Question

Prove that Sinx + Sin3x + Sin5x =0

Answer 1

Answer:

Explanation:

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Answer 2

ANSWER:-

To prove:

sinx + sin3x +sin5x = 0

Proof:

The given trigonometric equation is;

sinx + sin3x + sin5x =0.

$= > sin 3x + (sin5x + sinx) = 0 \\ \\ = > sin3x + 2sin( \frac{5x + x}{2} )cos( \frac{5x - x)}{2} = 0 \: \: \: \: \: \: \: [sinA+ sinB = 2sin( \frac{A + B}{2} )cos( \frac{A - B}{2} )]\\ \\ = > sin3x + 2sin \frac{6x}{2} cos \frac{4x}{2} = 0 \\ \\ = > sin3x + 2sin3xcos2x = 0 \\ \\ = > sin3x(1 + 2cos2x) = 0 \\ \\ = > sin3x = 0 \\ \\ = > 3x = n\pi \\ \\ = > x = \frac{n\pi}{3} , n∈z \\ \\And ,\\ \\ 1 + 2cos2x = 0 \\ \\ = > cos2x = - \frac{1}{2} \\ We \: know \: that \: cos( \frac{2\pi}{3} ) = - \frac{1}{2} \\ \\ = > 2x = 2m\pi ± \frac{2\pi}{3} \\ \\ = > x = m\pi ± \frac{\pi}{3} , \: m∈z \\ Thus, \\ The \: solution \: of \: the \: given \:equation \: is, \: x = \frac{n\pi}{3} \: \: o r \: \: x = m\pi ±\frac{\pi}{3} \: where \: m ,n∈z$

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