrow I = \frac{1}{5}\int\frac{1}{(t + 2)} - \frac{2}{(2t - 1)}(\mathrm{d}t)\) or equivalent	A1	1.1b
\(\displaystyle = \frac{1}{5}\int\frac{1}{(t + 2)} - \frac{2}{(2t - 1)}\,\mathrm{d}t = \frac{1}{5}\ln(t + 2) - \frac{1}{5}\ln(1 - 2t)\,(+k)\)	A1	1.1b
\(= \dfrac{1}{5}\ln\left(\dfrac{2 + t}{1 - 2t}\right)(+k) = \dfrac{1}{5}\ln\left(\dfrac{2 + \tan\left(\tfrac{x}{2}\right)}{1 - 2\tan\left(\tfrac{x}{2}\right)}\right) + k\ *\)	A1*	2.1
	(8)
	(8 marks)

Notes

B1: Uses the correct formulae to express \(4\cos x - 3\sin x\) in terms of \(t\)

B1(M1 on ePEN): Correct equation in terms of \(\mathrm{d}x\), \(\mathrm{d}t\) and \(t\) – can be implied if seen as part of their substitution.

M1: Makes a complete substitution to obtain an integral in terms of \(t\) only. Allow slips with the substitution of “\(\mathrm{d}x\)” but must be \(\mathrm{d}x = \mathrm{f}(t)\mathrm{d}t\) where \(\mathrm{f}(t) \neq 1\). This mark is also available if the candidate makes errors when attempting to simplify \(4\left(\dfrac{1 - t^2}{1 + t^2}\right) - 3\left(\dfrac{2t}{1 + t^2}\right)\) before attempting the substitution.

A1: For obtaining a fully correct simplified integral with a constant in the numerator and a 3 term quadratic expression in the denominator. (“\(\mathrm{d}t\)” not required)

M1: Realises the need to express the integrand in terms of partial fractions in order to attempt the integration. Must have a 3 term quadratic expression in the denominator and a constant in the numerator.

A1: Correct integral in terms of partial fractions – allow any equivalent correct integral. (“\(\mathrm{d}t\)” not required)

A1: Fully correct integration in terms of \(t\)

A1*: Correct solution with no errors including “\(+\,k\)” (allow “\(+\,c\)”) and with the constant dealt with correctly if necessary. The denominator must also be dealt with correctly. E.g. if it appears as \(2t - 1\) initially and becomes \(1 - 2t\) without justification, this final mark should be withheld.

Alternative for final 4 marks

Scheme	Marks	AO
\(\displaystyle = \int\frac{2}{4 - 4t^2 - 6t}(\mathrm{d}t) = -\frac{1}{2}\int\frac{1}{t^2 + \frac{3}{2}t - 1}(\mathrm{d}t) = -\frac{1}{2}\int\frac{1}{\left(t + \frac{3}{4}\right)^2 - \frac{25}{16}}(\mathrm{d}t)\) or e.g. \(\displaystyle\int\frac{1}{\frac{25}{8} - 2\left(t + \frac{3}{4}\right)^2}(\mathrm{d}t)\)	M1 A1	3.1a 1.1b
\(-\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{4}{5}\ln\left(\dfrac{t + \frac{3}{4} - \frac{5}{4}}{t + \frac{3}{4} + \frac{5}{4}}\right)(+c)\)	A1	1.1b
\(\begin{aligned}&-\dfrac{1}{5}\ln\left\|\dfrac{\tan\left(\tfrac{x}{2}\right) - \frac{1}{2}}{\tan\left(\tfrac{x}{2}\right) + 2}\right\| + c = \dfrac{1}{5}\ln\left\|\dfrac{\tan\left(\tfrac{x}{2}\right) + 2}{\tan\left(\tfrac{x}{2}\right) - \frac{1}{2}}\right\| + c = \dfrac{1}{5}\ln\left(\dfrac{\tan\left(\tfrac{x}{2}\right) + 2}{\frac{1}{2} - \tan\left(\tfrac{x}{2}\right)}\right) + c\\[6pt] &= \dfrac{1}{5}\ln\left(\dfrac{2\left(\tan\left(\tfrac{x}{2}\right) + 2\right)}{1 - 2\tan\left(\tfrac{x}{2}\right)}\right) + c = \dfrac{1}{5}\ln\left(\dfrac{\left(\tan\left(\tfrac{x}{2}\right) + 2\right)}{1 - 2\tan\left(\tfrac{x}{2}\right)}\right) + \dfrac{1}{5}\ln 2 + c\\[6pt] &= \dfrac{1}{5}\ln\left(\dfrac{\left(\tan\left(\tfrac{x}{2}\right) + 2\right)}{1 - 2\tan\left(\tfrac{x}{2}\right)}\right) + k\end{aligned}\)	A1*	2.1

M1: Realises the need to express the integrand in completed square form in order to attempt the integration. Must have a 3 term quadratic expression in the denominator and a constant in the numerator.

