Is there an easy way to quickly prove these formulas?

If not, is there any easy mnemonic way to memorize them fast?

$$\begin{align} \arcsin(a) &= \arctan\left(\frac{a}{\sqrt{1-a^2}}\right) \\[4pt] \arccos(a) &= \operatorname{arccot}\left(\frac{a}{\sqrt{1-a^2}}\right) \\[4pt] \arctan(a) = \arcsin\left(\frac{a}{\sqrt{1+a^2}}\right) &= \arccos\left(\frac{1}{\sqrt{1+a^2}}\right) = \operatorname{arccot}\left(\frac{1}{a}\right) \\[4pt] \operatorname{arccot}(a) &= \arccos\left(\frac{a}{\sqrt{1+a^2}}\right) \end{align}$$

P.S. Wikipedia desribes it here

enter image description here

  • 1
    $\begingroup$ remember $\tan=\dfrac{\sin}{\cos}$ and $sin^2+cos^2=1$ $\endgroup$ – J. W. Tanner Aug 16 '19 at 16:57

For the first one draw a right angled triangle as below.enter image description here

Now $\sin x = a \implies x = \sin^{-1}a = \tan^{-1}\frac{a}{\sqrt{1-a^2}}$

Do similarly for the other cases.

enter image description here


For the last one, if $\theta=\operatorname{arccot}(a)$ then $a=\cot\theta,$

so $a^2+1=\left(\dfrac{\cos\theta}{\sin\theta}\right)^2+\left(\dfrac{\sin\theta}{\sin\theta}\right)^2=\dfrac1{\sin^2\theta}=\dfrac1{1-\cos^2\theta}.$

Can you take it from here?

The others are similar.


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.