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 \(\dfrac{\theta}{2} = \{1.249\ldots, 2.553\ldots\} \Rightarrow \theta = \{2.498\ldots, 5.107\ldots\}\)