If θ =(π /2n+1), then the value of 2n cos θ cos 2θ cos 22θ ..... cos 2n-1θ is

Question

If θ =(π /2n+1), then the value of 2n cos θ cos 2θ cos 22θ ..... cos 2n-1θ is (A)  sin θ  (B)  (π /2)  (C)  0  (D)  1

Tardigrade Admin · Accepted Answer

2n. cos θ . cos 21θ . cos 22θ ...... cos 2n-1θ =2n( sin 2nθ /2n. sin θ )=( sin 2nθ / sin θ ) ∵ θ =(π /2n+1) (∵ 2n.θ =π -θ ) =( sin (π -θ )/ sin θ )=( sin θ / sin θ )=1

Q. If $θ = \frac{π}{2 ^{n} + 1},$ then the value of $2^{n} cos θ cos 2 θ cos 2^{2} θ ..... cos 2^{n - 1} θ$ is

$sin θ$

$\frac{π}{2}$

$0$

$1$

$2^{n} . cos θ . cos 2^{1} θ . cos^{2^{2}} θ ...... cos 2^{n - 1} θ$
$= 2^{n} \frac{s i n 2 ^{n} θ}{2 ^{n} . s i n θ} = \frac{s i n 2 ^{n} θ}{s i n θ}$
${∵ θ = \frac{π}{2 ^{n} + 1}}$
$(∵ 2^{n} . θ = π - θ)$
$= \frac{s i n ( π - θ )}{s i n θ} = \frac{s i n θ}{s i n θ} = 1$

Q. If θ=2n+1π​, then the value of 2ncosθcos2θcos22θ.....cos2n−1θ is

sinθ

2π​

0

1

2n.cosθ.cos21θ.cos22θ......cos2n−1θ =2n2n.sinθsin2nθ​=sinθsin2nθ​ {∵θ=2n+1π​} (∵2n.θ=π−θ) =sinθsin(π−θ)​=sinθsinθ​=1

Q. If $θ = \frac{π}{2 ^{n} + 1},$ then the value of $2^{n} cos θ cos 2 θ cos 2^{2} θ ..... cos 2^{n - 1} θ$ is

$sin θ$

$\frac{π}{2}$

$0$

$1$

$2^{n} . cos θ . cos 2^{1} θ . cos^{2^{2}} θ ...... cos 2^{n - 1} θ$
$= 2^{n} \frac{s i n 2 ^{n} θ}{2 ^{n} . s i n θ} = \frac{s i n 2 ^{n} θ}{s i n θ}$
${∵ θ = \frac{π}{2 ^{n} + 1}}$
$(∵ 2^{n} . θ = π - θ)$
$= \frac{s i n ( π - θ )}{s i n θ} = \frac{s i n θ}{s i n θ} = 1$