IB DP Math AA: Topic : AHL 3.10 :Compound angle identities: IB Style Questions HL Paper 2

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]

a.

\(\arctan \left( {\frac{1}{7}} \right) = \arcsin (x) \Rightarrow x = \sin \left( {\arctan \frac{1}{7}} \right) \approx 0.141\)     (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]

a.

\(\sin A = \sin \left( {180^\circ  – (B + C)} \right)\) 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]

b.

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.

a.

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.

b.

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]

Examiners report

[N/A]
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