Prove that sin theta minus 2 Sin cube theta upon 2 cos cube theta minus cos theta equal to 10 theta
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(sinθ- 2 sin^3θ)/(2cos^3θ-cosθ)
=sinθ(1-2sin^2θ)/cosθ(2cos^2 θ-1)
= sin θ [1- 2(1- cos^2 θ)/cos θ(2cos ^2 θ -1)
= sin θ [1-2+ 2cos ^2 θ]/ cos θ (2 cos^2θ -1)
= sin θ(2cos^2 θ -1)/ cos θ(2 cos ^2θ-1)
= sin θ/ cos θ
= tanθ
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=sinθ(1-2sin^2θ)/cosθ(2cos^2 θ-1)
= sin θ [1- 2(1- cos^2 θ)/cos θ(2cos ^2 θ -1)
= sin θ [1-2+ 2cos ^2 θ]/ cos θ (2 cos^2θ -1)
= sin θ(2cos^2 θ -1)/ cos θ(2 cos ^2θ-1)
= sin θ/ cos θ
= tanθ
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