Question

u=(1+cosx)(1+cos2x)-sinx sin2x and v=sinx(1+cos2x)+sin2x(1+cosx) find u^2 +v^2=?​

Answer 1

2(cosx+2cos2x−1)+2sinxcosx.(1+2cosx)−2sinx=0

or 2(2cos2x+cosx−1)+2sinx(2cos2x+cosx−1)=0

2(1+sinx)(cosx+1)(2cosx−1)=0

We have to determine values of x s.t. −π≤x≤π

1+sinx=0 ∴sinx=−1

∴x=2nπ+23π  ∴x=−2π,

for n=−1 ∵−π≤x<π

cosx=−1=cosπ ∴cosx=1/2=cos(π/3)

∴x=2nπ±π/3

∴x=π/3, −π/3

Hence the values of x s.t. −π≤x≤π are

−π,−π/2,−π/3,π/3,π.

Answer 2

Answer:

u^2+v^2=4(1+cosx) (1+cos2x