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Question: For any angle, prove that \[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \]....

For any angle, prove that cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta .

Explanation

Solution

We have to find the general solution cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta for this we have to apply the trigonometric identities. We can prove the result by equating the left-hand side is equal to the right-hand side. There are basic six trigonometric functions like sin, cosine, tangent, cotangent, secant and cosecant. These are related to each other. Trigonometric identities are to be used to find the general solution of the given function.

Complete step by step answer:
We have given that cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta to prove the result consider cos⁡3θ\cos 3\theta as L.H.S.
cos⁡3θ\cos 3\theta Can be written as cos⁡(θ+2θ)\cos (\theta + 2\theta )
Now we have a trigonometric identity cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos (a + b) = \cos a\cos b - \sin a\sin b
Replacing aa from θ\theta and bb from 2θ2\theta in the identity we get cos⁡(θ+2θ)=cos⁡θcos⁡2θ−sin⁡θsin⁡2θ−−−−−−−−−−(a)\cos (\theta + 2\theta ) = \cos \theta \cos 2\theta - \sin \theta \sin 2\theta - - - - - - - - - - (a)

We have a trigonometric identity cos⁡2x=2cos⁡2x−1\cos 2x = 2{\cos ^2}x - 1, apply this on the right-hand side of equation (aa) replacing xx from θ\theta it become cos⁡(θ+2θ)=cos⁡θ(2cos⁡2θ−1)−sin⁡θsin⁡2θ−−−−−−−−−−(b)\cos (\theta + 2\theta ) = \cos \theta (2{\cos ^2}\theta - 1) - \sin \theta \sin 2\theta - - - - - - - - - - (b)
We have a trigonometric identity sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x, apply this on the right-hand side of the equation bbit will become cos⁡(θ+2θ)=cos⁡θ(2cos⁡2θ−1)−sin⁡θ(2sin⁡θcos⁡θ)\cos (\theta + 2\theta ) = \cos \theta (2{\cos ^2}\theta - 1) - \sin \theta (2\sin \theta \cos \theta )
As θ+2θ\theta + 2\theta is equal to 3θ3\theta so cos⁡(θ+2θ)\cos (\theta + 2\theta ) becomes cos⁡3θ\cos 3\theta in the above equation.
cos⁡3θ=cos⁡θ(2cos⁡2θ−1)−sin⁡θ(2sin⁡θcos⁡θ)\cos 3\theta = \cos \theta (2{\cos ^2}\theta - 1) - \sin \theta (2\sin \theta \cos \theta )
Solving R.H.S. of the equation we get
cos⁡3θ=2cos⁡3θ−cos⁡θ−2sin⁡2θcos⁡θ\cos 3\theta = 2{\cos ^3}\theta - \cos \theta - 2{\sin ^2}\theta \cos \theta

As the trigonometric identity of sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x so replacing xx from θ\theta substitute the value in the above equation
⇒cos⁡3θ=2cos⁡3θ−cos⁡θ−2(1−cos⁡2x)cos⁡θ\Rightarrow \cos 3\theta = 2{\cos ^3}\theta - \cos \theta - 2(1 - {\cos ^2}x)\cos \theta
⇒cos⁡3θ=2cos⁡3θ−cos⁡θ−2cos⁡θ(1−cos⁡2x)\Rightarrow \cos 3\theta = 2{\cos ^3}\theta - \cos \theta - 2\cos \theta (1 - {\cos ^2}x)
⇒cos⁡3θ=2cos⁡3θ−cos⁡θ−2cos⁡θ+2cos⁡3θ\Rightarrow \cos 3\theta = 2{\cos ^3}\theta - \cos \theta - 2\cos \theta + 2{\cos ^3}\theta

Now in R.H.S. adding the terms according to their power we get
cos⁡3θ=2cos⁡3θ+2cos⁡3θ−cos⁡θ−2cos⁡θ\cos 3\theta = 2{\cos ^3}\theta + 2{\cos ^3}\theta - \cos \theta - 2\cos \theta
cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta
Hence the result is proved.

Note: The above equation is based upon the trigonometric functions. To solve the equation trigonometric identities are to be used. Trigonometric functions are widely used in all science that are related to geometry such as navigations and solid mechanics.