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Question: Write \[\sin 4x\] in terms of \[\sin x\]....

Write sin⁡4x\sin 4x in terms of sin⁡x\sin x.

Explanation

Solution

In the given question sin⁡4x\sin 4x needs to be written in terms of sin⁡x\sin x. So we have to use a multiple angle formula. Use the formula sin⁡2a\sin 2a considering a=2xa = 2x, and then convert the “cos⁡\cos ” terms to “sin⁡\sin ” terms using the identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1.

Complete step by step solution:
Given expression is sin⁡4x\sin 4x, this can be rewritten as sin⁡2(2x)\sin 2\left( {2x} \right).
Let 2x=a2x = a.
∴sin⁡4x=sin⁡2(a)\therefore \sin 4x = \sin 2(a)
Use the formula sin⁡2a=2sin⁡a⋅cos⁡a\sin 2a = 2\sin a \cdot \cos a
∴sin⁡2(2x)=2sin⁡2x⋅cos⁡2x\therefore \sin 2\left( {2x} \right) = 2\sin 2x \cdot \cos 2x
Again use the formula sin⁡2a=2sin⁡a⋅cos⁡a\sin 2a = 2\sin a \cdot \cos a:
∴sin⁡2(2x)=2⋅(2sin⁡x⋅cos⁡x)⋅(cos⁡2x)\therefore \sin 2\left( {2x} \right) = 2 \cdot \left( {2\sin x \cdot \cos x} \right) \cdot \left( {\cos 2x} \right)
Use from the formula : cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x,
⇒cos⁡2x=1−2sin⁡2x\Rightarrow \cos 2x = 1 - 2{\sin ^2}x…..(1) [∵cos⁡2x=1−sin⁡2x][\because {\cos ^2}x = 1 - {\sin ^2}x]
Substitute the value of cos⁡2x\cos 2x from (1):
∴sin⁡2(2x)=2⋅(2sin⁡x⋅cos⁡x)⋅(1−2sin⁡2x)\therefore \sin 2\left( {2x} \right) = 2 \cdot \left( {2\sin x \cdot \cos x} \right) \cdot \left( {1 - 2{{\sin }^2}x} \right)
⇒sin⁡2(2x)=4sin⁡xcos⁡x−8sin⁡3xcos⁡x\Rightarrow \sin 2(2x) = 4\sin x\cos x - 8{\sin ^3}x\cos x
Take cos⁡x\cos x common:
⇒sin⁡2(2x)=(4sin⁡x−8sin⁡3x)cos⁡x\Rightarrow \sin 2(2x) = \left( {4\sin x - 8{{\sin }^3}x} \right)\cos x
Again, cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x
⇒cos⁡x=1−sin⁡2x\Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x}
Substitute the value of cos⁡x\cos x:
⇒sin⁡2(2x)=(4sin⁡x−8sin⁡3x)(1−sin⁡2x)\Rightarrow \sin 2(2x) = \left( {4\sin x - 8{{\sin }^3}x} \right)\left( {\sqrt {1 - {{\sin }^2}x} } \right)

Hence, sin⁡4x=(4sin⁡x−8sin⁡3x)(1−sin⁡2x)\sin 4x = \left( {4\sin x - 8{{\sin }^3}x} \right)\left( {\sqrt {1 - {{\sin }^2}x} } \right)

Note:
Students must memorise the following identities used in this solution:
sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A
cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = {\cos ^2}A - {\sin ^2}A
sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1
Some other trigonometric identities and formulas of multiple and sub multiple angles, sums and products, must also be memorized.
While converting angles is the quadrant of the angle is mentioned then the signs of the trigonometric ratios in different quadrants must also be taken care of. For that remember the following:
1st{1^{st}} quadrant: All trigonometric functions are positive.
2nd{2^{nd}} quadrant: Sine functions are positive.
3rd{3^{rd}} quadrant: Tan functions are positive.
4th{4^{th}} quadrant: Cos functions are positive.