Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
1  
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 at 17:01
    
You do acknowledge that sin always return values less than 1 for real numbers? –  Awesome Mar 29 at 17:29

2 Answers 2

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|improve this answer
1  
Thank you. I mean solving it with algebra. –  user3346607 Mar 29 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.

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

Your Answer

 
discard

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.