Q. 26 C3.9( 9 Votes )

# If Tn

Given Tn = 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 cos2x)] + 10 (1 – 3 sin2x cos2x) – 1

= 6 (1 – 5 sin2x cos2x + 5 sin4x cos4x) – 15 (1 – 4 sin2x cos2x + 2 sin4x 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.

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