Practise Compound and double angles; R cos(θ ± α). 4 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Build a paper on this topicWritten for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy8 marks
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
(a) Using the formula for \(\sin(A + B)\), show that \[\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}.\][3]
(b) Without using a calculator, find the exact value of \[\sin 12^\circ\cos 33^\circ + \cos 12^\circ\sin 33^\circ.\][2]
(c) Find the exact value of \(\tan 105^\circ\), giving your answer in the form \(a + b\sqrt{3}\), where \(a\) and \(b\) are integers.[3]
Show the answer and mark scheme
(a) Answer: \(\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}\) (shown)
- B1 for writing \(\sin 75^\circ\) as \(\sin\left(45^\circ + 30^\circ\right)\) (or an equivalent split)
- M1 for a correct expansion with exact values, e.g. \(\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2}\)
- A1* for \(\frac{\sqrt{6} + \sqrt{2}}{4}\) with no errors
Worked solution: \(\sin 75^\circ = \sin\left(45^\circ + 30^\circ\right) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ\)
\(= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}\)
(b) Answer: \(\frac{\sqrt{2}}{2}\)
- M1 for recognising the expression as \(\sin(12^\circ + 33^\circ) = \sin 45^\circ\)
- A1 for \(\frac{\sqrt{2}}{2}\) oe
Worked solution: \(\sin 12^\circ\cos 33^\circ + \cos 12^\circ\sin 33^\circ = \sin(12^\circ + 33^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2}\)
(c) Answer: \(\tan 105^\circ = -2 - \sqrt{3}\)
- M1 for \(\tan\left(135^\circ - 30^\circ\right) = \frac{\tan 135^\circ - \tan 30^\circ}{1 + \tan 135^\circ\tan 30^\circ}\) with exact values
- M1 for rationalising the denominator, e.g. multiplying by \(\frac{3 + \sqrt{3}}{3 + \sqrt{3}}\)
- A1 for \(-2 - \sqrt{3}\)
Worked solution: \(\tan 105^\circ = \tan\left(135^\circ - 30^\circ\right) = \frac{-1 - \frac{\sqrt{3}}{3}}{1 + \left(-1\right) \times \frac{\sqrt{3}}{3}} = \frac{-3 - \sqrt{3}}{3 - \sqrt{3}}\)
\(= \frac{-3 - \sqrt{3}}{3 - \sqrt{3}} \times \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{-12 - 6\sqrt{3}}{6} = -2 - \sqrt{3}\)
Question 2Medium7 marks
The angles \(A\) and \(B\) are such that \[\sin A = \frac{21}{29}, \quad \cos B = -\frac{12}{13},\] where \(\frac{\pi}{2} \lt A \lt \pi\) and \(\frac{\pi}{2} \lt B \lt \pi\).
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
(a) Find the exact value of \(\cos(A - B)\).[3]
(b) Find the exact value of \(\sin(A + B)\).[2]
(c) Find the exact value of \(\tan(A - B)\).[2]
Show the answer and mark scheme
(a) Answer: \(\cos(A - B) = \frac{345}{377}\)
- B1 for \(\cos A = -\frac{20}{29}\) and \(\sin B = \frac{5}{13}\) (with the correct signs for the quadrants of \(A\) and \(B\))
- M1 for \(\cos(A - B) = \cos A\cos B + \sin A\sin B\) with their values
- A1 for \(\frac{345}{377}\)
Worked solution: \(\cos A = -\frac{20}{29}\) (negative, as \(A\) is obtuse) and \(\sin B = \frac{5}{13}\) (\(B\) is in the second quadrant).
\(\cos(A - B) = \cos A\cos B + \sin A\sin B = \left(-\frac{20}{29}\right) \times \left(-\frac{12}{13}\right) + \frac{21}{29} \times \frac{5}{13} = \frac{345}{377}\)
(b) Answer: \(\sin(A + B) = -\frac{352}{377}\)
- M1 for \(\sin(A + B) = \sin A\cos B + \cos A\sin B\) with their values
- A1 for \(-\frac{352}{377}\)
Worked solution: \(\sin(A + B) = \sin A\cos B + \cos A\sin B = \frac{21}{29} \times \left(-\frac{12}{13}\right) + \left(-\frac{20}{29}\right) \times \frac{5}{13} = -\frac{352}{377}\)
(c) Answer: \(\tan(A - B) = -\frac{152}{345}\)
- M1 for \(\frac{\tan A - \tan B}{1 + \tan A\tan B}\) with \(\tan A = -\frac{21}{20}\) and \(\tan B = -\frac{5}{12}\), or dividing their \(\sin(A - B)\) by \(\cos(A - B)\)
- A1 for \(-\frac{152}{345}\)
Worked solution: \(\tan A = -\frac{21}{20}, \ \tan B = -\frac{5}{12}\), so \(\tan(A - B) = \frac{-\frac{21}{20} - \left(-\frac{5}{12}\right)}{1 + \left(-\frac{21}{20}\right) \times \left(-\frac{5}{12}\right)} = -\frac{152}{345}\).
