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For any $x \in [0,1]$ show that $$\arcsin(x)+\arccos(x)=\frac{\pi}{2}$$ Please note that this is not a homework problem, this is something I came across that appears to be true.

Wintermute
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    Draw a diagram. – Lost1 Jan 01 '14 at 16:23
  • http://www.proofwiki.org/wiki/Sum_of_Arcsine_and_Arccosine – lab bhattacharjee Jan 01 '14 at 16:34
  • I might not answer an elementary question for which several answers are already here, but I noticed none of them told you to draw the right triangle. So I posted that with an explanation. I see someone's mentioned it in a comment above. The same argument works for $\arctan x + \operatorname{arccot} x$ and for $\operatorname{arcsec} x+\operatorname{arccsc} x$. – Michael Hardy Jan 01 '14 at 17:08
  • @senshin Thank you for pointing that out. As you can guess, I completly missed that. – Git Gud Jan 01 '14 at 20:29
  • @mtiano Read the above comment, please. My apologies. I will remove the comments. – Git Gud Jan 01 '14 at 20:30

5 Answers5

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Let $\displaystyle\arcsin x=\phi\implies$

$\displaystyle (i)x=\sin\phi $

and $\displaystyle (ii)-\frac\pi2\le \phi\le\frac\pi2$ based on the definition of principal value of inverse sine function

Now, $\displaystyle x=\sin\phi=\cos\left(\frac\pi2-\phi\right)$ the angle complies with the principal value of inverse cosine function $[0, \pi]$

$\displaystyle\implies \arccos x=\frac\pi2-\phi$

Remember this identity holds true for $x\in[-1,1]$

lab bhattacharjee
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3

Let $$f(x)=\arcsin(x)+\arccos(x)$$ and since we have $$f'(x)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0$$ then $$f(x)=f(0)=\frac \pi 2$$

Edit: I assume that the expressions of the derivative of $\arcsin$ and $\arccos$ functions are obtained using the derivative of inverse function.

Second edit: To clarify my answer after some comments: We know that the function $f$ is differentaible only on the interval $[0,1)$ so $f$ is a constant equal to $\frac \pi 2$ on this interval but we can deduce the desired result since $f$ is continuous on the interval $[0,1]$.

2

Draw a right triangle. Let $x$ be the sine of one of the (non-right) angles. Then it's the cosine of the other. If you know that the three angles must add up to $180^\circ$ and one of the three is $90^\circ$, then that does it.

Michael Hardy
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Setting $$\arcsin(x)= a,$$ we get $$\sin(a)= x.$$

Also, we have $$\sin(a)=\cos(π/2-a)=x.$$

So we have $$\arccos (x)=\pi/2-a.$$

mathlove
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As $x\in[0,1],$ $\displaystyle0\le \arcsin x,\arccos x\le\frac\pi2$

If $\displaystyle\arcsin x=A,\arccos x=B,$

$\displaystyle \sin A=x\implies \cos A=+\sqrt{1-x^2},\cos B=x\implies\sin B=+\sqrt{1-x^2}$

$\displaystyle \sin(A+B)=\sin A\cos B+\sin B\cos A=x\cdot x+\sqrt{1-x^2}\cdot\sqrt{1-x^2}=1$

As $\displaystyle 0\le A,B\le\frac\pi2, 0\le A+B\le\pi, A+B=\frac\pi2$

lab bhattacharjee
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