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$$ \tan x = -\frac{2}{3} $$

when $\dfrac{5\pi}{2} < x < 3\pi$.

I understand this, but I don't know how to calculate the two other functions' values, $\cos x$, $\sin x$, using $\tan x$

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Please ask a clear question. – Rasmus Mar 1 '13 at 16:37
@Back in a Flash: Please don't edit questions into the imperative form if they were not originally asked that way. Many people find it impolite, and may even downvote for it, which is unfair on the OP. – Tara B Mar 2 '13 at 13:32
Indeed, someone did down-vote the question. – Michael Hardy Mar 2 '13 at 13:42
Why are there votes to close this question? – Michael Hardy Mar 2 '13 at 13:54
@Back in a Flash: I'm very sorry! I just discovered that it was not you but the OP who edited the question into the imperative form. Shall I remove my previous comment? I don't like it being there accusing you of something you didn't do, but on the other hand it might make the whole discussion going on on meta confusing if I remove it. – Tara B Mar 2 '13 at 16:29

So, $$\frac{\sin x}{-2}=\frac{\cos x}3=\pm\frac{\sqrt{\sin^2x+\cos^2x}}{\sqrt{(-2)^2+3^2}}=\pm\frac1{\sqrt{13}}$$

So, if $\sin x=\mp\frac2{\sqrt{13}},\cos x=\pm\frac3{\sqrt{13}}$

Now as $\frac{5\pi}2<x<3\pi,$ x lies in the second Quadrant.

Using "All Sin Tan Cos" formula, $\sin x>0$ and $\cos x<0$

So, $\sin x=\frac2{\sqrt{13}}, \cos x=-\frac3{\sqrt{13}}$

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Could anybody please tell me how it has become "Community Wiki" as the revision shows: "Post Made Community Wiki by lab bhattacharjee" – lab bhattacharjee Mar 2 '13 at 14:12
Did you edit the answer repeatedly? – Hagen von Eitzen Mar 2 '13 at 15:36
There is only one revision, one can check. – lab bhattacharjee Mar 2 '13 at 15:39
Then maybe you accidentally ticked the 'community wiki' box? – Tara B Mar 2 '13 at 16:11
@lab: If you want "Community Wiki" removed, just flag this post, select "other" and type a note to that effect in the text box. Your message will be sent (privately) to the mods (and only the mods). – cardinal Mar 2 '13 at 17:48

$$ \frac{\sin x}{\cos x} = \frac{-2}{3} = \frac{-2/\sqrt{(-2)^2+3^2}}{3/\sqrt{(-2)^2+3^2}} = \underbrace{\frac{-2/\sqrt{13}}{3/\sqrt{13}} = \frac{2/\sqrt{13}}{-3/\sqrt{13}}}_{\text{Which one of these?}}. $$

One divides by $\sqrt{(-2)^2+3^2}$ in order to make the squares of the numerator and denominator add up to $1$, since one must have $\sin^2 x+\cos^2 x = 1$.

One chooses between the two alternatives over the $\underbrace{\text{underbrace}}$ by looking at which quadrant one is in. In that quadrant, the sine is positive and the cosine is negative.

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From $\frac{5\pi}2<x<3\pi$, we conclude that $\sin x>0$, $\cos x<0$. Now from $$\frac49=\tan^2x =\frac{\sin^2 x}{\cos^2 x}=\frac{\sin^2 x}{1-\sin^2x}$$ we obtain $$\frac49(1-\sin^2x)=\sin^2x$$ hence $$\sin^2 x=\frac4{13},\qquad \cos^2 x=1-\sin^2x=\frac9{13}$$ and ultimately (using the above remark about the signs) $$\sin x=\frac2{\sqrt{13}},\qquad \cos x=-\frac3{\sqrt{13}}.$$

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This answer seems more complicated than it needs to be. – Michael Hardy Mar 2 '13 at 13:56

Consider the trigonometric identity:

$$\sec^2 x = \tan^2 x + 1.$$

Since $\sec x = \frac{1}{\cos x},$ this implies that

$$|\cos x \,| = \frac{1}{\sqrt{\tan^2 x + 1}}.$$

Note that $\tan (x + \pi) = \tan x$, but $\cos (x + \pi) = -\cos x,$ so we really do need the absolute value function (or a $\pm$ sign) in the equation above. Additional information is needed in order to decide whether to use the positive or negative value of $\cos x$; if we are given that $\frac52 \pi < x < 3\pi$ then we should use the negative value (since $\cos x < 0$ for every such $x$), hence

$$\cos x = -\frac{1}{\sqrt{\tan^2 x + 1}} = -\frac{3}{\sqrt{13}}.$$

For $\sin x$, simply take

$$\sin x = (\cos x)(\tan x) = \left(-\frac{3}{\sqrt{13}}\right)\left(-\frac23\right) = \frac{2}{\sqrt{13}}. $$

If we did not already know that $\cos x < 0$ from the restriction $\frac52 \pi < x < 3\pi,$ we would have to take other measures to deal with the sign of $\cos x$. For example, if we know that $2\pi < x < 3\pi,$ then that alone does not tell us whether $\cos x$ should be positive or negative; but since $\sin x > 0$ for every $x$ in that range and we are given that $\tan x = -\frac23 < 0$, then $\cos x = \frac{\sin x}{\tan x} < 0.$

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