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Question: Show that: \(\cos 6x=32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1\)...

Show that:
cos⁡6x=32cos⁡6x−48cos⁡4x+18cos⁡2x−1\cos 6x=32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1

Explanation

Solution

Hint : To prove the given expression, we can write cos⁡6x\cos 6xascos⁡2(3x)\cos 2\left( 3x \right). Now, apply the identity ofcos⁡2θ\cos 2\theta on cos⁡2(3x)\cos 2\left( 3x \right) which will be equal to 2cos⁡23x−12{{\cos }^{2}}3x-1. Then apply the identity of cos⁡3x\cos 3x which is equal to 4cos⁡3x−3cos⁡x4{{\cos }^{3}}x-3\cos x. And simplify.

Complete step by step solution :
The given equation that we have to prove is:
cos⁡6x=32cos⁡6x−48cos⁡4x+18cos⁡2x−1\cos 6x=32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1
We are going to expand cos⁡6x\cos 6x in the above equation as follows.
We can write cos⁡6x\cos 6x as cos⁡2(3x)\cos 2\left( 3x \right) and then we will apply the identity of cos⁡2θ\cos 2\theta on cos⁡2(3x)\cos 2\left( 3x \right).
cos⁡2(3x)=2cos⁡23x−1\cos 2\left( 3x \right)=2{{\cos }^{2}}3x-1
From the trigonometric double angle identities we know that, cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x=4{{\cos }^{3}}x-3\cos x. Substituting this value ofcos⁡3x\cos 3xin the above equation we get,
2(4cos⁡3x−3cos⁡x)2−1 =2(16cos⁡6x+9cos⁡2x−24cos⁡4x)−1 =32cos⁡6x+18cos⁡4x−48cos⁡4x−1 \begin{aligned} & 2{{\left( 4{{\cos }^{3}}x-3\cos x \right)}^{2}}-1 \\\ & =2\left( 16{{\cos }^{6}}x+9{{\cos }^{2}}x-24{{\cos }^{4}}x \right)-1 \\\ & =32{{\cos }^{6}}x+18{{\cos }^{4}}x-48{{\cos }^{4}}x-1 \\\ \end{aligned}
From the above simplification, we get the L.H.S=32cos⁡6x+18cos⁡4x−48cos⁡4x−1L.H.S=32{{\cos }^{6}}x+18{{\cos }^{4}}x-48{{\cos }^{4}}x-1
And R.H.S of the given expression is equal to 32cos⁡6x−48cos⁡4x+18cos⁡2x−132{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1
As L.H.S = R.H.S Hence, we have proved the given expression.

Note : The other way of showing the equality of the given expression:
cos⁡6x=32cos⁡6x−48cos⁡4x+18cos⁡2x−1\cos 6x=32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1
We can also write cos⁡6x\cos 6x as cos⁡3(2x)\cos 3(2x) and then apply the identity of cos⁡3θ\cos 3\theta on cos⁡3(2x)\cos 3(2x) whereθ=2x\theta =2x.
cos⁡6x=4cos⁡32x−3cos⁡2x\cos 6x=4{{\cos }^{3}}2x-3\cos 2x
Now, substituting cos⁡2x=2cos⁡2x−1\cos 2x=2{{\cos }^{2}}x-1 in the above equation we get,
cos⁡6x=4(2cos⁡2x−1)3−3(2cos⁡2x−1)\cos 6x=4{{\left( 2{{\cos }^{2}}x-1 \right)}^{3}}-3\left( 2{{\cos }^{2}}x-1 \right)
Applying (a−b)3{{\left( a-b \right)}^{3}} identity in the above equation we get,
cos⁡6x=4(8cos⁡6x−1−3.4cos⁡4x+3.2cos⁡2x)−6cos⁡2x+3 ⇒cos⁡6x=32cos⁡6x−4−48cos⁡4x+24cos⁡2x−6cos⁡2x+3 ⇒cos⁡6x=32cos⁡6x−48cos⁡4x+18cos⁡2x−1 \begin{aligned} & \cos 6x=4\left( 8{{\cos }^{6}}x-1-3.4{{\cos }^{4}}x+3.2{{\cos }^{2}}x \right)-6{{\cos }^{2}}x+3 \\\ & \Rightarrow \cos 6x=32{{\cos }^{6}}x-4-48{{\cos }^{4}}x+24{{\cos }^{2}}x-6{{\cos }^{2}}x+3 \\\ & \Rightarrow \cos 6x=32{{\cos }^{6}}x-48{{\cos }^{4}}x+18{{\cos }^{2}}x-1 \\\ \end{aligned}
Now, the simplification of cos⁡6x\cos 6x yields the same result as given in the R.H.S of the given expression.
Hence, we have proved L.H.S = R.H.S by making a little change in writing the trigonometric functioncos⁡6x\cos 6x.
You must know the double and triple angle identities of cosine and sine as it has many applications in proving the trigonometric expressions just as we have shown above.