2n-1sin-nsin2-n-sinn-1-n- equals

The correct option is B n From #160;( x α #160;) ( x α #160; #160;) = x ^ 2 2xcosπ #160;/ n #160; +1 and #160; x ^ 2n 1 / x ^ 2 1 ...

You visited us 8 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Integration by Substitution

2n-1sinπ/nsin...

Question

$2^{n - 1} sin \frac{π}{n} sin \frac{2 π}{n} . . . sin \frac{n - 1}{n} π$ equals

n

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $n$
From $(x - α) (x - ¯ ¯¯ ¯ α) = x^{2} - 2 x cos \frac{π}{n} + 1$
and $\frac{x^{2 n} - 1}{x^{2} - 1} = x^{2 n - 2} + x^{2 n + 3} + . . . + x + 1$
We get
$(x^{2} - 2 x cos \frac{π}{n} + 1) (x^{2} - 2 x cos \frac{2 π}{n} + 1) . . . (x^{2} - 2 x cos \frac{n - 1}{n} π + 1) = 1 + x + . . . + x^{2 n - 2}$
Divide both sides by $x^{n - 1}$ to obtain
$(x + \frac{1}{x} - 2 cos \frac{π}{n}) (x + \frac{1}{x} - 2 cos \frac{2 π}{n}) . . . (x + \frac{1}{x} - 2 cos \frac{n - 1}{n} π) = \frac{x^{n} - x^{- n}}{x - x^{- 1}}$
Now put $x = cos θ + i sin θ$
we get
$2^{n - 1} (cos θ - cos \frac{π}{n}) (cos θ - cos \frac{2 π}{n}) . . . (cos θ - cos \frac{n - 1}{n} π) = \frac{sin n θ}{sin θ}$
Take $lim θ \to 0$
$n = 2^{n - 1} (2 {sin}^{2} \frac{π}{2 n}) (2 {sin}^{2} \frac{2 π}{2 n}) . . . (2 {sin}^{2} \frac{n - 1}{n} π)$
$= 2^{n - 1} (2 sin \frac{π}{2 n} sin \frac{n - 1}{2 n} π) (2 sin \frac{2 π}{2 n} sin \frac{n - 2}{2 n} π) . . . (2 sin \frac{n - 1}{2 n} π sin \frac{π}{n})$
Now, use
$2 sin \frac{k π}{2 n} sin (\frac{n - k}{2 n} π) = sin (\frac{k π}{n})$
we get
$2^{n - 1} sin \frac{π}{n} sin \frac{2 π}{n} . . . sin \frac{n - 1}{n} π = n$

Suggest Corrections

Similar questions

If $a_{0}, a_{1}, a_{2}, . . . . . . . . . . . ., a_{n - 1}$ are nth roots of unity, then $a_{k}$ = cos $\frac{2 k π}{n}$ + i sin $\frac{2 k π}{n}$ where $0 \leq k \leq n - 1$ .

Also then Value of $2^{n - 1}$ sin $(\frac{π}{n})$ sin $(\frac{2 π}{n})$ .........sin $(\frac{n - 1}{n} π)$ is

s i n \frac{π}{2 n} s i n \frac{2 π}{2 n} s i n \frac{3 π}{2 n} . . . . . . . . . s i n \frac{(n - 1) π}{2 n} = \frac{\sqrt{n}}{2^{n - 1}}

2^{n - 1} (cos θ - cos \frac{π}{n}) (cos θ - cos \frac{2 π}{n}) . . . (cos θ - cos \frac{n - 1}{n} π)

equals

L e t f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} (1 + | c o s x |)^{\frac{a b}{| c o s x |}}, & n π < x < (2 n + 1) \frac{π}{2} e^{a} . e^{b}, & x = (2 n + 1) \frac{π}{2} e^{\frac{c o t 2 x}{c o t 8 x}}, & (2 n + 1) \frac{π}{2} < x < (n + 1) π \end{matrix} I f f (x) i s c o n t i n u o u s i n ((n π), (n + 1) π, t h e n)

Q. If

a = \infty \sum n = 1 \frac{2 n}{(2 n - 1)!}, b = \infty \sum n = 1 \frac{2 n}{(2 n + 1)!}

, then

a b

equals

Join BYJU'S Learning Program

2n−1sinπnsin2πn...sinn−1nπ equals

$2^{n - 1} sin \frac{π}{n} sin \frac{2 π}{n} . . . sin \frac{n - 1}{n} π$ equals