# If we know $\sin \alpha$ and $\cos \alpha.$ can we find the angle $\alpha$?

For example $$\sin \alpha=\frac{8}{\sqrt{65}}, \cos \alpha=\frac{1}{\sqrt{65}}$$

Can we analytically find $$\alpha$$ ? The only thing I did was calculate $$\tan$$; it isn't helpful.

It turns out my question is if we know $$\cos \alpha$$ and $$\sin \alpha$$ is there a formula for $$\arcsin$$ and $$\arccos$$ ? The only way of finding arcsin or arccos that i know is using a calculator

There are infinitely many such $$\alpha$$'s. However, you can get one such $$\alpha$$ taking$$\alpha=\arcsin\left(\frac8{\sqrt{65}}\right)=\int_0^{\frac8{\sqrt{65}}}\frac{\operatorname dx}{\sqrt{1-x^2}}.$$

• I can make $\alpha$ unique by restricting the domain. But how to find value of arcsin without using calculator – Milan May 5 '19 at 14:02
• You can get good approximations to compute it, like I did here for $\arcsin\left(\frac12\right)$, using the Taylor series of $\arcsin$. – José Carlos Santos May 5 '19 at 14:05
• So we cannot find alpha ˝analytically˝ because sin and cos are transcedental functions? is that right ? by analytically i mean an exact value – Milan May 5 '19 at 14:09
• No, that is not right. The expression $\arcsin\left(\frac8{\sqrt{65}}\right)$ is an analytical expression that gives $\alpha$. And so is $\int_0^{\frac8{\sqrt{65}}}\frac{\operatorname dx}{\sqrt{1-x^2}}$. – José Carlos Santos May 5 '19 at 14:12

My calculator gives $$1.4464...$$

From Taylor series $$\alpha=\arccos(\dfrac 1 {\sqrt{65}})=\dfrac\pi2-\dfrac 1{\sqrt{65}}...\approx\dfrac\pi2-\dfrac18\approx1.4458.$$

• I am looking for an exact solutions if possible – Milan May 5 '19 at 14:10

There is a Taylor series expansion for $$\arcsin(x)$$ as described here:

Then $$\arccos(x) = {\pi \over 2} - \arcsin(x)$$ in the first quadrant.