C3 January 2012 Q8

Notes

M1 Uses the identity \(\left\{\tan(A + B) = \dfrac{\sin(A + B)}{\cos(A + B)}\right\} = \dfrac{\sin A\cos B \pm \cos A\sin B}{\cos A\cos B \mp \sin A\sin B}\). Accept incorrect signs for this.
Just the right hand side is acceptable.

A1 Fully correct statement in terms of cos and sin \(\{\tan(A + B)\} = \dfrac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}\)

M1 Divide both numerator and denominator by cosAcosB. This can be stated or implied by working. If implied you must have seen at least one term modified on both the numerator and denominator.

A1* This is a given solution. The last two principal’s reports have highlighted lack of evidence in such questions. Both sides of the identity must be seen or implied. Eg lhs=
The minimum expectation for full marks is
\(\tan(A + B) = \dfrac{\sin(A + B)}{\cos(A + B)} = \dfrac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B} = \dfrac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{1 - \frac{\sin A\sin B}{\cos A\cos B}} = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}\)
The solution \(\tan(A + B) = \dfrac{\sin(A + B)}{\cos(A + B)} = \dfrac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B} = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}\) scores M1A1M0A0
The solution \(\tan(A + B) = \dfrac{\sin(A + B)}{\cos(A + B)} = \dfrac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}\ (\div \cos A\cos B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}\) scores M1A1M1A0

Alternative to (a) starting from rhs

M1 Uses correct identities for both tanA and tanB in the rhs expression. Accept only errors in signs

A1 \(\dfrac{\tan A + \tan B}{1 - \tan A\tan B} = \dfrac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{1 - \frac{\sin A\sin B}{\cos A\cos B}}\)

M1 Multiplies both numerator and denominator by cosAcosB. This can be stated or implied by working. If implied you must have seen at least one term modified on both the numerator and denominator

A1 This is a given answer. Correctly completes proof. All three expressions must be seen or implied.
\(\dfrac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B} = \dfrac{\sin(A + B)}{\cos(A + B)} = \tan(A + B)\)

Alternative to (a) starting from both sides

The usual method can be marked like this

M1 Uses correct identities for both tanA and tanB in the rhs expression. Accept only errors in signs

A1 \(\dfrac{\tan A + \tan B}{1 - \tan A\tan B} = \dfrac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{1 - \frac{\sin A\sin B}{\cos A\cos B}}\)

M1 Multiplies both numerator and denominator by cosAcosB. This can be stated or implied by working. If implied you must have seen at least one term modified on both the numerator and denominator

A1 Completes proof. Starting now from the lhs writes \(\tan(A + B) = \dfrac{\sin(A + B)}{\cos(A + B)} = \dfrac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}\)
And then states that the lhs is equal to the rhs Or hence proven. There must be a statement of closure

Notes

M1 An attempt to use part (a) with A=\(\theta\) and B=\(\frac{\pi}{6}\). Seeing \(\dfrac{\tan\theta + \tan\frac{\pi}{6}}{1 - \tan\theta\tan\frac{\pi}{6}}\) is enough evidence. Accept sign slips

M1 Uses the identity \(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\) or \(\frac{\sqrt{3}}{3}\) in the rhs of the identity on both numerator and denominator

A1* cso. This is a given solution. Both sides of the identity must be seen. All steps must be correct with no unreasonable jumps. Accept
\(\tan\left(\theta + \frac{\pi}{6}\right) = \dfrac{\tan\theta + \tan\frac{\pi}{6}}{1 - \tan\theta\tan\frac{\pi}{6}} = \dfrac{\tan\theta + \frac{1}{\sqrt{3}}}{1 - \tan\theta\frac{1}{\sqrt{3}}} = \dfrac{\sqrt{3}\tan\theta + 1}{\sqrt{3} - \tan\theta}\)
However the following is only worth 2 out of 3 as the last step is an unreasonable jump without further explanation
\(\tan\left(\theta + \frac{\pi}{6}\right) = \dfrac{\tan\theta + \tan\frac{\pi}{6}}{1 - \tan\theta\tan\frac{\pi}{6}} = \dfrac{\tan\theta + \frac{\sqrt{3}}{3}}{1 - \tan\theta\frac{\sqrt{3}}{3}} = \dfrac{\sqrt{3}\tan\theta + 1}{\sqrt{3} - \tan\theta}\)

