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Question: How do you prove \(\sin 3x = 3{\cos ^2}x\sin x - {\sin ^3}x\)?...

How do you prove sin⁡3x=3cos⁡2xsin⁡x−sin⁡3x\sin 3x = 3{\cos ^2}x\sin x - {\sin ^3}x?

Explanation

Solution

In order to proof the above statement ,first take the left hand side of the equation and write 3x3x as 2x+x2x + x and apply the formula of sum of angles of sine as sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B.Now use the formula of sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x and cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x to simplify the left-hand side and combine all the like terms, you will get left-hand side equal to right hand side .

Complete step by step solution:
To prove: sin⁡3x=3cos⁡2xsin⁡x−sin⁡3x\sin 3x = 3{\cos ^2}x\sin x - {\sin ^3}x
Proof: In order to prove the above equation, we will be first taking the left-hand side of the equation and do some operations
Taking Left-hand Side of the equation,
⇒sin⁡3x\Rightarrow \sin 3x
writing 3x3x as 2x+x2x + x,we get
⇒sin⁡(2x+x)\Rightarrow \sin \left( {2x + x} \right)
Now applying the sum of angle formula of sine sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B, we get
⇒sin⁡2xcos⁡x+cos⁡2xsin⁡x\Rightarrow \sin 2x\cos x + \cos 2x\sin x
As we know that sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos xand cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x,putting these values in above
⇒(2sin⁡xcos⁡x)cos⁡x+(cos⁡2x−sin⁡2x)sin⁡x\Rightarrow \left( {2\sin x\cos x} \right)\cos x + \left( {{{\cos }^2}x - {{\sin }^2}x} \right)\sin x
Simplifying further,
⇒2sin⁡xcos⁡2x+sin⁡xcos⁡2x−sin⁡3x\Rightarrow 2\sin x{\cos ^2}x + \sin x{\cos ^2}x - {\sin ^3}x
Combining like terms,
⇒3sin⁡xcos⁡2x−sin⁡3x\Rightarrow 3\sin x{\cos ^2}x - {\sin ^3}x
∴LHS=3sin⁡xcos⁡2x−sin⁡3x\therefore LHS = 3\sin x{\cos ^2}x - {\sin ^3}x
Taking Right-hand Side part of the equation
RHS=3sin⁡xcos⁡2x−sin⁡3xRHS = 3\sin x{\cos ^2}x - {\sin ^3}x
∴LHS=RHS\therefore LHS = RHS
Hence, proved.

Additional Information:
1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2. Even Function – A function f(x)f(x) is said to be an even function ,if f(−x)=f(x)f( - x) = f(x)for all x in its domain.
Odd Function – A function f(x)f(x) is said to be an even function ,if f(−x)=−f(x)f( - x) = - f(x)for all x in its domain.
We know that sin⁡(−θ)=−sin⁡θ.cos⁡(−θ)=cos⁡θ and tan⁡(−θ)=−tan⁡θ\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta
Therefore, sin⁡θ\sin \theta and tan⁡θ\tan \theta and their reciprocals, cosec⁡θ\cosec\theta and cot⁡θ\cot \theta are odd functions whereas cos⁡θ\cos \theta and its reciprocal sec⁡θ\sec \theta are even functions.

Note:
1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2.Fromula should be correctly used at every point.