A1: Correct integral with the square completed – allow any equivalent correct integral (“\(\mathrm{d}t\)” not required)

A1: Fully correct integration in terms of \(t\)

A1*: Correct solution with no errors including “\(+\,k\)” (allow “\(+\,c\)”) and with the constant dealt with correctly if necessary as shown in the scheme and with the denominator dealt with correctly if necessary.

Note that it is acceptable for the “\(\mathrm{d}t\)” to appear and disappear throughout the proof as long as the intention is clear.

AS June 2019 Q1

EdexcelAS paperCurrent spec9 markst-Formulae

(a) Write down the \(t\)-formula for \(\sin x\). (1)

(b) Use the answer to part (a)

(i) to find the exact value of \(\sin x\) when\[\tan\left(\frac{x}{2}\right) = \sqrt{2}\]

(ii) to show that\[\cos x = \frac{1 - t^2}{1 + t^2}\] (4)

(c) Use the \(t\)-formulae to solve for \(0 \lt \theta \leqslant 360^\circ\)\[7\sin\theta + 9\cos\theta + 3 = 0\]giving your answers to one decimal place. (4)

Scheme	Marks	AO
\(\{\sin x =\}\ \dfrac{2t}{1 + t^2}\)	B1	1.2
	(1)

Mark scheme (b)

Scheme	Marks	AO
(b)(i) \(\left\{\tan\left(\dfrac{x}{2}\right) = \sqrt{2} \Rightarrow t = \sqrt{2} \Rightarrow\right\}\ \sin x = \dfrac{2(\sqrt{2})}{1 + (\sqrt{2})^2}\) or \(\dfrac{2(\sqrt{2})}{1 + 2}\)	M1	1.1b
\(\sin x = \dfrac{2}{3}\sqrt{2}\) or \(\dfrac{1}{3}\sqrt{8}\) or \(\sqrt{\dfrac{8}{9}}\)	A1	1.1b
	(2)
(b)(ii) Way 1 \(\left\{\cos x \equiv \dfrac{\sin x}{\tan x} \Rightarrow\right\}\ \cos x = \dfrac{\;\dfrac{2t}{1 + t^2}\;}{\dfrac{2t}{1 - t^2}}\ ;\ = \dfrac{1 - t^2}{1 + t^2}\ *\ \text{cso}\)	M1; A1*	1.1b 2.1
	(2)

Notes

(b)(i)

M1: Complete substitution of \(t = \sqrt{2}\) into their expression from part (a)

A1: Correct exact answer. See scheme.

Note: Give M0 A0 for writing down the correct exact answer without any evidence of substituting \(t = \sqrt{2}\) into \(\sin x = \dfrac{2t}{1 + t^2}\)

Note: For reference, \(\sin x = \dfrac{2}{3}\sqrt{2} = 0.9428\ldots\)

(b)(ii) Way 1, Way 2 and Way 3

M1: Uses a correct trigonometric identity (or correct trigonometric identities) to find a correct expression which connects only \(\cos x\) (or \(\cos^2 x\)) and \(t\)

A1*: Correct proof

(b)(ii) Way 4

M1: Uses \(\sin x = \dfrac{o}{h}\) and a correct Pythagoras method to express the adjacent edge of a triangle in terms of \(t\).

A1*: Correct proof

(b)(ii) Way 2

Scheme	Marks	AO
\(\left\{\tan x \equiv \dfrac{\sin x}{\cos x} \Rightarrow\right\}\ \dfrac{2t}{1 - t^2} = \dfrac{\;\dfrac{2t}{1 + t^2}\;}{\cos x}\ ;\ \Rightarrow \cos x = \dfrac{1 - t^2}{1 + t^2}\ *\ \text{cso}\)	M1; A1*	1.1b 2.1
	(2)

(b)(ii) Way 3

Scheme	Marks	AO
\(\left\{\sin^2 x + \cos^2 x \equiv 1 \Rightarrow\right\}\ \left(\dfrac{2t}{1 + t^2}\right)^2 + \cos^2 x = 1\)	M1	1.1b
\(\cos^2 x = 1 - \left(\dfrac{2t}{1 + t^2}\right)^2 = \dfrac{(1 + t^2)^2 - 4t^2}{(1 + t^2)^2} = \dfrac{1 - 2t^2 + t^4}{(1 + t^2)^2} = \dfrac{(1 - t^2)^2}{(1 + t^2)^2}\) \(\Rightarrow \cos x = \dfrac{1 - t^2}{1 + t^2}\ *\ \text{cso}\)	A1	2.1
	(2)

