How do you prove sin^-1x+cos^-1x=pi/2?

Oct 7, 2016

See the Explanation.

Explanation:

Let ${\sin}^{-} 1 x = \theta \text{, where, } | x | \le 1.$

Then, by the Defn. of ${\sin}^{-} 1$ fun.,

sintheta=x, &, theta in [-pi/2,pi/2].

$\theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] \Rightarrow - \frac{\pi}{2} \le \theta \le \frac{\pi}{2.}$

Multiplying the inequality by $- 1$, hence, reversing its order,

$\frac{\pi}{2} \ge - \theta \ge - \frac{\pi}{2.}$ Now, adding, $\frac{\pi}{2}$,

$\pi \ge \frac{\pi}{2} - \theta \ge 0 , i . e . , \left(\frac{\pi}{2} - \theta\right) \in \left[0 , \pi\right] .$

Further, $\cos \left(\frac{\pi}{2} - \theta\right) = \sin \theta = x .$

Thus, $\cos \left(\frac{\pi}{2} - \theta\right) = x , w h e r e , \left(\frac{\pi}{2} - \theta\right) \in \left[0 , \pi\right] .$

This, together with the Defn. of ${\cos}^{-} 1$ fun., allow us to say that,

$\left(\frac{\pi}{2} - \theta\right) = {\cos}^{-} 1 x .$

Replacing $\theta$ by ${\sin}^{-} 1 x ,$ we finally arrive at,

$\frac{\pi}{2} - {\sin}^{-} 1 x = {\cos}^{-} 1 x , \mathmr{and} , {\sin}^{-} 1 x + {\cos}^{-} 1 x = \frac{\pi}{2.}$

Enjoy maths.!