Question

Prove that : sin5 = 5 − 20
³ + 16
⁵​

Answer 1

★ Correct Qúestion :-

Prove that sin 5x = 5 sin x - 20 sin³ x + 16 sin⁵ x

★ Concept ::

Here the concept of Trigonometry has been used. We see that we are given an equation to prove that LHS = RHS. So firstly we can take the LHS. There we shall apply one Trigonometric formula and simplify it. Then again we can apply some formulas to simplify it more. Then on solving it, we can prove the terms.

Let's do it !!

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★ Formulas Used :-

$\;\boxed{\sf{\pink{\sin(A\:+\:B)\;=\;\sin A\cos B\;+\;\cos A\sin B}}}$

$\;\boxed{\sf{\pink{\sin 3x\;=\;3\sin x\:-\:4\sin^{3}x}}}$

$\;\boxed{\sf{\pink{\cos 3x\;=\;4\cos^{3} x\:-\:3\sin x}}}$

$\;\boxed{\sf{\pink{\cos 2x\;=\;1\:-\:2\sin^{2}x}}}$

$\;\boxed{\sf{\pink{\sin 2x\;=\;2\sin x\cos x}}}$

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★ Solution :-

Given,

» sin 5x = 5 sin x - 20 sin³ x + 16 sin⁵ x

Here ::

• LHS = sin 5x

• RHS = 5 sin x - 20 sin³ x + 16 sin⁵ x

Firstly let's start simplifying LHS.

→ LHS = sin 5x

This can be written as ,

→ LHS = sin (3x + 2x)

We know that,

$\;\tt{\rightarrow\;\;\sin(A\:+\:B)\;=\;\sin A\cos B\;+\;\cos A\sin B}$

• Here A = 3x

• Here B = 2x

By applying values, we get

→ LHS = sin(3x).cos(2x) + cos(3x).sin(2x)

Now we know that,

$\;\tt{\rightarrow\;\;\sin 3x\;=\;3\sin x\:-\:4\sin^{3}x}$

By applying values, we get

→ LHS = [3 sin(x) - 4 sin³(x)] × cos(2x) + cos(3x).sin(2x)

From formula we see that,

$\;\tt{\rightarrow\;\;\cos 2x\;=\;1\:-\:2\sin^{2}x}$

→ LHS = (3 sin x - 4 sin³x)(1 - 2 sin² x) + (cos 3x × sin 2x)

→ LHS = [ 3 sin x - 6 sin³ x - 4 sin³ x + 8 sin⁵ x ] + (cos 3x × sin 2x)

We know that,

$\;\sf{\rightarrow\;\;\cos 3x\;=\;4\cos^{3} x\:-\:3\cos x}$

By applying this, we get

→ LHS = [3 sin x - 10 sin³x + 8 sin⁵x] + [ (4 cos³x - 3 cos x) × (sin 2x) ]

Also we see that,

$\;\tt{\rightarrow\;\;\sin 2x\;=\;2\sin x\cos x}$

By applying this in the equation, we get

→ LHS = [3 sin x - 10 sin³x + 8 sin⁵ x] + [ (4 cos³ x - 3 cos x) × (2 sin x . cos x) ]

Taking sin x common from second part, we get

→ LHS = [3 sin x - 10 sin³x + 8 sin⁵ x] + [(8 cos⁴ x - 6 cos² x) × sin x]

We know that :: cos² x = 1 - sin²x

So :: cos⁴x = (1 - sin²x)²

By using this, we get

→ LHS = [3 sin x - 10 sin³x + 8 sin⁵x] + [8 (1 - sin²x)² - 6 (1 - sin²x)] × sin x

→ LHS = [3 sin x - 10 sin³x + 8 sin⁵x] + [8 sin x (1 - sin²x)² - 6 sin x (1 - sin²x)]

Now squaring the term given in the equation,

→ LHS = [3 sin x - 10 sin³x + 8 sin⁵x] + [8 sin x (1 + sin⁴x - 2 sin²x) - 6 sin x (1 - sin²x)]

→ LHS = [3 sin x - 10 sin³x + 8 sin⁵x] + [8 sin x + 8 sin⁵ x - 16 sin³x - 6 sin x + 6 sin³x]

Now opening both brackets, we get

→ LHS = 3 sin x - 10 sin³x + 8 sin⁵x + 8 sin x + 8 sin⁵ x - 16 sin³x - 6 sin x + 6 sin³ x

On rearranging we get,

→ LHS = 3 sin x + 2 sin x - 26 sin³x + 6 sin³x + 8 sin⁵ x + 8 sin⁵ x

→ LHS = 5 sin x - 20 sin³x + 16 sin⁵ x

This gives us that,

→ LHS = RHS = 5 sin x - 20 sin³x + 16 sin⁵x

Hence proved .

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★ More to know :-

• cos (A + B) = cos A cos B - sin A sin B

• cos (A - B) = cos A cos B + sin A sin B

• sin (A - B) = sin A cos B - cos A sin B

• cos 2x = cos²x - sin²x = 2 cos²x - 1

• sin² A - sin² B = sin(A + B) . sin (A - B)

Answer 2

Let theta = x

LHS sin5x

=sin(3x+2x)

=sin3x.cos2x+cos3x.sin2x

=(3sinx-4sin^3x).(1–2sin^2x)+(4cos^3x-3cosx).(2sinx.cosx)

=(3sinx-4sin^3x)(1–2sin^2x)+(4cos^4x-3cos^2x)(2sinx).

=(3sinx-4sin^3x)(1–2sin^2x)+cos^2x.(4cos^2x-3).(2sinx)

=3sinx-4sin^3x-6sin^3x+8sin^5x+(1-sin^2x).(4–4sin^2x-3).(2sinx)

=8sin^5x-10sin^3x+3sinx+(2sinx-2sin^3x)(1–4sin^2x).

=8sin^5x-10sin^3x+3sinx+2sinx-2sin^3x-8sin^3x+8sin^5x

=16sin^5x-20sin^3x+5sinx proved