# Sketching Inverse Trig Functions of Functions (1 Viewer)

#### SunnyScience

##### Member

Sketch the graphs of the following functions:
i. y = cos^-1(cosx)
ii.y = cos(cos^-1x)
iii.y= tan^-1(tanx)

How do you determine the domain and range of these functions, as well as there shape?

thank you.

#### Sanjeet

##### Member
they're all y=x ?

#### zeebobDD

##### Member
they're all y=x ?
nope, there y=x from a specific point to another

#### Skeptyks

##### Member
II: Draw the straight line y = x with domain -1 <= x <= 1, range -1 <= x <=1. <= means less than equal to btw.
III: Draw the straight line y = x with domain -pi/2 <= x <= pi/2 and range -pi/2 <= x < pi/2.

I don't know if this explanation is 100% correct but here it goes... y=cos^-1x has a domain of -1 <= x <= 1 and if you go cos^-1x(cosx), they simply cancel each other out and leave you with x, giving you the equation y =x and a domain as stated above. Similarly true for tan^-1x.

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#### SpiralFlex

##### Well-Known Member
This may or may not help you. (Sketching other graphs) This was posted by me a long time ago. I sent this to someone so they could explain it to a friend.

SpiralFlex said:
Hello, sorry I was having a fun day out. Sorry for late reply. This is how I would explain it.

$\bg_white \tan^{-1} (\cos x)$

$\bg_white Differentiate it,$

$\bg_white -\frac{\sin x}{1+\cos^2 x}$

$\bg_white For stationary points,$

$\bg_white \sin x = 0$

$\bg_white x=0, \pi, 2\pi...$

$\bg_white So in general,$

$\bg_white \therefore x = 2kn,\;Where n is an element of\; \mathbb{Z}$

$\bg_white (It is noteworthy, our curve is monotonically decreasing between \;0-\pi,\; monotonically increasing\; \pi-2\pi\; etc...$

$\bg_white By inspecting \;\tan^{-1} (\cos x), \; we can see if we substitute the general solution of sine into the equation, we get an alternation of \; \frac{\pi}{4}, -\frac{\pi}{4}$

$\bg_white That is, maximum/minimum is at (2kn, \frac{(-1)^{k+1}*\pi}{4}), \;I have define k as the number of alternations. (Ie. the first alternation is a maximum point at \;(0, \frac{\pi}{4}).\; second is a minimum at\;(\pi, -\frac{\pi}{4})$

$\bg_white With this in mind, let's investigate the intercepts,$

$\bg_white Let \;y=0\;. \therefore \tan^{-1}(\cos x)=\tan^{-1}(0)$

$\bg_white \cos x = 0$

$\bg_white x=\frac{-\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \;etc...$

$\bg_white Hence the intercepts will have a general solution of \; x=\frac{\pi}{2}, \frac{3\pi}{2} +2\pi n\; Where n is an element of\; \mathbb{Z}$

$\bg_white For further confirmation, we can test quickly to see if this curve is an even, odd or neither.$

$\bg_white f(-x)=\tan^{-1} (\cos (-x))=\tan^{-1} (\cos (x))=f(x)$

$\bg_white Since cosine function is even, the whole curve will be even too. So our calculations should be right at this stage.$

$\bg_white Now all you need to consider is the restrictions,$

$\bg_white Domain: All real x \; [-\infty, \infty]\; as inverse tangent can take hold of any value of x.$

$\bg_white Now, the range at this stage, we need to see if it is correct. Maybe the curve does something funny.$

$\bg_white We'll test with limits.$

$\bg_white \lim_{x \rightarrow 0} \tan^{-1}(\cos x)=\frac{\pi}{4}$

$\bg_white You can also investigate the others near the turning points.$

$\bg_white Conclusively, our curve is not dodgy, so our range takes the extremity values of \; -\frac{\pi}{4}\leq y \leq \frac{\pi}{4}$

$\bg_white The curve should look like a cosine curve a bit. Too lazy to draw it out haha. Sorry.$