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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)

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Hint Compute $\sin(\arccos(12/13) + \arccos(56/65))$ and prove that the result is $4/5$. You can use \begin{equation} \sin(a+b) = \sin a \cos b + \cos a \sin b \end{equation} Also note that $65^2-56^2 = 33^2$ and $13^2-12^2=5^2$.

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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 $\:\arccos(56/65)$.

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Hint:

Use Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ to convert all the inverse ratios to arcsin

and then utilize Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $

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