Alternative to (b) from sin and cos

M1 Writes \(\tan\left(\theta + \frac{\pi}{6}\right) = \dfrac{\sin\left(\theta + \frac{\pi}{6}\right)}{\cos\left(\theta + \frac{\pi}{6}\right)} = \dfrac{\sin\theta\cos\frac{\pi}{6} + \cos\theta\sin\frac{\pi}{6}}{\cos\theta\cos\frac{\pi}{6} - \sin\theta\sin\frac{\pi}{6}}\)

M1 Uses the identities \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\) and \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\) oe in the rhs of the identity on both numerator and denominator and divides both numerator and denominator by \(\cos\theta\) to produce an identity in \(\tan\theta\)

A1 As in original scheme

Notes

M1 Use the given identity in (b) to obtain \(\tan\left(\theta + \frac{\pi}{6}\right) = \tan(\pi - \theta)\). Accept sign slips

dM1 Writes down an equation that will give one value of \(\theta\), usually \(\theta + \frac{\pi}{6} = \pi - \theta\). This is dependent upon the first M mark. Follow through on slips

ddM1 Attempts to solve their equation in \(\theta\). It must end \(\theta\)= and the first two marks must have been scored.

A1 Cso \(\theta = \frac{5}{12}\pi\) or \(\frac{11}{12}\pi\)

dddM1 Writes down an equation that would produce a second value of \(\theta\). Usually \(\theta + \frac{\pi}{6} = 2\pi - \theta\)

A1 cso \(\theta = \frac{5}{12}\pi\) (accept \(\frac{\pi}{2.4}\)) and \(\frac{11}{12}\pi\) with no extra solutions in the range. Ignore extra solutions outside the range.

Note that under this method one correct solution would score 4 marks. A small number of candidates find the second solution only. They would score 1,1,1,1,0,0

Alternative solution for c.

Starting with \(1 + \sqrt{3}\tan\theta = \left(\sqrt{3} - \tan\theta\right)\tan(\pi - \theta)\)

Let \(\tan\theta = t\)

\[\begin{aligned} 1 + \sqrt{3}t &= \left(\sqrt{3} - t\right)(-t) \\ t^2 - 2\sqrt{3}t - 1 &= 0 \\ t &= \frac{2\sqrt{3} \pm \sqrt{(12 + 4)}}{2} \\ &= \sqrt{3} \pm 2 \quad \text{Must find an exact surd} \\ \theta &= \frac{5\pi}{12}, \frac{11\pi}{12} \end{aligned}\]

Accept the use of a calculator for the A marks as long as there is an exact surd for the solution of the quadratic and exact answers are given.

M1 Starting with \(1 + \sqrt{3}\tan\theta = \left(\sqrt{3} - \tan\theta\right)\tan(\pi - \theta)\) expand \(\tan(\pi - \theta)\) by the correct compound angle identity (or otherwise) and substitute \(\tan\pi = 0\) to produce an equation in \(\tan\theta\)

dM1 Collect terms and produce a 3 term quadratic in \(\tan\theta\)

ddM1 Correct use of quadratic formula to produce exact solutions to \(\tan\theta\). All previous marks must have been scored.

dddM1 All 3 previous marks must have been scored. This is for producing two exact values for \(\theta\)

A1 One solution \(\frac{5}{12}\pi\) (accept \(\frac{\pi}{2.4}\)) or \(\frac{11}{12}\pi\)

A1 Both solutions \(\frac{5}{12}\pi\) (accept \(\frac{\pi}{2.4}\)) and \(\frac{11}{12}\pi\) and no extra solutions inside the range. Ignore extra solutions outside the range.

Special case: Watch for candidates who write \(\tan(\pi - \theta) = \tan(\pi) - \tan(\theta) = -\tan(\theta)\) and proceed correctly. They will lose the first mark but potentially can score the others.

Solutions in degrees
Apply as before. Lose the first correct mark that would have been scored-usually \(75^0\)