Question

Plz solve Question no. 84

Answer 1

Answer:

Step-by-step explanation:

Given: sinθ+sin2θ=1

Let x=cos12θ+3cos10θ+3cos8θ+cos6θ+2cos4θ+2cos2θ−2?

Because, cos2θ+sin2θ=1;

From given equation we can get

cos2θ=sinθ;

Now,

x=sin6θ+3sin5θ+3sin4θ+sin3θ+2sin2θ+2sinθ−2

=sin6θ+3sin4θ+3sin5θ+sin3θ+2(sin2θ+sinθ)−2

=sin6θ+sin5θ+2sin5θ+2sin4θ+sin4θ+sin3θ+2(1)−2

=sin4θ(sin2θ+sinθ)+2sin3θ(sin2θ+sinθ)+sin2θ(sin2θ+sinθ)

=sin4θ(1)+2sin3θ(1)+sin2θ(1)

=sin4θ+sin3θ+sin3θ+sin2θ

=sin2θ(sin2θ+sinθ)

+sinθ(sin2θ+sinθ)

=sin2θ(1)+sinθ(1)

=1