show that $\displaystyle 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.
For your problem:
In other words, you want to prove that:
$\displaystyle sin^{-1}(x) = \frac{\pi}{2} - cos^{-1}(x)$
Let $\displaystyle f(x)=sin^{-1}(x)$, and $\displaystyle g(x)=\frac{\pi}{2} - cos^{-1}(x)$
If you proved that $\displaystyle f^{'}(x)=g^{'}(x)$
then, $\displaystyle f(x)=g(x)+C$.
By substituting x = anynumber $\displaystyle \in [-1,1]$ in the last equation you will get the value of $\displaystyle C$.
Try it.
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
$\displaystyle \frac{opposite}{hypotenuse}$ and $\displaystyle \frac{adjacent}{hypotenuse}$
For this triangle
$\displaystyle SinA=x$ and $\displaystyle CosB=x$
$\displaystyle Sin^{-1}x=A$ and $\displaystyle Cos^{-1}x=B$
Since the sum of the acute angles in a right-angled triangle is $\displaystyle 90^o=\frac{\pi}{2}$
your statement is proven.