(b)(ii) Way 4

Scheme	Marks	AO
\(\{o^2 + a^2 = h^2 \Rightarrow\}\ (2t)^2 + a^2 = (1 + t^2)^2\)	M1	1.1b
\(a^2 = (1 + t^2)^2 - (2t)^2 = 1 - 2t^2 + t^4 = (1 - t^2)^2\) \(a = 1 - t^2 \Rightarrow \cos x = \dfrac{1 - t^2}{1 + t^2}\ *\ \text{cso}\)	A1	2.1
	(2)

Mark scheme (c)

Scheme	Marks	AO
\(\{7\sin\theta + 9\cos\theta + 3 =\}\ 7\left(\dfrac{2t}{1 + t^2}\right) + 9\left(\dfrac{1 - t^2}{1 + t^2}\right) + 3\)	M1	1.1b
\(7\left(\dfrac{2t}{1 + t^2}\right) + 9\left(\dfrac{1 - t^2}{1 + t^2}\right) + 3 = 0 \Rightarrow 14t + 9 - 9t^2 + 3 + 3t^2 = 0\) \(\Rightarrow 6t^2 - 14t - 12 = 0 \Rightarrow 3t^2 - 7t - 6 = 0 \Rightarrow (t - 3)(3t + 2) = 0 \Rightarrow t = \ldots\)	M1	1.1b
Either \(\left\{t = 3 \Rightarrow \dfrac{\theta}{2} = \arctan(3) \Rightarrow\right\}\ \theta = 2\arctan(3)\) or \(\left\{t = -\dfrac{2}{3} \Rightarrow \dfrac{\theta}{2} = 180^\circ + \arctan\left(-\dfrac{2}{3}\right) \Rightarrow\right\}\ \theta = 2\left(180^\circ + \arctan\left(-\dfrac{2}{3}\right)\right)\)	M1	1.1b
\(\dfrac{\theta}{2} = \{71.5650\ldots, 146.3099\ldots\} \Rightarrow \theta = \{143.1301\ldots, 292.6198\ldots\}\)
\(\theta = 143.1^\circ, 292.6^\circ\ (1\text{dp})\)	A1	1.1b
	(4)
	(9 marks)

Notes

M1: Uses at least one of \(\sin\theta = \dfrac{2t}{1 + t^2}\) or \(\cos\theta = \dfrac{1 - t^2}{1 + t^2}\) to express \(7\sin\theta + 9\cos\theta + 3\) in terms of \(t\) only

M1: Uses both correct formula \(\sin\theta = \dfrac{2t}{1 + t^2}\) and \(\cos\theta = \dfrac{1 - t^2}{1 + t^2}\) in \(7\sin\theta + 9\cos\theta + 3 = 0\), multiplies both sides by \(1 + t^2\), forms a 3TQ and uses a correct method (e.g. using the quadratic formula, completing the square or a calculator approach) for solving their 3TQ to give \(t = \ldots\)

M1: Uses both correct formula \(\sin\theta = \dfrac{2t}{1 + t^2}\) and \(\cos\theta = \dfrac{1 - t^2}{1 + t^2}\) in \(7\sin\theta + 9\cos\theta + 3 = 0\), adopts a correct applied strategy to find at least one value of \(\theta\) within the range \(0 \lt \theta \leqslant 360^\circ\) (or in radians \(0 \lt \theta \leqslant 2\pi\)) such that either

\(\theta = 2\arctan(\text{their found } t)\), where their found \(t \gt 0\)
\(\theta = 2(180^\circ + \arctan(\text{their found } t))\), where their found \(t \lt 0\)
\(\theta = 2(180^\circ - \arctan|\text{their found } t|)\), where their found \(t \lt 0\)

A1: Correct answer only of \(\theta = 143.1^\circ, 292.6^\circ\)

Note: Give A0 for extra solutions given within the range \(0 \lt \theta \leqslant 360^\circ\)

Note: Ignore extra solutions outside the range \(0 \lt \theta \leqslant 360^\circ\) for the A mark

Note: Give 3^rd M0 for \(\dfrac{\theta}{2} = \{71.565\ldots, 146.309\ldots\}\) without attempting to find \(\theta\)

Note Give 3^rd M0 for \(\dfrac{\theta}{2} = \{71.565\ldots, 146.309\ldots\} \Rightarrow \theta = \{35.782\ldots, 73.154\ldots\}\)

