# Solve sin−1 x+cos−1 x=π 2

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Solve `sin^(-1)x+cos^(-1)x=pi/2 ` :

The inverse sine (or arcsin) has domain `-1<=x<=1 ` and a range of `-pi/2<= alpha <= pi/2 ` .

The inverse cosine (or arccos) has domain `-1<=x<=1 ` and a range of `0<=alpha<=pi ` .

Consider a right triangle `Delta ABC, m/_ C=90^@ ` with legs a and b, hypotenuse c.

First note that `m/_A+m/_B=90^@ ` .

Now consider: `sin A=a/c ` ==> `sin^(-1)(a/c)=m/_A ` .

Also `cosB=a/c ` ==> `cos^(-1)(a/c)=m/_B ` .

So `sin^(-1)(a/c)+cos^(-1)(a/c)=m/_A+m/_B=90^@ ` . (Of course, in radians this is `pi/2 ` .)

In terms of your problem, this shows that for all `0<=x<=1 ` , `sin^(-1)x+cos^(-1)x=pi/2 ` . (The ratio of a leg of a right triangle to the hypotenuse will always be positive and smaller than 1, as the hypotenuse is the longest side of the triangle.)

We can extend this for `-1<=x<=0 ` : Note that for x<0, the arcsin is negative. The arccos will be an angle `pi/2<=alpha<=pi ` . Again the sum is `pi/2 ` .

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`sin^(-1)x+cos^(-1)x=pi/2 ` for all `-1<=x<=1 `

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