Is this an identity?

Geometry Level pending

If cos ⁡ − 1 x + cos ⁡ − 1 y = π 2 \cos^{-1} x + \cos^{-1} y = \dfrac \pi 2 , what is the value of x 2 + y 2 x^2+y^2 ?

-1 0 2 1

César Castro by César Castro , 20, USA

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1 solution

Chew-Seong Cheong
Apr 18, 2018

x 2 + y 2 = cos ⁡ 2 ( cos ⁡ − 1 x ) + cos ⁡ 2 ( cos ⁡ − 1 y ) Since cos ⁡ − 1 x + cos ⁡ − 1 y = π 2 = cos ⁡ 2 ( cos ⁡ − 1 x ) + cos ⁡ 2 ( π 2 − cos ⁡ − 1 x ) = cos ⁡ 2 ( cos ⁡ − 1 x ) + sin ⁡ 2 ( cos ⁡ − 1 x ) = 1 \begin{aligned} x^2 + y^2 & = \cos^2 \left(\cos^{-1} x\right) + \cos^2 \left({\color{#3D99F6}\cos^{-1} y}\right) & \small \color{#3D99F6} \text{Since } \cos^{-1} x + \cos^{-1} y = \frac \pi 2 \\ & = \cos^2 \left(\cos^{-1} x\right) + \cos^2 \left({\color{#3D99F6}\frac \pi 2 - \cos^{-1} x}\right) \\ & = \cos^2 \left(\cos^{-1} x\right) + \sin^2 \left(\cos^{-1} x\right) \\ & = \boxed{1} \end{aligned}

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It's cos-1(x)+cos-2(y)=pi/2,not cos-1(x)+cos-1(y)=pi/2

X X - 3 years, 1 month ago

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Sorry, I have amended the problem wrongly.

Chew-Seong Cheong - 3 years, 1 month ago

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It's OK now,the problem is now cos-1(y).

X X - 3 years, 1 month ago

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