How would I prove the following.
arcsin(4/5)+arctan(3/4)=pi/2
Any help is appreciated.
You could draw a 3-4-5 right triangle and:
a) find that $\displaystyle \arcsin \left( \frac{4}{5} \right)= \text{arccot} \left( \frac{3}{4} \right)$
b) find that $\displaystyle \arctan \left( \frac{3}{4} \right)= \arccos \left( \frac{4}{5} \right)$
Then use the fact that the sum of two complementary inverse trig functions having the same argument is always $\displaystyle \frac{\pi}{2}$.
What you mean is:
$\displaystyle \arcsin\left(\frac{3}{5} \right)+\arcsin\left(\frac{15}{17} \right)=\arccos\left(-\frac{13}{85} \right)$
There are two ways we could proceed:
a) Take the sine of both sides:
$\displaystyle \sin\left(\arcsin\left(\frac{3}{5} \right)+\arcsin\left(\frac{15}{17} \right) \right)=\sin\left(\arccos\left(-\frac{13}{85} \right) \right)$
$\displaystyle \sin\left(\arcsin\left(\frac{3}{5} \right) \right)\cos\left(\arcsin\left(\frac{15}{17} \right) \right)+\cos\left(\arcsin\left(\frac{3}{5} \right) \right)\sin\left(\arcsin\left(\frac{15}{17} \right) \right)=\frac{\sqrt{85^2-(-13)^2}}{85}$
$\displaystyle \frac{3}{5}\cdot\frac{\sqrt{17^2-15^2}}{17}+\frac{\sqrt{5^2-3^2}}{5}\cdot\frac{15}{17}=\frac{84}{85}$
$\displaystyle \frac{3}{5}\cdot\frac{8}{17}+\frac{4}{5}\cdot\frac {15}{17}=\frac{84}{85}$
$\displaystyle \frac{24+60}{85}=\frac{84}{85}$
$\displaystyle \frac{84}{85}=\frac{84}{85}$
b) Take the cosine of both sides:
See if you can do this. Post your working, please.