0

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

Milan
  • 1,631

3 Answers3

1

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

José Carlos Santos
  • 427,504
  • I can make $\alpha$ unique by restricting the domain. But how to find value of arcsin without using calculator – Milan May 05 '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 05 '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 05 '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 05 '19 at 14:12
0

My calculator gives $1.4464...$

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

J. W. Tanner
  • 60,406
0

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

Maclaurin expansion of $\arcsin x$

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

Zarrax
  • 44,950