I am trying to solve $\tan(\arccos(-5/8))$. I thought that $\cos(-x)$ is the same as $\cos(x)$, so I got a positive answer, $\sqrt{39}/5$. But when I looked up the answer it said it was $-\sqrt{39}/5$. Why is it negative?
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A trick $\sin (\arccos X)$ is always positive. $\arccos$ is always in first or second quandrant and the $\sin$ of anything in the first or third is always positive. And $\cos(\arcsin x)$ is always positive. $\arcsin$ is always in the first or fourth quadrant and $\cos$ of anything int first of fourth quadrant is always positive. And $sin(\arcsin x) = x$ and $\cos(\arccos x) = x$ so $\tan (\arccos (x)) = \frac {\sin (arcos x)}{\cos(\arccos x)} =\frac {+\sqrt{1- x^2}}{x}$. And as $x$ is negative.... – fleablood Mar 11 '20 at 00:09
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Also note that $\sin(\arcsin x) = x$ be definition gut $\arcsin(\sin \theta)=\pm x = |x|$. – fleablood Mar 11 '20 at 00:12
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https://math.stackexchange.com/questions/1224415/how-do-i-prove-that-arccosx-arccos-x-pi-when-x-in-1-1 – lab bhattacharjee Mar 11 '20 at 01:42
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Trig functions are not injective. If $\cos x = \cos y$ it is not the case that $x = y$. So if we define $\arccos k$ is the angle $\theta$ so that $\cos \theta = k$, then we have to ask which angle $\theta$ so that $\cos \theta = k$. There are an infinite number of such angles. And there are two different such angles between $0$ and $2\pi$. Which one do we choose for $\arccos$?
We have to choose one and not the other. Which one do we choose?
The answer is completely arbitrary. But it is consistant.
If $M > 0$ and $\cos \theta = M$ then either $\theta$ is in the first quadrant $0\le \theta \le \frac {\pi}2$, or $\theta$ is in the fourth quadrat $\frac {3\pi}2 \le \theta \le 2\pi$. Which one is the answer to $\arccos M$?
The answer is... the one in the first quadrant. Why do we choose that one and not the one in the fourth quadrant? "Because" that's why. I'd say we flipped a coin but we had to pick one and ... the positive value would make things easier.
And likewise if $M < 0$ and $\cos \theta =M$ then either $\theta$ is in the second or third quadrant. Which one is $\arccos M$? The one in the third quadrant.
Arbitrarily: $0 \le \arccos M \le \pi$. And $-\frac {\pi}2 \le \arcsin M \le \frac {\pi}2$.
That's just convention.
....
So
We have $\arcsin(-\frac 58) = \theta$. We have $\cos \theta = -\frac 58<0$. There are two possible values for $\theta$ where that is true. One is between $\frac \pi 2$ and $\pi$, and the other between $\pi$ and $\frac {3\pi}2$. We choose the one that is between $0$ and $\pi$ so that is the one between $\frac {\pi}2$ and $\pi$. Why that one? "Because".
And if $\frac {\pi}2 \le \theta \le \pi$, then we know that $\sin \theta > 0$.
Notice IF we had choose then other possible value then we would have had $\sin \theta < 0$.
But we always have $0 \le \arccos \le \pi$ and we never have $\pi < \arccos < 2\pi$. So we didn't pick it.
So $\cos \theta = -\frac 58$.
Then $\sin \theta = \pm \sqrt{1- (\frac 58)^2} =\pm \frac {\sqrt{39}}8$. But because $\arccos -\frac 58\le \pi$ we have $\sin \theta > 0$.
So $\sin(arccos(-\frac 58)) = \frac {\sqrt{39}}8$.
And $\tan (\arccos(-\frac 58)) = \frac {\sin(\arccos(-\frac 58))}{\sin(\arccos(-\frac 58))}=\frac {\frac {\sqrt{39}}8}{-\frac 58}=-\frac {\sqrt{39}}5$
fleablood
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The trick is to look at the ratio. It is negative, so the angle formed by -5 and 8 is in the second or fourth quadrant. What is tangent of any angle in these two quadrants? Negative. Hope this helps.
Debbie
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$$\tan(\arccos(-x))=\tan(\pi/2-\arcsin(-x))=\tan(\pi/2+\arcsin x)=-\cot(\arcsin x)$$
Now if $\arcsin x=y,x=\sin y$ and using Principal Value of Inverse Trig Function
$\cos y=+\sqrt{1-x^2}$
$\cot(\arcsin x)=\cot y=?$
lab bhattacharjee
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