Sketching Inverse Trig Functions of Functions (1 Viewer)

SunnyScience · May 2, 2012

Can someone please explain to me how to answer the following questions please

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 · May 4, 2012

they're all y=x ?

zeebobDD · May 4, 2012

Sanjeet said:
they're all y=x ?

nope, there y=x from a specific point to another

Skeptyks · May 5, 2012

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.

SpiralFlex · May 5, 2012

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.

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

$Differentiate it,$

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

$For stationary points,$

$\sin x = 0$

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

$So in general,$

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

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

$$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}$

$$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})$

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

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

$\cos x = 0$

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

$$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}$

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

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

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

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

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

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

$We'll test with limits.$

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

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

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

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

Sketching Inverse Trig Functions of Functions (1 Viewer)

SunnyScience

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Sanjeet

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zeebobDD

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Skeptyks

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SpiralFlex

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