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Question: How do you prove \[\sin 3x = 3\sin x - 4{\sin ^3}x\]?...

How do you prove sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4{\sin ^3}x?

Explanation

Solution

In the above question, the concept is based on the concept of trigonometry. The main approach to solve this expression is by applying angle sum formula and also double angle formula in trigonometric functions. Using this identity, we will get the above equation for multiple angle identity.

Complete step by step solution:
Trigonometric function means the function of the angle between the two sides. It tells us the relation between the angles and sides of the right-angle triangle.

If we can rewrite the given angle in terms of two angles that have known trigonometric values by finding the exact value of the sine, cosine, or tangent of an angle is often easier.

So, to find the angle of sine we break it up into a sum of two angles which is called special angles or alpha and beta.

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta

We will also use double angle formula for cosine to substitute in above formula,
cos⁡(2α)=cos⁡2α−sin⁡2α=2cos⁡2α−1=1−2sin⁡2α\cos \left( {2\alpha } \right) = {\cos ^2}\alpha - {\sin ^2}\alpha = 2{\cos ^2}\alpha - 1 = 1 - 2{\sin ^2}\alpha

So, starting with sin3x we need to split the angle in 2x and x. Therefore, it is written as we get,
sin⁡3x=sin⁡(2x+x)\sin 3x = \sin \left( {2x + x} \right)

Since, we have got two angles with angles 2x and x we can apply the angle sum formula,

sin⁡(2x+x)=sin⁡(2x)cos⁡(x)+cos⁡(2x)sin⁡x =(2sin⁡xcos⁡x)cos⁡x+(1−2sin⁡2x)sin⁡x \sin \left( {2x + x} \right) = \sin (2x)\cos (x) + \cos (2x)\sin x \\\ = \left( {2\sin x\cos x} \right)\cos x + \left( {1 - 2{{\sin }^2}x} \right)\sin x \\\

(∵sin⁡(2x)=2sin⁡xcos⁡x)\left( {\because \sin \left( {2x} \right) = 2\sin x\cos x} \right)

Further we get,

sin⁡(2x+x)=2sin⁡xcos⁡2x+sin⁡x−2sin⁡3x =2sin⁡x(1−sin⁡2x)+sin⁡x−2sin⁡3x =3sin⁡x−4sin⁡3x \sin (2x + x) = 2\sin x{\cos ^2}x + \sin x - 2{\sin ^3}x \\\ = 2\sin x(1 - {\sin ^2}x) + \sin x - 2{\sin ^3}x \\\ = 3\sin x - 4{\sin ^3}x \\\

Hence the expression is proved.
Note: An important thing to note is that angle sum formulas allow you to express the exact value of trigonometric expressions that you could not express. Splitting the above angle 3x into 2x and x makes it easier to solve the equation using the angle sum formula.