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Question: Verify the identity \[\cos 3t = {\cos ^3}t - 3{\sin ^2}t.\cos t\]....

Verify the identity cos⁡3t=cos⁡3t−3sin⁡2t.cos⁡t\cos 3t = {\cos ^3}t - 3{\sin ^2}t.\cos t.

Explanation

Solution

Given identity is related to trigonometric function only. We have to verify whether it is an identity or not. For that we will use some formulas from the trigonometry itself. We will start from LHS and will try to conclude it to RHS.

Complete step by step answer:
Given the identity to be verified is cos⁡3t=cos⁡3t−3sin⁡2t.cos⁡t\cos 3t = {\cos ^3}t - 3{\sin ^2}t.\cos t
Now we will start with LHS,
cos⁡3t=cos⁡(t+2t)\cos 3t = \cos (t + 2t)
We can express the angle in the bracket as the above form. Now as we know that,
cos⁡(x+y)=cos⁡x.cos⁡y−sin⁡x.sin⁡y\cos (x + y) = \cos x.\cos y - \sin x.\sin y
We will use this formula, cos⁡3t=cos⁡t.cos⁡2t−sin⁡t.sin⁡2t\cos 3t = \cos t.\cos 2t - \sin t.\sin 2t
Now we will rearrange the double angles of sin and cos,
=cos⁡t(cos⁡2t−sin⁡2t)−sin⁡t(2sin⁡t.cos⁡t)= \cos t({\cos ^2}t - {\sin ^2}t) - \sin t(2\sin t.\cos t)
As, sin⁡2t=2sin⁡t.cos⁡t\sin 2t = 2\sin t.\cos t and cos⁡2t=cos⁡2t−sin⁡2t\cos 2t = {\cos ^2}t - {\sin ^2}t
On multiplying we get,
=cos⁡3t−cos⁡t.sin⁡2t−2sin⁡2t.cos⁡t= {\cos ^3}t - \cos t.{\sin ^2}t - 2{\sin ^2}t.\cos t
=cos⁡3t−3sin⁡2t.cos⁡t= {\cos ^3}t - 3{\sin ^2}t.\cos t
This is nothing but the RHS
=RHS= RHS
Hence proved.

Note:
Note that, we are about to verify whether it is an identity or not. So we used to prove LHS is equal to RHS. Sometimes we are asked to prove a certain identity. generally all the trigonometric identities are proved with the help of a right angle triangle. Note that, cos⁡3t\cos 3t is a triple angle and cos⁡3t{\cos ^3}t is a cubic function.