Question

Prove the following:

3(sin x – cos x)⁴ + 6(sin x + cos x)² + 4(sin⁶x + cos⁶x) = 13

Answer 1

(sin x – cos x)⁴ 

= [(sin x – cos x)²]²

= (sin²x + cos²x – 2 sin x cos x)²

= (1 – 2 sin x cos x)²

= 1 – 4 sin x cos x + 4 sin²x cos²x

(sin x + cos x)²

= sin²x + cos²x + 2 sin x cos x

= 1 + 2 sin x cos x

sin⁶x + cos⁶x

= (sin²x)³ + (cos²x)³

= (sin²x + cos²x)³ – 3 sin²x cos²x (sin²x + cos²x) ...[∵ a³ + b³ = (a + b)³ – 3ab (a + b)]

= 1³ – 3 sin²x cos²x (1)

= 1 – 3 sin²x cos²x

 L.H.S. = 3 (sin x – cos x)⁴ + 6 (sin x + cosx)² + 4 (sin⁶x + cos⁶x)

= 3 (1 – 4 sin x cos x + 4 sin²x cos²x) + 6 (1 + 2 sin x cos x) + 4 (1 – 3 sin²x cos²x)

=  3 – 12 sin x cos x + 12 sin²x cos²x + 6 + 12 sin x cos x + 4 – 12 sin²x cos²x

= 13

= R.H.S.