IB DP Maths Topic 3.3 Double angle identities for sine and cosine SL Paper 2

 

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Question

The diagram below shows a plan for a window in the shape of a trapezium.


Three sides of the window are \(2{\text{ m}}\) long. The angle between the sloping sides of the window and the base is \(\theta \) , where \(0 < \theta  < \frac{\pi }{2}\) .

Show that the area of the window is given by \(y = 4\sin \theta + 2\sin 2\theta \) .

[5]
a.

Zoe wants a window to have an area of \(5{\text{ }}{{\text{m}}^2}\). Find the two possible values of \(\theta \) .

[4]
b.

John wants two windows which have the same area A but different values of \(\theta \) .

Find all possible values for A .

[7]
c.
Answer/Explanation

Markscheme

evidence of finding height, h     (A1)

e.g. \(\sin \theta = \frac{h}{2}\) , \(2\sin \theta \)

evidence of finding base of triangle, b     (A1)

e.g. \(\cos \theta = \frac{b}{2}\) , \(2\cos \theta \)

attempt to substitute valid values into a formula for the area of the window     (M1)

e.g. two triangles plus rectangle, trapezium area formula

correct expression (must be in terms of \(\theta \) )     A1

e.g. \(2\left( {\frac{1}{2} \times 2\cos \theta \times 2\sin \theta } \right) + 2 \times 2\sin \theta \) , \(\frac{1}{2}(2\sin \theta )(2 + 2 + 4\cos \theta )\)

attempt to replace \(2\sin \theta \cos \theta \) by \(\sin 2\theta \)     M1

e.g. \(4\sin \theta + 2(2\sin \theta \cos \theta )\)

\(y = 4\sin \theta + 2\sin 2\theta \)     AG     N0

[5 marks]

a.

correct equation     A1

e.g. \(y = 5\) , \(4\sin \theta + 2\sin 2\theta = 5\)

evidence of attempt to solve     (M1)

e.g. a sketch, \(4\sin \theta + 2\sin \theta – 5 = 0\)

\(\theta = 0.856\) \(({49.0^ \circ })\) , \(\theta = 1.25\) \(({71.4^ \circ })\)     A1A1     N3

[4 marks]

b.

recognition that lower area value occurs at \(\theta = \frac{\pi }{2}\)     (M1)

finding value of area at \(\theta  = \frac{\pi }{2}\)     (M1)

e.g. \(4\sin \left( {\frac{\pi }{2}} \right) + 2\sin \left( {2 \times \frac{\pi }{2}} \right)\) , draw square

\(A = 4\)     (A1)

recognition that maximum value of y is needed     (M1)

\(A = 5.19615 \ldots \)     (A1)

\(4 < A < 5.20\) (accept \(4 < A < 5.19\) )     A2      N5

[7 marks]

c.
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