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Question
Prove that :
2(sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) + 1 = 0
Solution
LHS = 2(sin6θ + cos6θ) - 3(sin4θ + cos4θ) + 1
Simplifying the expression 2(sin6θ + cos6θ) - 3(sin4θ + cos4) we have,
2(sin6θ + cos6θ) - 3(sin4θ + cos4θ)
= 2sin6θ + 2cos6θ - 3sin4θ - 3cos4θ
= (2sin6θ - 3sin4θ) + (2cos6θ - 3cos4θ)
= sin4θ(2sin2θ - 3) + cos4θ(2cos2θ - 3)
= sin4θ{2( 1 - cos2θ) - 3} + cos4θ{ 2( 1 - sin2θ) - 3}
= sin4θ( 2 - 2cos2θ - 3) + cos4θ( 2 - 2sin2θ - 3)
= sin4θ( -1 - 2cos2θ) + cos4θ( - 1 - 2sin2θ)
= - sin4θ - 2sin4θcos2θ - cos4θ - 2cos4θsin2θ
= - sin4θ - cos4θ - 2cos4θsin2θ - 2sin4θcos2θ
= - sin4θ - cos4θ - 2cos2θsin2θ( cos2θ + sin2θ )
= - sin4θ - cos4θ - 2cos2θsin2θ(1)
= - sin4θ - cos4θ - 2cos2θsin2θ
= - (sin4θ + cos4θ + 2cos2θsin2θ)
= - {(sin2θ)2 + (cos2θ)2 + 2sin2θcos2θ }
= - (sin2θ + cos2θ)2
= - (1)2
= - 1
2(sin6θ + cos6θ) - 3(sin4θ + cos4θ) + 1 = −1 + 1 = 0 = RHS
Hence proved.
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