Prove the following identities:
3(sin x – cos x)4+6(sin x + cos x)2+4(sin6 x + cos6 x) = 13
To prove: 3(sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x) = 13
Proof:
Take LHS:
3(sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x)
Identities used:
(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)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)}
{∵ sin2 x + cos2 x = 1}
= 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
Hence Proved