Question

If cos theta sin theta minus sin theta cos theta show by the induction that n equal to bracket cos n theta sin n theta minus sin theta cos sin theta

Answer 1

Hlw mate


To Prove 
8cos2(θ)+4cos4(θ)+cos6(θ)=48cos2(θ)+4cos4(θ)+cos6(θ)=4

Given that
sin(θ)+sin2(θ)+sin3(θ)=1sin(θ)+sin2(θ)+sin3(θ)=1

Therefore 
sin(θ)+sin3(θ)=1−sin2(θ)sin(θ)+sin3(θ)=1−sin2(θ)

sin(θ)+sin3(θ)=cos2(θ)sin(θ)+sin3(θ)=cos2(θ)

sin(θ)[1+1−cos2(θ)]=cos2(θ)sin(θ)[1+1−cos2(θ)]=cos2(θ)

sin(θ)[2−cos2(θ)]=cos2(θ)sin(θ)[2−cos2(θ)]=cos2(θ)

sin(θ)=cos2(θ)/[2−cos2(θ)]sin(θ)=cos2(θ)/[2−cos2(θ)]

(1−cos2θ)−−−−−−−−−√=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)](1−cos2θ)=cos4(θ)/[4+cos4θ−4cos2(θ)]

4−4cos2θ+cos4θ−cos6θ−4cos2θ+4cos4θ=cos44−4cos2θ+cos4θ−cos6θ−4cos2θ+4cos4θ=cos4

4−8cos2θ−cos6θ+4cos4θ=04−8cos2θ−cos6θ+4cos4θ=0

cos6θ−4cos4θ+8cos2θ=4


Hope it helpful