Q. 214.2( 5 Votes )

Prove that the following identities :

(sin8θ – cos8θ) = (sin2θ – cos2θ)(1 – 2sin2θ .cos2θ)

Answer :

Taking LHS

= sin8 θ – cos8 θ


= (sin4 θ)2 – (cos4 θ)2


= (sin4 θ – cos4 θ)(sin4 θ+ cos4 θ )


[ (a2 – b2) = (a + b) (a – b)]


= {(sin2 θ)2 – (cos2 θ)2}{(sin2 θ)2 + (cos2 θ)2}


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


[ (a2 + b2) = (a +b)2 – 2ab]


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


= (sin2 θ – cos2 θ)(1 – 2 sin2 θ cos2 θ)


=RHS


Hence Proved


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