Inverse trig (1 Viewer)

Pinkled · May 6, 2013

Hey guys, can anybody please help me with q9 as well as 10b? Thanks.

http://puu.sh/2NXy6.png

Drongoski · May 6, 2013

Q10

tan^-1x + tan^-1(1/x) = pi/2 - a constant

So the derivative should be zero.

Sy123 · May 6, 2013

$y = \cos^{-1}(\sin x) \ \ \ -\pi \leq x \leq \pi$

$y '= \frac{-1}{\sqrt{1-\sin^2 x}} \cdot \cos x$

$y' = \frac{- \cos x}{|\cos x|}$

$y' = -1 \ \ $for$ \ \ \cos x \geq 0$

$y'= 1 \ \ $for$ \ \ \cos x \leq 0$

This is because the cos x on top cancels out the cos x at the bottom, but for different values of x, the numerator cos x could either be - cos x, or cos x. And if this cancels out with the always positive denominator, depending on the value of x, the derivative will have different sign. We are only asked to analyse it in the domain given, so the answer is:

$y' = \begin{cases}-1 \ \ \ \frac{-\pi}{2} \leq x \leq \frac{\pi}{2} \\ 1 \ \ \ -\pi \leq x \leq \frac{\pi}{2} , \ \frac{\pi}{2} \leq x \leq \pi \end{cases}$

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For 10b, notice that in part a, you should of found that $\frac{d}{dx} (\tan^{-1}(x) + \tan^{-1}(1/x)) = 0$

This means that the integral should be simply a constant right?
That means, $y = \tan^{-1}(x) + \tan^{-1}(1/x) = C$ for some constant C, this constant is always constant no matter what value of x. So we can simply put in a value of x and see what we get, we will pick x=1.

$C = \tan^{-1}(1) + \tan^{-1}(1/1) = \frac{\pi}{2}$

That means its the line $y = \frac{\pi}{2}$ and that is all.......
nope

We also need to consider that x =/= 0, so the answer is a straight line y=pi/2 with an open circle at the y-intercept.

Inverse trig (1 Viewer)

Pinkled

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Drongoski

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Sy123

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