1.

Prove the identity: 3(sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x) = 13

Answer»

Let us consider the LHS

3(sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x)

As we know, (a + b)2 = a2 + b2 + 2ab

(a – b)2 = a2 + b2 – 2ab

a3 + b3 = (a + b) (a2 + b2 – ab)

Therefore,

3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x) = 3{(sin x – cos x) 2}2 + 6 {(sin x)2 + (cos x)2 + 2 sin x cos x)} + 4{(sin2 x)3 + (cos2 x)3}

= 3{(sin x)2 + (cos x)2 – 2 sin x cos x)}2 + 6(sin2 x + cos2 x + 2 sin x cos x) + 4{(sin2 x + cos2 x) (sin4 x + cos4 x – sin2 x cos2 x)}

= 3(1 – 2 sin x cos x)2 + 6 (1 + 2 sin x cos x) + 4{(1) (sin4 x + cos4 x – sin2 x cos2 x)}

As we know, sin2 x + cos2 x = 1

Therefore,

= 3{12 + (2 sin x cos x)2 – 4 sin x cos x} + 6(1 + 2 sin x cos x) + 4{(sin2 x)2 + (cos2 x)2 + 2 sin2 x cos2 x – 3 sin2 x cos2 x)}

= 3{1 + 4 sin2 x cos2 x – 4 sin x cos x} + 6(1 + 2 sin x cos x) + 4{(sin2 x + cos2 x) 2 – 3 sin2 x cos2 x)}

= 3 + 12 sin2 x cos2 x – 12 sin x cos x + 6 + 12 sin x cos x + 4{(1)2 – 3 sin2 x cos2 x)}

= 9 + 12 sin2 x cos2 x + 4(1 – 3 sin2 x cos2 x)

= 9 + 12 sin2 x cos2 x + 4 – 12 sin2 x cos2 x

= 13

= RHS

Thus proved.


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