How would you suggest I go about solving this question? I've been thinking about it for ages and nothing comes to mind.
$$\arcsin x + \arccos x = \frac{\pi}{2}$$
Asked
Active
Viewed 276 times
3
-
Has OP already accepted an answer? I answered there. – Narasimham Sep 29 '16 at 18:23
6 Answers
7
METHODOLOGY $1$: Calculus Based
Let $f(x)=\arcsin(x)+\arccos(x)$. Note that $f'(x)=0$. Therefore, $f(x)$ is constant.
Since $f(0)=\pi/2$, $f(x)=\pi/2$ for all $x$.
METHODOLOGY $2$: Non-Calculus Based
Alternatively, $\sin(f(x))=x^2+\sqrt{1-x^2}\sqrt{1-x^2}=1$. Therefore, $f(x)=\pi/2+2n\pi$ for some $n$.
Noting that $|f(x)|\le \pi$, we conclude $f(x)=\pi/2$.
Mark Viola
- 179,405
-
-
-
-
-
Just curious ... but why did you change the "best vote?" I was the first to post both solutions and even answered a question you asked of another user. – Mark Viola Sep 29 '16 at 17:21
6
Hint: Define $f(x)=\arcsin(x)+\arccos(x)$. Find $f'(x)$, see that it is $0$ and conclude that $f(x)=\text{const.}=c$. Plug in x=0 to find the value of $c$, which happens to be $\pi/2$.
Another method goes like this: Let $g(x)=\arcsin(x)$ and $h(x)=\pi/2-\arccos(x)$. Now take the sine of both equations:
$\sin(g)=x$ and $\sin(h)=\sin(\pi/2-\arccos(x))$. The second equation can be transformed by using the complementary angle formula for $\sin(\pi/2-x)=\cos(x)$ to result in $\sin(h)=\cos(\arccos(x))=x$. Hence, $\sin(g)=\sin(h)$ or $g=h+2\pi k$. By plugging in a specific value for $x=0$ you will see that $k=0$.
MrYouMath
- 15,833
-
-
No worry. That's happened to me a number of times. I've added a second way forward also. ;-)) – Mark Viola Sep 29 '16 at 16:24
-
The next chapter in my textbook is differentiation, which covers the concepts: Chain rule, product rule and quotient rule. So, I'm currently unable to find f'(x) with my limited knowledge of just the power rule. If it's not a hassle, could you think of another way to solve this without calc? – Anonymous Sep 29 '16 at 16:25
-
-
-
-
There is a funny trap: $f'(x)=0$ everywhere in the domain of $f$ does not necessarilty mean the function is constant. But here, you are right, of course. If you try the same method on $\arctan x+\arctan\dfrac1x$, it won't work (well, it will, but you will have to change something) – Jean-Claude Arbaut Sep 29 '16 at 17:26
2
$$\arccos x=\frac\pi2-\arcsin x$$
Then, taking the cosine,
$$\cos(\arccos x)=\cos\left(\frac\pi2-\arcsin x\right)=\sin(\arcsin x),$$
and by the definition of these functions, for all $-1\le x\le1$,
$$x=x.$$
Alternatively, take the cosine, and by the addition formula we get an identity,
$$\cos(\arccos x+\arcsin x)=\cos(\arccos x)\cos(\arcsin x)-\sin(\arccos x)\sin(\arcsin x)=x\sqrt{1-x^2}-\sqrt{1-x^2}x=0=\cos\left(\frac\pi2\right).$$
-
-
For the first method, how did cos(pi/2 - arcsin x) = sin(arcsin x) @Yves Daoust – Anonymous Sep 29 '16 at 16:39
-
@Anonymous Note that $\cos(\pi/2-y)=\cos(\pi/2)\cos(y)+\sin(\pi/2)\sin(y)=\sin(y)$. Hence, for $y=\arcsin(x)$, we have $\cos(\pi/2-y)=x$ – Mark Viola Sep 29 '16 at 16:43
2
$$\sin^{-1}x=\theta $$ $$x=\sin\theta$$ $$x=\cos\left(\frac{\pi}{2}-\theta\right) $$ $$-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$$
$$ 0 \le \frac{\pi}{2} -\theta \le \pi$$ $$\frac{\pi}{2} -\theta=\cos^{-1}x$$ $$\frac{\pi}{2} -\sin^{-1}x=\cos^{-1}x$$ $$\sin^{-1}x + \cos^{-1}x=\frac{\pi}{2}$$
Aakash Kumar
- 3,480
2
You started with an identity which means it is satisfied for all $x$. Stated in words what you said is
"In a right triangle if the two acute angles sum up to $ \pi/2, $ what is one of them? Answer is of course "Any angle !"
If you had not a priori known the identity,then taking $cos$ of arguments on both sides of equation you get
$$ \cos(\, \sin^{-1} x + \cos^{-1} x )= \sqrt{1-x^2 }\, x -x \sqrt{1-x^2 } =0 $$
So it can make your infer that angle can be any value. The same holds if on RHS $\pi/2 $ or $ 3 \pi/2 $ or those obtained by adding $ 2n\pi$ to them are given in the problem at start.
Narasimham
- 40,495
