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NDA Math Trigonometric Ratios of Angles and Equations

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Given that

16\sin^5x=p\sin5x+q\sin3x+rsinx

[NDA II 2014]

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Question : 87 of 100

Marks: +1, -0

What is the value of p ?

1
2
-1
-2

Solution:

Given that

16\sin^5 x = p\sin 5x + q\sin 3x + r\sin x

16\sin^5 x = p(16\sin^5 x + 5\sin 3x - 10\sin x) + q\sin 3x + r\sin x

\Rightarrow 16\sin^5 x = p(16\sin^5 x + 5(3\sin x - 4\sin^3 x) - 10\sin x) + q(3\sin x - 4\sin^3) + r\sin x

\Rightarrow 16\sin^5 x = p(16\sin^5 x + 15\sin x - 20\sin^3 x) + q(3\sin x - 4\sin^3 x) + r\sin x

\Rightarrow p = 1

and

-20p - 4q = 0 \Rightarrow q = -5

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