Question
a.Given that \(\arctan \frac{1}{2} – \arctan \frac{1}{3} = \arctan a,{\text{ }}a \in {\mathbb{Q}^ + }\), find the value of a.[3]
b.Hence, or otherwise, solve the equation \(\arcsin x = \arctan a\).[2]
▶️Answer/Explanation
Markscheme
\(\tan \left( {\arctan \frac{1}{2} – \arctan \frac{1}{3}} \right) = \tan (\arctan a)\) (M1)
\(a = 0.14285 \ldots = \frac{1}{7}\) (A1)A1
[3 marks]
\(\arctan \left( {\frac{1}{7}} \right) = \arcsin (x) \Rightarrow x = \sin \left( {\arctan \frac{1}{7}} \right) \approx 0.141\) (M1)A1
Note: Accept exact value of \(\left( {\frac{1}{{\sqrt {50} }}} \right)\).
[2 marks]
Question
In triangle \(ABC\),
\(3\sin B + 4\cos C = 6\) and
\(4\sin C + 3\cos B = 1\).
a.Show that \(\sin (B + C) = \frac{1}{2}\).[6]
b.Robert conjectures that \({\rm{C\hat AB}}\) can have two possible values.
Show that Robert’s conjecture is incorrect by proving that \({\rm{C\hat AB}}\) has only one possible value.[5]
▶️Answer/Explanation
Markscheme
METHOD 1
squaring both equations M1
\(9{\sin ^2}B + 24\sin B\cos C + 16{\cos ^2}C = 36\) (A1)
\(9{\cos ^2}B + 24\cos B\sin C + 16{\sin ^2}C = 1\) (A1)
adding the equations and using \({\cos ^2}\theta + {\sin ^2}\theta = 1\) to obtain \(9 + 24\sin (B + C) + 16 = 37\) M1
\(24(\sin B\cos C + \cos B\sin C) = 12\) A1
\(24\sin (B + C) = 12\) (A1)
\(\sin (B + C) = \frac{1}{2}\) AG
METHOD 2
substituting for \(\sin B\) and \(\cos B\) to obtain
\(\sin (B + C) = \left( {\frac{{6 – 4\cos C}}{3}} \right)\cos C + \left( {\frac{{1 – 4\sin C}}{3}} \right)\sin C\) M1
\( = \frac{{6\cos C + \sin C – 4}}{3}\;\;\;\)(or equivalent) A1
substituting for \(\sin C\) and \(\cos C\) to obtain
\(\sin (B + C) = \sin B\left( {\frac{{6 – 3\sin B}}{4}} \right) + \cos B\left( {\frac{{1 – 3\cos B}}{4}} \right)\) M1
\( = \frac{{\cos B + 6\sin B – 3}}{4}\;\;\;\)(or equivalent) A1
Adding the two equations for \(\sin (B + C)\):
\(2\sin (B + C) = \frac{{(18\sin B + 24\cos C) + (4\sin C + 3\cos B) – 25}}{{12}}\) A1
\(\sin (B + C) = \frac{{36 + 1 – 25}}{{24}}\) (A1)
\(\sin (B + C) = \frac{1}{2}\) AG
METHOD 3
substituting \(\sin B\) and \(\sin C\) to obtain
\(\sin (B + C) = \left( {\frac{{6 – 4\cos C}}{3}} \right)\cos C + \cos B\left( {\frac{{1 – 3\cos B}}{4}} \right)\) M1
substituting for \(\cos B\) and \(\cos B\) to obtain
\(\sin (B + C) = \sin B\left( {\frac{{6 – 3\sin B}}{4}} \right) + \left( {\frac{{1 – 4\sin C}}{3}} \right)\sin C\) M1
Adding the two equations for \(\sin (B + C)\):
\(2\sin (B + C) = \frac{{6\cos C + \sin C – 4}}{3} + \frac{{6\sin B + \cos B – 3}}{4}\;\;\;\)(or equivalent) A1A1
\(2\sin (B + C) = \frac{{(18\sin B + 24\cos C) + (4\sin C + 3\cos B) – 25}}{{12}}\) A1
\(\sin (B + C) = \frac{{36 + 1 – 25}}{{24}}\) (A1)
\(\sin (B + C) = \frac{1}{2}\) AG
[6 marks]
\(\sin A = \sin \left( {180^\circ – (B + C)} \right)\) so \(\sin A = \sin (B + C)\) R1
\(\sin (B + C) = \frac{1}{2} \Rightarrow \sin A = \frac{1}{2}\) A1
\( \Rightarrow A = 30^\circ \) or \(A = 150^\circ \) A1
if \(A = 150^\circ \), then \(B < 30^\circ \) R1
for example, \(3\sin B + 4\cos C < \frac{3}{2} + 4 < 6\), ie a contradiction R1
only one possible value \((A = 30^\circ )\) AG
[5 marks]
Total [11 marks]
Examiners report
Most candidates found this a difficult question with a large number of candidates either not attempting it or making little to no progress. In part (a), most successful candidates squared both equations, added them together, used \({\cos ^2}\theta + {\sin ^2}\theta = 1\) and then simplified their result to show that \(\sin (B + C) = \frac{1}{2}\). A number of candidates started with a correct alternative method (see the markscheme for alternative approaches) but were unable to follow them through fully.
In part (b), a small percentage of candidates were able to obtain \(B + C = 30^\circ {\text{ }}(A = 150^\circ )\) or \(B + C = 150^\circ {\text{ }}(A = 30^\circ )\) but were then unable to demonstrate or explain why \(A = 30^\circ \) is the only possible value for triangle ABC.
Question
Let \(f\left( x \right) = {\text{tan}}\left( {x + \pi } \right){\text{cos}}\left( {x – \frac{\pi }{2}} \right)\) where \(0 < x < \frac{\pi }{2}\).
Express \(f\left( x \right)\) in terms of sin \(x\) and cos \(x\).
▶️Answer/Explanation
Markscheme
\({\text{tan}}\left( {x + \pi } \right) = \tan x\left( { = \frac{{{\text{sin}}\,x}}{{{\text{cos}}\,x}}} \right)\) (M1)A1
\({\text{cos}}\left( {x – \frac{\pi }{2}} \right) = {\text{sin}}\,x\) (M1)A1
Note: The two M1s can be awarded for observation or for expanding.
\({\text{tan}}\left( {x + \pi } \right) = {\text{cos}}\left( {x – \frac{\pi }{2}} \right) = \frac{{{\text{si}}{{\text{n}}^2}\,x}}{{{\text{cos}}\,x}}\) A1
[5 marks]