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Prove that:

  1. $$\cos^{-1}x= \tan^{-1}\Big(\frac{\sqrt{1-x^2}}{x}\Big)\quad\text{for}\quad 0<x\leq 1$$

  2. $$\cos^{-1}x= k+ \tan^{-1}\Big(\frac{\sqrt{1-x^2}}{x}\Big) \quad\text{for}\quad\mathop{}-1\leq x <0$$ and find the value of $k$.

My turn:

1. Let $\cos^{-1}x = y$ , $x =\cos y$ , $\tan y = \frac{\sqrt{1-x^2}}{x}$ , $y = \tan^{-1}(\frac{\sqrt{1-x^2}}{x})$

2.

From part 1 we note that $\cos^{-1}x \in [0,\frac{\pi}{2})$ For part 2 $\cos^{-1}x \in (\frac{\pi}{2}, \pi]$, so the curve is translated $\frac{\pi}{2}$ units vertical so $k= \frac{\pi}{2}$ Is the solution correct ?

Nathanael Skrepek
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Hussien Mohamed
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2 Answers2

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You are almost there, just recall the relation $ \sin^2(x) + \cos^2(x)=1$. So setting $y=\cos^{-1}(x)$, $x=\cos(y)$ one has

$$ \tan^{-1}(\frac{\sqrt{1-x^2}}{x})=\tan^{-1}\left(\frac{\sqrt{1-\cos^2(y)}}{\cos(y)}\right) =\tan^{-1}\left(\frac{\sin(y)}{\cos(y)}\right) =\tan^{-1}\left( \tan(y) \right) =y =\cos^{-1}(x) $$

Concerning point (2): I think intuitive understanding is sufficient. For a negative $x<0$ you add to the function value at zero $\cos^{-1}(0)=\frac{\pi}{2}$ the number $\frac{\pi}{2}-\cos^{-1}(-x)$, look at the graph: acos_chart. One so has:

$$ \cos^{-1}(x)|_{x<0}= \frac{\pi}{2} + \left[\frac{\pi}{2}-\cos^{-1}(-x) \right] $$ $$\cos^{-1}(x)|_{x<0}=\pi-\tan^{-1}\left(\frac{\sqrt{1-x^2}}{-x}\right)$$

Arcustangent is an odd function: changing the sign of the argument changes the sign before the function

$$\cos^{-1}(x)|_{x<0}=\pi+\tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$$

EDIT: more details about point (2). Looking at the graph acos_chart one sees that $f(x)= \cos^{-1}(x)-\frac{\pi}{2}$ is an odd function (you translate the graph downwards by $\pi/2$). One has $f(x)=-f(-x)$. So now:

$$ \cos^{-1}(x)|_{x<0} = f(x) + \frac{\pi}{2} = -f(-x) + \frac{\pi}{2} = -(\cos^{-1}(-x)-\frac{\pi}{2}) + \frac{\pi}{2}$$

$$ = -\cos^{-1}(-x)+\frac{\pi}{2} + \frac{\pi}{2} = -\tan^{-1}(\frac{\sqrt{1-x^2}}{-x})+\pi =\pi + \tan^{-1}(\frac{\sqrt{1-x^2}}{x})$$

F. Jatpil
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1

Another way

Using https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values

$\sin^{-1}y,\tan^{-1}y$ have the same ranges

We can prove $$\sin^{-1}y=\tan^{-1}\dfrac y{\sqrt{1-y^2}}$$

Now for $x\ge0,\cos^{-1}x=\sin^{-1}\sqrt{1-x^2}$

Set $y=\sqrt{1-x^2}$

For $x<0,$ using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?

$\cos^{-1}x=\pi-\cos^{-1}(-x)=\pi-\arcsin\sqrt{1-x^2}=\pi+\arcsin(-\sqrt{1-x^2})$

Set $y=-\sqrt{1-x^2}$

lab bhattacharjee
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