Question

If T_n = sinⁿx + cosⁿx, prove that

6 T₁₀ – 15 T₈ + 10 T₆ – 1 = 0

Answer 1

Given T_n = sinⁿx + cosⁿx

LHS = 6T₁₀ – 15 T₈ + 10T₆ – 1

= 6 (sin¹⁰x + cos¹⁰x) – 15 (sin⁸x + cos⁸x) + 10 (sin⁶x + cos⁶x) – 1

= 6 (sin⁶x + cos⁶x) (sin⁴x + cos⁴x) – cos⁴x sin⁴x (sin²x + cos²x) - 15 (sin⁶x + cos⁶x) (sin²x + cos²x) – cos²x sin²x (sin⁴x + cos⁴x) + 10 (sin²x + cos²x) (sin⁴x + cos⁴x – cos²x sin²x) – 1

We know that sin²x + cos²x = 1.

= 6 [(1 – 3 sin²x cos²x) (1 – 2 sin²x cos²x) – sin⁴x cos⁴x] - 15 [(1 – 3 sin²x cos²x) – sin²x cos²x (1 – 2 sin²x cos 2x)] + 10 (1 – 3 sin²x cos²x) – 1

= 6 (1 – 5 sin²x cos²x + 5 sin⁴x cos⁴x) – 15 (1 – 4 sin²x cos²x + 2 sin ⁴x cos⁴x) + 10 (1 – 3 sin²x cos²x) – 1

= 6 – 30 sin²x cos²x + 30 sin⁴x cos⁴x – 15 + 60 sin²x cos²x - 30 sin⁴x cos⁴x + 10 – 30 sin²x cos²x – 1

= 6 – 15 + 10 – 1

= 0

= RHS

Hence proved.