# How to evaluate inverse trig functions without a calculator or trig tables?

I am trying to understand how to evaluate the following equation without using a calculator or trig tables:

arcsin(4/5) - arccos(12/13) = arccos(56/65)

Hint Compute $$\sin(\arccos(12/13) + \arccos(56/65))$$ and prove that the result is $$4/5$$. You can use $$$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$$$ Also note that $$65^2-56^2 = 33^2$$ and $$13^2-12^2=5^2$$.
Let $$\theta=\arcsin(4/5)$$, $$\:\varphi =\arccos(12/13)$$, and compute $$\cos(\theta-\varphi)$$.
First note that $$\theta\in \bigl[\frac \pi 3,\frac\pi 2\bigr]$$, so $$\cos\theta >0$$ and $$\varphi\in\bigl[0,\frac\pi 6\bigr]$$, so $$\sin \varphi \ge 0$$.
Therefore $$\cos\theta=\sqrt{1-\frac{16}{25}}=\frac 35$$ and $$\sin\varphi=\sqrt{1-\frac{144}{169}}=\frac 5{13}$$, and consequently, by the addition formula, $$\cos(\theta-\varphi)=\frac35\frac{12}{13}+\frac45\frac5{13}=\frac{56}{65}.$$ Furthermore, $$\theta-\varphi\:$$ lives in the interval $$\bigl[\frac\pi3-\frac\pi 6,\frac\pi 2\bigr]\subset[0,\pi]$$, so it is exactly $$\:\arccos(56/65)$$.
Use Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ to convert all the inverse ratios to arcsin