3

Verify that: $\arcsin \theta+\arccos \theta=\frac{\pi}{2}.$ (1)

How can one verify (1) when it is not generally true? We can rewrite (1) as

Verify that if $\sin u = \cos v$, then $u+v=\frac{\pi}{2}$.

What about $\sin \frac{2\pi}{3}$ and $\cos \frac{\pi}{6}$? $\sin \frac{2\pi}{3} = \cos \frac{\pi}{6}$, but $\frac{2\pi}{3}+\frac{\pi}{6} \neq \frac{\pi}{2}$.We may try to verify (1) as:

Let $\arcsin \theta=u$ and $\arccos \theta=v$, then $\sin u=\cos v = \theta.$ We can write

$$\sin(u+v)=\sin u \cos v+ \cos u \sin v,$$

which becomes

$$\sin(u+v)=\theta^2+ \cos u \sin v.$$

Now, $\cos u=\pm \sqrt{1-\theta^2}$, and $\sin v=\pm \sqrt{1-\theta^2}$. For $\cos u= \sqrt{1-\theta^2}$ and $\sin v= \sqrt{1-\theta^2}$, we have

$$\sin(u+v)= \theta^2+(1-\theta^2)=1,$$

$$\therefore u+v=\arcsin \theta + \arccos \theta = \arcsin(1)=\frac{\pi}{2}.$$ But $\arcsin(1)$ is also equal to $\frac{\pi}{2}+2n\pi, n \in \mathbb{Z}.$ What about that? And what about the cases when $\cos u = -\sqrt{1-\theta^2}$ and $\sin u = \sqrt{1-\theta^2}$ and the other way? Then $\sin(u+v)=2\theta^2-1=1\ \mbox{iff}\ \theta =\pm 1.$

I don't know where I am making a mistake, what assumptions are wrong, or where my reasoning goes wrong. Please help.

Samama Fahim
  • 1,459
  • 4
  • 24
  • 39

3 Answers3

7

The error is in thinking that $\arcsin$ and $\arccos$ are multivalued functions that give all possible values that will yield the given sine or cosine. Instead the range of $\arcsin$ is $[-\pi/2,\pi/2]$ and the range of $\arccos$ is $[0,\pi]$, and the functions take no values outside these ranges.

Matt Samuel
  • 58,164
4

@MattSamuel has already found your mistake, hence only a different proof.

An alternative way of verifying the equation in question is differentiating $\arcsin x+\arccos x$. We obtain 0, hence the function is constant.

Przemysław Scherwentke
  • 13,668
  • 5
  • 35
  • 56
1

here is a geometric argument to show that $$\sin^{-1}(b) + \cos^{-1}b = \pi/2 \tag 1$$ using the unit circle. i will make may argument for the case $0 \le b \le 1.$ the same should go through for $-1 \le b \le 0.$

we will interpret $\sin^{-1} b$ as the arc length $t$ between $-\pi/2$ and $\pi/2$ such that $\sin t$ is the $y$-coordinate of the terminal point $(a,b)$ on the unit circle. and interpret $\cos^{-1} a$ as the arc length $t$ between $0$ and $\pi$ such that $\cos t$ is the $x$-coordinate of the terminal point $(a,b)$ on the unit circle.

pick a point $P = (a,b), a \ge 0, b \ge 0$ on the unit circle. reflect this point $P$ on the line $y = x$ is $Q = (b, a)$ let $A = (1,0), B = (0,1).$ verify that the $$arc \,AP = arc\, BQ, arc\, AQ + arc \, BQ = \pi/2. \tag 2$$

now observe that $$\sin^{-1}b = arc \, AP, \cos^{-1}(b)= arc\, AQ$$ together with $(2)$ establishes $(1).$

abel
  • 29,170