0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I can make $\alpha$ unique by restricting the domain. But how to find value of arcsin without using calculator $\endgroup$ – Milan May 5 '19 at 14:02
  • $\begingroup$ You can get good approximations to compute it, like I did here for $\arcsin\left(\frac12\right)$, using the Taylor series of $\arcsin$. $\endgroup$ – José Carlos Santos May 5 '19 at 14:05
  • $\begingroup$ So we cannot find alpha ˝analytically˝ because sin and cos are transcedental functions? is that right ? by analytically i mean an exact value $\endgroup$ – Milan May 5 '19 at 14:09
  • $\begingroup$ No, that is not right. The expression $\arcsin\left(\frac8{\sqrt{65}}\right)$ is an analytical expression that gives $\alpha$. And so is $\int_0^{\frac8{\sqrt{65}}}\frac{\operatorname dx}{\sqrt{1-x^2}}$. $\endgroup$ – José Carlos Santos May 5 '19 at 14:12
0
$\begingroup$

My calculator gives $1.4464...$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I am looking for an exact solutions if possible $\endgroup$ – Milan May 5 '19 at 14:10
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.