Question

please solve this sum...............

Answer 1

written not so good
but hope helps

Answer 2

To Prove

8cos2(θ)+4cos4(θ)+cos6(θ)=4

Given that

sin(θ)+sin2(θ)+sin3(θ)=1

Therefore

sin(θ)+sin3(θ)=1−sin2(θ)

sin(θ)+sin3(θ)=cos2(θ)

sin(θ)[1+1−cos2(θ)]=cos2(θ)

sin(θ)[2−cos2(θ)]=cos2(θ)

sin(θ)=cos2(θ)/[2−cos2(θ)]

(1−cos2θ)−−−−−−−−−√=cos2(θ)/[2−cos2(θ)]

Squaring both sides we get

(1−cos2θ)=cos4(θ)/[4+cos4θ−4cos2(θ)]

4−4cos2θ+cos4θ−cos6θ−4cos2θ+4cos4θ=cos4

4−8cos2θ−cos6θ+4cos4θ=0

cos6θ−4cos4θ+8cos2θ=4

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