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Question: Prove the following Trigonometric expression: \(\cos 4x=1-8{{\sin }^{2}}x{{\cos }^{2}}x\)...

Prove the following Trigonometric expression:
cos⁡4x=1−8sin⁡2xcos⁡2x\cos 4x=1-8{{\sin }^{2}}x{{\cos }^{2}}x

Explanation

Solution

Hint:To prove the given expression, apply the identity of cos⁡2θ\cos 2\theta on cos⁡4x\cos 4x then you will get1−2sin⁡22x1-2{{\sin }^{2}}2x. Now, apply sin⁡2x\sin 2x as 2sin⁡xcos⁡x2\sin x\cos x in the expression 1−2sin⁡22x1-2{{\sin }^{2}}2x and hence, simplify the expression to make it equal to R.H.S of the given expression.

Complete step-by-step answer:
The equation that we have to prove is:
cos⁡4x=1−8sin⁡2xcos⁡2x\cos 4x=1-8{{\sin }^{2}}x{{\cos }^{2}}x
We are going to apply the identity of cos⁡2θ\cos 2\theta on cos⁡4x\cos 4x as:
cos⁡2θ=1−2sin⁡2θ cos⁡4x=1−2sin⁡22x \begin{aligned} & \cos 2\theta =1-2{{\sin }^{2}}\theta \\\ & \cos 4x=1-2{{\sin }^{2}}2x \\\ \end{aligned}
We know that sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x. So, applying this value of sin⁡2x\sin 2x in the above equation we get,
cos⁡4x=1−2(2sin⁡xcos⁡x)2 ⇒cos⁡4x=1−8sin⁡2xcos⁡2x \begin{aligned} & \cos 4x=1-2{{\left( 2\sin x\cos x \right)}^{2}} \\\ & \Rightarrow \cos 4x=1-8{{\sin }^{2}}x{{\cos }^{2}}x \\\ \end{aligned}
From the above simplification, we have got cos⁡4x\cos 4x as 1−8sin⁡2xcos⁡2x1-8{{\sin }^{2}}x{{\cos }^{2}}x which is equal to R.H.S of the given expression.
Hence, we have proved that cos⁡4x=1−8sin⁡2xcos⁡2x\cos 4x=1-8{{\sin }^{2}}x{{\cos }^{2}}x.

Note: In spite of solving the L.H.S of the given equation:
cos⁡4x=1−8sin⁡2xcos⁡2x\cos 4x=1-8{{\sin }^{2}}x{{\cos }^{2}}x
We can also resolve R.H.S to L.H.S. Let’s check out how we are going to do this.
The R.H.S of given equation is:
1−8sin⁡2xcos⁡2x1-8{{\sin }^{2}}x{{\cos }^{2}}x
Rewriting the above expression as:
1−2(2sin⁡xcos⁡x)21-2{{(2\sin x\cos x)}^{2}}
Now, we know that sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x so we can write 2sin⁡xcos⁡x2\sin x\cos x in the above equation as sin⁡2x\sin 2x.
1−2(sin⁡2x)21-2{{\left( \sin 2x \right)}^{2}}
From the trigonometric double angles identity we know that cos⁡2θ=1−2sin⁡2θ\cos 2\theta =1-2{{\sin }^{2}}\theta . In the above expression the value of θ is equal to 2x so substituting the value of θ in cos⁡2θ=1−2sin⁡2θ\cos 2\theta =1-2{{\sin }^{2}}\theta we get, cos⁡4x=1−2sin⁡22x\cos 4x=1-2{{\sin }^{2}}2x. Hence, we can write 1−2sin⁡22x1-2{{\sin }^{2}}2x as cos⁡4x\cos 4x.
From the above simplification the R.H.S has come out as cos⁡4x\cos 4x which is equal to L.H.S.
Hence, we have proved L.H.S = R.H.S of the given expression by resolving R.H.S to L.H.S.