1. ## inverse trig

show that $arcsinx + arccosx = \frac{\pi}{2}$

2. Originally Posted by Tweety
show that $arcsinx + arccosx = \frac{\pi}{2}$
I think I can use calculus to solve your problem, since you wrote some threads in calculus sub-forum, which means that you know the derivatives.

In other words, you want to prove that:
$sin^{-1}(x) = \frac{\pi}{2} - cos^{-1}(x)$
Let $f(x)=sin^{-1}(x)$, and $g(x)=\frac{\pi}{2} - cos^{-1}(x)$
If you proved that $f^{'}(x)=g^{'}(x)$
then, $f(x)=g(x)+C$.
By substituting x = anynumber $\in [-1,1]$ in the last equation you will get the value of $C$.

Try it.

3. Hi Tweety,

Draw a right-angled triangle.
Label one acute angle A and the other B.
Label the side opposite A as x and the hypotenuse 1.

Inverse Sine and Inverse Cosine give you angle from the ratios of

$\frac{opposite}{hypotenuse}$ and $\frac{adjacent}{hypotenuse}$

For this triangle

$SinA=x$ and $CosB=x$

$Sin^{-1}x=A$ and $Cos^{-1}x=B$

Since the sum of the acute angles in a right-angled triangle is $90^o=\frac{\pi}{2}$