Question

if alpha =π/3 then πr=1to 6 cosπalpha​

Answer 1

Answer:

Let A=Π

r=1

6

cos(rα). Then

A=cosα.cos2α.cos3α.cos4α.cos5α.cos6α.....(i)

Where α=

13

π

, then 13α=π, and α=π−12α,5α=π−8α

⇒cosα=−cos12α,cos(5α)=−cos8α

Now (1) ⇒A=cos12α.cos2α.cos3α.cos4α.cos8α.cos6α

⇒A=(cos2α.cos4α.cos8α)(cos3α.cos6α.cos12α)

=(cosθ.cos2θ.cos4θ)(cosϕ.cos2ϕ.cos4ϕ), where θ=2α,ϕ=3α

=

2

3

sinθ)

sin(2

3

θ)

.

2

3

sinϕ

sin(2

3

ϕ)

=

64

1

.

sin(2α).sin(3/alpha)

sin(16α).sin(24α)

....(ii)

But sin(6α)=sin(π+3α+=−sin3α,sin(24α)=sin(2π−2α)=−2α

⇒sin(16α)sin(24α)=(sin2α).(sin3α) putting in (ii), we get A=

64

1

.

Option A

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