Question 3Hard11 marks
The diagram shows the right-angled triangles \(OQP\), \(ORQ\), \(OTP\) and \(PSQ\). The points \(O\), \(T\) and \(R\) lie on a horizontal line, \(PT\) is vertical and \(SQ\) is horizontal.
Angle \(QOR = A\), angle \(POQ = B\), where \(A\) and \(B\) are acute and \(A + B \lt 90^\circ\), and \(OP = 1\).
Diagram not drawn to scale
[object Object]
(a) Write down expressions for the lengths \(OQ\) and \(PQ\) in terms of \(B\).[1]
(b) Explain why angle \(QPS = A\).[2]
(c) By finding the lengths \(PS\) and \(QR\), prove that \[\sin(A + B) = \sin A\cos B + \cos A\sin B.\][3]
(d) Use a similar method to prove that \(\cos(A + B) = \cos A\cos B - \sin A\sin B\).[3]
(e) Hence prove that \[\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}.\][2]
Show the answer and mark scheme
(a) Answer: \(OQ = \cos B, \ PQ = \sin B\)
- B1 for both \(OQ = \cos B\) and \(PQ = \sin B\)
Worked solution: In the right-angled triangle \(OQP\) with hypotenuse \(OP = 1\): \(OQ = \cos B\) and \(PQ = \sin B\).
(b) Answer: Angle \(SQO = A\) (alternate angles), so angle \(PQS = 90^\circ - A\) and angle \(QPS = A\).
- B1 for angle \(SQO = A\) (alternate angles, since \(SQ\) is parallel to \(OR\))
- B1 for angle \(PQS = 90^\circ - A\), so in the right-angled triangle \(PSQ\), angle \(QPS = 90^\circ - (90^\circ - A) = A\)
Worked solution: \(SQ\) is parallel to \(OR\), so angle \(SQO\) = angle \(QOR = A\) (alternate angles). Angle \(OQP = 90^\circ\), so angle \(PQS = 90^\circ - A\). The angles of triangle \(PSQ\) add up to \(180^\circ\) and angle \(PSQ = 90^\circ\), so angle \(QPS = A\).
(c) Answer: Proof
- M1 for \(QR = OQ\sin A = \sin A\cos B\)
- M1 for \(PS = PQ\cos A = \cos A\sin B\)
- A1* for \(PT = PS + ST = PS + QR\) and \(PT = OP\sin(A + B) = \sin(A + B)\), completing the proof
Worked solution: In triangle \(ORQ\): \(QR = OQ\sin A = \sin A\cos B\). In triangle \(PSQ\): \(PS = PQ\cos A = \cos A\sin B\).
\(SQRT\) is a rectangle, so \(ST = QR\). In triangle \(OTP\): \(PT = OP\sin(A + B) = \sin(A + B)\).
So \(\sin(A + B) = PT = PS + ST = \sin A\cos B + \cos A\sin B\).
(d) Answer: Proof
- M1 for \(OR = OQ\cos A = \cos A\cos B\)
- M1 for \(TR = SQ = PQ\sin A = \sin A\sin B\)
- A1* for \(OT = OR - TR\) and \(OT = \cos(A + B)\), completing the proof
Worked solution: \(OR = OQ\cos A = \cos A\cos B\) and \(TR = SQ = PQ\sin A = \sin A\sin B\). In triangle \(OTP\): \(OT = OP\cos(A + B) = \cos(A + B)\).
So \(\cos(A + B) = OT = OR - TR = \cos A\cos B - \sin A\sin B\).
(e) Answer: Proof
- M1 for \(\tan(A + B) = \frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}\) and dividing numerator and denominator by \(\cos A\cos B\)
- A1* for the given result
Worked solution: \(\tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)} = \frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}\). Dividing the numerator and denominator by \(\cos A\cos B\) gives \(\frac{\tan A + \tan B}{1 - \tan A\tan B}\).