Q. 26 B4.1( 7 Votes )

# If Tn

Given Tn = 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.

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