 # value of 3(Sinx-Cosx)^4 + 6(Sinx + Cosx)^2 + 4(Sin^6x + Cos^6x)

5 years ago
Dear student

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(sin2x + cos2x + 2 sin x cos x) + 4 [(sin2x)3 + (cos2x)3]

= 3 [sin2x + cos2x – 2 sin x cos x]2 + 6(1 + 2 sin x cos x) + 4 [(sin2x + cos2x) (sin4x + cos4x – sin2x cos2x)]

= 3 [1 – 2 sin x cosx)2 + 6 + 12 sin x cos x + 4 [(sin2x)2 + (cos2 x)2 + 2 sin2x cos2x – 3 sin2 x cos2x]

[ using: sin 2 x + cos 2 x = 1]

= 3 [1 + 4 sin2x cos2x – 4 sin x cos x)] + 6 + 12 sin x cos x + 4 [(sin2x + cos x)2 – 3 sin2x cos2x]

= 3 + 12 sin2 x cos2x – 12 sin x cos x + 6 + 12 sin x cos x + 4 – 12 sin2x cos2x

= 13, which is independent of x.

Regards

one year ago

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(sin2x + cos2x + 2 sin x cos x) + 4 [(sin2x)3 + (cos2x)3]

= 3 [sin2x + cos2x – 2 sin x cos x]2 + 6(1 + 2 sin x cos x) + 4 [(sin2x + cos2x) (sin4x + cos4x – sin2x cos2x)]

= 3 [1 – 2 sin x cosx)2 + 6 + 12 sin x cos x + 4 [(sin2x)2 + (cos2 x)2 + 2 sin2x cos2x – 3 sin2 x cos2x]

[ using: sin 2 x + cos 2 x = 1]

= 3 [1 + 4 sin2x cos2x – 4 sin x cos x)] + 6 + 12 sin x cos x + 4 [(sin x + cos x)2 – 3 sin2x cos2x]

= 3 + 12 sin2 x cos2x – 12 sin x cos x + 6 + 12 sin x cos x + 4 – 12 sin2x cos2x

= 13, which is independent of x.