wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

cos2(θ+ϕ)+4cos(θ+ϕ)sinθsinϕ+2sin2ϕ=

Open in App
Solution

cos2(θ+ϕ)+4cos(θ+ϕ)sinθsinϕ+2sin2ϕ=cos2(θ+ϕ)+2cos(θ+ϕ)[2sinθsinϕ]+2sin2ϕ=cos2(θ+ϕ)+2cos(θ+ϕ)[cos(θϕ)cos(θ+ϕ)]+2sin2ϕ=cos2(θ+ϕ)+2cos(θ+ϕ)cos(θϕ)cos2(θ+ϕ)+2sin2ϕ=[2cos22(θ+ϕ)1]+cos(θ+ϕ+θϕ)+cos(θ+ϕθ+ϕ)2cos2(θ+ϕ)+2sin2ϕ=1+cos2θ+cos2ϕ+(1cos2θ)=cos2ϕ

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Transformations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon