# Why is $\arcsin(\sqrt 2/2) = 45^{\circ}$?

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?

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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

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?)

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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 http://www.khanacademy.org/math/trigonometry/less-basic-trigonometry/trig_iden_tutorial/v/proof--cos-a-b-----cos-a--cos-b---sin-a--sin-b.

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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