Prove the given trigonometric identity:
cos6β−sin6β=116(15cos2β+cos6β)
My Approach:
L.H.S.=cos6β−sin6β=(cos2β)3−(sin2β)3=cos32β+3cos2β⋅sin2β⋅cos2β
Please help me to continue further.
Answer
(cos2β)3−(sin2β)3=(cos2β−sin2β)(cos4β+sin4β+cos2βsin2β)
=cos2β(1−sin2βcos2β)
=cos2β(1−sin22β4)
=cos2β(1+cos4β−18)
Now use 2cosA.cosB=cos(A+B)+cos(A−B)
=18(7cos2β+cos2βcos4β)
=18(7cos2β+cos6β+cos2β2)
=116(15cos2β+cos6β)
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