Question

cos(n+1)α⋅cos(n−1)α−sin(n+1)α⋅sin(n−1)α=

a.
cos2n alpha

B.
sin2n alpha

C.
cos2 alpha

D.
sin2 alpha

Answer 1

Step-by-step explanation:

ANSWER

sin(θ+α)

cos(θ−α)

=

m−1

m+1

Applying componendo and dividendo

(m+1)−(m−1)

(m+1)+(m−1)

=

cos(θ−α)−sin(θ+α)

cos(θ−α)+sin(θ+α)

⇒m=

cosθcosα+sinθsinα−sinθsinα−cosθsinα

cosθcosα+sinθsinα+sinθsinα+cosθsinα

=

cosθ(cosα−sinα)+sinθ(sinα−cosα)

cosθ(cosα+sinα)+sinθ(sinα+cosα)

=

(cosθ−sinθ)(cosα−sinα)

(cosθ+sinθ)(cosα+sinα)

=

(cosθ−sinθ)

(cosθ+sinθ)

⋅

(cosα−sinα)

(cosα+sinα)

=

1−tanθ

1+tanθ

⋅

1−tanα

1+tanα

=tan(

4

π

+θ)⋅tan(

4

π

+α)