Question

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

Answer 1

Hint: We have to find the general solution \[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \]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 \[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \] to prove the result consider \[\cos 3\theta \] as L.H.S.
\[\cos 3\theta \] Can be written as \[\cos (\theta + 2\theta )\]
Now we have a trigonometric identity \[\cos (a + b) = \cos a\cos b - \sin a\sin b\]
Replacing \[a\] from \[\theta \] and \[b\] from \[2\theta \]in the identity we get \[\cos (\theta + 2\theta ) = \cos \theta \cos 2\theta - \sin \theta \sin 2\theta - - - - - - - - - - (a)\]

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

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

Now in R.H.S. adding the terms according to their power we get
\[\cos 3\theta = 2{\cos ^3}\theta + 2{\cos ^3}\theta - \cos \theta - 2\cos \theta \]
\[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \]
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.