Problem on ITF

Geometry Level 3

sin ⁡ − 1 ( x − x 2 2 + x 3 4 − ⋯ ) + cos ⁡ − 1 ( x 2 − x 4 2 + x 6 4 ⋯ ) = π 2 \sin^{-1}\left ( x - \frac{x^2}{2} + \frac{x^3}{4} - \cdots \right ) + { \cos^{-1}\left ( { x^2 - \frac{x^4}{2} + \frac{x^6}{4} \cdots} \right ) } = { \frac{\pi }{2}}

Given the equation above for 0 < ∣ x ∣ < 2 0.5 0<|x|<2^{0.5} , then x x equals to:

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The answer is 1.

Vatsalya Tandon by Vatsalya Tandon , 21, India

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1 solution

Vatsalya Tandon
Jan 21, 2016

s i n − 1 ( x − x 2 2 + x 3 4 − … ) + c o s − 1 ( x 2 − x 4 2 + x 6 4 … ) = π 2 sin^{-1}\left ( x - \frac{x^2}{2} + \frac{x^3}{4} - \dots \right ) + { cos^{-1}\left ( { x^2 - \frac{x^4}{2} + \frac{x^6}{4} \dots} \right ) } = { \frac{\pi }{2}}

= s i n − 1 ( x 1 + x 2 ) + c o s − 1 ( x 2 1 + x 2 2 ) = π 2 =sin^{-1} (\frac{x}{1+\frac{x}{2}}) + cos^{-1}(\frac{x^2}{1+\frac{x^2}{2}}) = \frac{\pi}{2}

= s i n − 1 ( 2 x 2 + x ) + c o s − 1 ( 2 x 2 2 + x 2 ) = π 2 =sin^{-1} (\frac{2x}{2+x}) + cos^{-1}(\frac{2x^2}{2+x^2}) = \frac{\pi}{2}

= s i n − 1 ( 2 x 2 + x ) = s i n − 1 ( 2 x 2 2 + x 2 ) =sin^{-1} (\frac{2x}{2+x}) = sin^{-1}(\frac{2x^2}{2+x^2})

2 x 2 + x = 2 x 2 2 + x 2 \frac{2x}{2+x} = \frac{2x^2}{2+ x^2}

1 + 2 x = 1 + 2 x 2 1+ \frac{2}{x} = 1 + \frac{2}{x^2}

x = x 2 x = x^2

x = 1 x = 1

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