Q. 22

# Prove that the following identities :2(sin6 θ – cos6 θ) – 3(sin4 θ + cos4 θ ) + (sin2 θ + cos2 θ)

Taking LHS

= 2(sin6 θ – cos6 θ) – 3(sin4 θ+ cos4 θ ) + (sin2 θ + cos2 θ)

= 2[(sin2 θ)3 – (cos2 θ)3 ] – 3[(sin2 θ)2 + (cos2 θ)2 ]+1 [ cos2 θ + sin2 θ = 1]

Now, we use these identities, (a3 – b3)= (a + b)3 – 3ab(a+b) and (a2 + b2) = (a +b)2 – 2ab]

= 2[(sin2 θ + cos2 θ)3 – 3sin2θ cos2θ (sin2 θ+ cos2 θ)] –3[(sin2 θ + cos2 θ) – 2 sin2 θ cos2 θ]+ 1

=2[(1) – 3sin2θ cos2θ (1)] – 3[(1) – 2 sin2 θ cos2 θ] + 1 [ cos2 θ + sin2 θ = 1]

=2(1 – 3 sin2 θ cos2 θ )– 3 + 6sin2 θ cos2 θ+ 1

= 2– 6sin2θ cos2θ – 2 + 6sin2 θ cos2 θ

=0

=RHS

Hence Proved

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