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Question
Prove that: \[\cos^3 2x + 3 \cos 2x = 4\left( \cos^6 x - \sin^6 x \right)\]
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Solution
Let us consider the RHS
4(cos6 x – sin6 x)
Now, upon expansion we get,
4(cos6 x – sin6 x) = 4[(cos2 x)3 – (sin2 x)3]
= 4(cos2 x – sin2 x) (cos4 x + sin4 x + cos2 x sin2 x)
On using the formula,
a3 – b3 = (a - b) (a2 + b2 + ab)
= 4 cos 2x (cos4 x + sin4 x + cos2 x sin2 x + cos2 x sin2 x – cos2 x sin
As we know, cos 2x = cos2 x – sin2 x
Therefore,
= 4 cos 2x (cos4 x + sin4 x + 2 cos2 x sin2 x – cos2 x sin2 x)
= 4 cos 2x [(cos2 x)2 + (sin2 x)2 + 2 cos2 x sin2 x – cos2 x sin2 x]
As we know, a2 + b2 + 2ab = (a + b)2
= 4 cos 2x [(1)2 – 1/4 (4 cos2 x sin2 x)]
= 4 cos 2x [(1)2 – 1/4 (2 cos x sin x)2]
Again as we know, sin 2x = 2sin x cos x
= 4 cos 2x [(12) – 1/4 (sin 2x)2]
= 4 cos 2x (1 – 1/4 sin2 2x)
As as we know, sin2 x = 1 – cos2 x
= 4 cos 2x [1 – 1/4 (1 – cos2 2x)]
= 4 cos 2x [1 – 1/4 + 1/4 cos2 2x]
= 4 cos 2x [3/4 + 1/4 cos2 2x]
= 4 (3/4 cos 2x + 1/4 cos3 2x)
= 3 cos 2x + cos3 2x
= cos3 2x + 3 cos 2x
= LHS
Thus proved.
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