First slide

Introduction to ITF

Question

If $\sum_{i = 1}^{2 n} \sin^{- 1} x_{i} = nπ$ , then $\sum_{i = 1}^{2 n} x_{i}$ is equal to

Moderate

A

n
B

2n
C

$\frac{n (n + 1)}{2}$
D

none of these

Solution

We have, $\sum_{i = 1}^{2 n} \sin^{- 1} x_{i} = nπ$
$\Rightarrow \sin^{- 1} x_{1} + \sin^{- 1} x_{2} + \dots + \sin^{- 1} x_{2 n} = nπ$
Let $\sin^{- 1} x_{1} = α_{1}, \sin^{- 1} x_{2} = α_{2}, \dots, \sin^{- 1} x_{n} = α_{n}$
$∴ α_{1} + α_{2} + \dots + α_{2 n} = nπ$ ……… (1)
$\Rightarrow x_{1} = \sin α_{1}, x_{2} = \sin α_{2}, \dots, x_{2 n} = \sin α_{2 n}$
$\begin{matrix} ∴ \sum_{i = 1}^{2 n} x_{i} = x_{1} + x_{2} + \dots + x_{2 n} \\ = \sin α_{1} + \sin α_{2} + \dots + \sin α_{2 n} \end{matrix}$
Clearly, from (1), $α_{1} = \frac{π}{2}, \forall i = 1, 2, \dots, 2 n$
$∴ \sum_{i = 1}^{2 n} x_{i} = \underset{2 n times}{\underset{⏟}{1 + 1 + \dots + 1}} = 2 n$

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