If T n = sin n x + cos n x, prove that 2 T 6 – 3 T 4 + 1 = 0

Question

Philoid · Accepted Answer

Given T_n = sinⁿx + cosⁿx

LHS = 2T₆ – 3T₄ + 1

= 2 (sin⁶x + cos⁶x) – 3 (sin⁴x + cos⁴x) + 1

= 2 (sin²x + cos²x) (sin⁴x + cos⁴x – cos²x sin²x) – 3 (sin⁴x + cos⁴x) + 1

We know that sin²x + cos²x = 1.

= 2 (1) (sin⁴x + cos⁴x – cos²x sin²x) – 3 (sin⁴x + cos⁴x) + 1

= 2sin⁴x + 2cos⁴x – 2sin²x cos²x – 3sin⁴x – 3cos⁴x + 1

= - (sin⁴x + cos⁴x) – 2sin²x cos²x + 1

= - (sin²x + cos²x) ² + 1

= - 1 + 1

= 0

= RHS

Hence proved.