Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Just found this out by playing around with my calculator. Does that mean that $ \arcsin(\sqrt 2) = 90^{\circ}$?

And then i wonder how you show that $\cos(90^{\circ}-v) = \sin(v)$ mathematically?

share|cite|improve this question
Depending on your situation, $\cos(90^\circ -v)=\sin v$ holds by definition (the name co-sine etymologically referes to the complementary angle) – Hagen von Eitzen Mar 29 '14 at 17:01
You do acknowledge that sin always return values less than 1 for real numbers? – evil999man Mar 29 '14 at 17:29
up vote 2 down vote accepted

If you draw up an isoceles right-angled triangle where two sides are 1, then the hypotenuse is $\sqrt{2}$. Since the triangle is isoceles we know that the non-right angles are equally large and since they must sum up to $90^{\circ}$, they must each be $45^\circ$. Thus we have $$\sin{45^\circ} = \frac{1}{\sqrt{2}} \Leftrightarrow \arcsin{\frac{1}{\sqrt{2}}}=45^\circ$$

Since $$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$we see that also $\arcsin{\frac{\sqrt{2}}{2}}=45^\circ$.

This does however not imply that $\arcsin{\sqrt{2}} = 90^\circ$.

(What do you mean by proving mathematically, is it OK with a geometric proof and recalling definitions, or do you mean by using formulas for addition like Sanath Devalapurkar did in his answer?)

share|cite|improve this answer
Thank you. I mean solving it with algebra. – user3346607 Mar 29 '14 at 17:50

We know by drawing an isosceles right triangle that $\sin 45^\circ=\dfrac{1}{\sqrt{2}}$, so $45^\circ=\arcsin \dfrac{1}{\sqrt{2}}$.

We have $$\cos(90^\circ-\theta)=\cos 90^\circ\cos\theta+\sin 90^\circ\sin\theta-\sin\theta$$ For a proof, see

share|cite|improve this answer
You should really explain where the second equation comes from. – Cameron Williams Mar 29 '14 at 17:07
@CameronWilliams See my edited answer. – Sanath K. Devalapurkar Mar 29 '14 at 17:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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