Trigonometry

Geometry Level 1

If $\color{#20A900}\sin^{-1}$ $\color{#3D99F6}x$ $+$ $\color{#20A900}\sin^{-1}$ $\color{#D61F06}y$ $=\dfrac{\pi}{2}$

then what is the value of

$\color{#3D99F6}x^2$ $+$ $\color{#D61F06}y^2$ ?

The answer is 1.

1 solution

Naren Bhandari
Oct 11, 2017

$\arcsin(x)+\arcsin(y)= \frac{\pi}{2}$ $\arcsin(x) =\frac{\pi}{2}-\arcsin(y)$ $\arcsin(x) = \arccos(y)$ $\arcsin(x) = \arcsin\sqrt{1-y^2}$ $x = \sqrt{1-y^2}\implies x^2+y^2=1$

Trigonometry

The answer is 1.

1 solution

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