Note: In degrees, \(\dfrac{\theta}{2} = \{71.565\ldots,\ 251.565\ldots,\ -33.690\ldots,\ 146.309\ldots\}\)

Note: Working in radians gives \(\dfrac{\theta}{2} = \{1.249\ldots, 2.553\ldots\} \Rightarrow \theta = \{2.498\ldots, 5.107\ldots\}\)

AS June 2018 Q1

EdexcelAS paperCurrent spec7 markst-Formulae

(a) Use the substitution \(t = \tan\left(\dfrac{x}{2}\right)\) to show that the equation\[5\sin x + 12\cos x = 2\]can be written in the form\[7t^2 - 5t - 5 = 0\] (3)

(b) Hence solve, for \(-180^\circ \lt x \lt 180^\circ\), the equation\[5\sin x + 12\cos x = 2\]giving your answers to one decimal place. (4)

Mark scheme (a)

Scheme	Marks	AO
\(t = \tan\left(\dfrac{x}{2}\right),\ 5\sin x + 12\cos x = 2 \Rightarrow 7t^2 - 5t - 5 = 0\)
\(\{5\sin x + 12\cos x =\}\ 5\left(\dfrac{2t}{1 + t^2}\right) + 12\left(\dfrac{1 - t^2}{1 + t^2}\right)\)	M1	1.1b
\(5\left(\dfrac{2t}{1 + t^2}\right) + 12\left(\dfrac{1 - t^2}{1 + t^2}\right) = 2 \Rightarrow 5(2t) + 12(1 - t^2) = 2(1 + t^2)\)	M1	1.1b
\(7t^2 - 5t - 5 = 0\ *\)	A1*	2.1
	(3)

Notes

M1: Uses at least one of \(\sin x = \dfrac{2t}{1 + t^2}\) or \(\cos x = \dfrac{1 - t^2}{1 + t^2}\) to express \(5\sin x + 12\cos x\) in terms of \(t\) only

M1: Uses both correct formula \(\sin x = \dfrac{2t}{1 + t^2}\) and \(\cos x = \dfrac{1 - t^2}{1 + t^2}\) in \(5\sin x + 12\cos x\), equates their expression to 2 and eliminates the fractions

A1*: Collects terms to one side and simplifies to obtain the printed answer

Mark scheme (b)

Scheme	Marks	AO
\(t = \dfrac{5 \pm \sqrt{5^2 - 4(7)(-5)}}{2(7)}\ \left\{= \dfrac{5 \pm \sqrt{165}}{14} = 1.2746\ldots, -0.5603\ldots\right\}\)	M1	1.1a
\(\dfrac{x}{2} = \arctan\left(\dfrac{5 + \sqrt{165}}{14}\right)\) or \(\dfrac{x}{2} = \arctan\left(\dfrac{5 - \sqrt{165}}{14}\right)\) and \(\Rightarrow x = \ldots\)	M1	3.1a
\(x = \text{awrt } 104^\circ\) or \(x = \) any answer in the range \([-58.6^\circ, -58^\circ]\)	A1	1.1b
\(x = 103.8^\circ\) and \(x = -58.5^\circ\)	A1	1.1b
	(4)
	(7 marks)

Notes

M1: Selects a correct process (e.g. using the quadratic formula, completing the square or calculator approach) to solve \(7t^2 - 5t - 5 = 0\)

Note: Allow 1^st M1 for at least one of awrt 1.3 or awrt \(-0.6\) or for a correct exact value of \(t\)

Note: Do not allow an attempt at factorisation of \(7t^2 - 5t - 5\) for the 1^st M1

M1: Adopts a correct applied strategy of taking \(\arctan(\text{their found } t)\) and multiplying the result by 2 to obtain at least one value for \(x\) within the range \(-180^\circ \lt x \lt 180^\circ\) (or in radians \(-\pi \lt x \lt \pi\))

A1: See scheme

A1: For both 103.8 and \(-58.5\)

Note: Give final A0 for extra solutions given within the range \(-180^\circ \lt x \lt 180^\circ\)

Note: Ignore extra solutions outside the range \(-180^\circ \lt x \lt 180^\circ\) for the final A mark

Note: In degrees, \(\dfrac{x}{2} = \{51.88\ldots,\ -128.11\ldots,\ -29.26\ldots,\ 150.73\ldots\}\)

Note: Working in radians gives \(\dfrac{x}{2} = \{0.905\ldots, -0.510\ldots\} \Rightarrow x = \{1.81\ldots, -1.02\ldots\}\)

Note: Give 2^nd M0 for \(\dfrac{x}{2} = \{51.88\ldots, -29.26\ldots\} \Rightarrow x = \{25.9, -14.6\}\)