First slide

Multiple and sub- multiple Angles

Question

If $\sin 5 θ = a \sin^{5} θ + b \sin^{3} θ + c \sin θ + d,$ then

Moderate

A

$a + b + c = 0$
B

$a + b + c + d = 0$
C

$5 a + 3 b - 4 c = 0$
D

$a - 3 c + d = 0$

Solution

$\begin{array}{l} Sin 5 θ = \sin (2 θ + 3 θ) \\ = \sin 2 θ \cos 3 θ + \cos 2 θ \sin 3 θ \\ = 2 \sin θ \cos θ \times \cos θ (4 \cos^{2} θ - 3) + (1 - 2 \sin^{2} θ) (3 \sin θ - 4 \sin^{3} θ) \\ = 2 \sin θ (1 - \sin^{2} θ) (1 - 4 \sin^{2} θ) + (1 - 2 \sin^{2} θ) (3 \sin θ - 4 \sin^{3} θ) \\ = 16 \sin^{5} θ - 20 \sin^{3} θ + 5 \sin θ \\ So a = 16, b = - 20, c = 5, d = 0 \\ \Rightarrow 5 a + 3 b - 4 c = 0 \end{array}$

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