The angle $\alpha$ is greater than $90$ and less than $360$ and $\cos(\alpha)=\frac23$. Find the exact value of $\tan(\alpha)$.

Question

This question is from an OCR Additional Mathematics paper from 2009. The answer is $-\frac{\sqrt{5}}{2}$.

I know how to solve this problem by using graphs but I do not understand how to find the EXACT value that contains surds.

I know that $\cos(\alpha)= \frac{adjacent}{hypotenuse}$, and so if the adjacent side is $2$ units and the hypotenuse is $3$ units, the opposite side must have a length of $\sqrt5$. And so I thought the answer would be $\frac{\sqrt5}{2}$. But the answer clearly states it is that, but negative.

Please explain.

Answer 1

From $cos(x)= 2/3$ we get the ratio of adjacent over hypotenuse is 2/3.

That makes the ratio of opposite over adjacent $\sqrt 5 /2.$

Since the angle x is in the fourth quadrant the tangent is negative.

Thus tan(x)= $ -\sqrt 5 /2.$

Answer 2

Using the fact that $\sin^2 x + \cos^2 x = 1$ you get that $\sin^2 x = \frac{5}{9}$ so $\sin x = \pm \frac{\sqrt5}{3}$.

The given angle range tells you that you need the negative one, because the the first quadrant is excluded by the given range, and the only other quadrant with positive cosine is the fourth quadrant, which has negative sine.

Then $\tan x = \frac{\sin x}{ \cos x} = -\frac{\sqrt5}{2}$

Answer 3

The point on the unit circle at this angle is $(\frac 23, -\frac {\sqrt 5}3)$. The $x$ value has to be positive to make the cosine positive and we are given that the angle is greater than $90^\circ$ so it can't be $(\frac 23, \frac {\sqrt 5}3)$. The tangent is then $\frac {-\frac {\sqrt 5}3}{\frac 23}=-\frac {\sqrt 5}2$

Answer 4

$$ \tan x = \sqrt{ \sec ^2 x -1 } = \sqrt{(3/2) ^2 -1 } = \pm \dfrac {\sqrt 5}{2}, $$ Sign depends on whether the originating angle is in the first quadrant or in the fourth.

Answer 5

Since $\cos\alpha>0$ and $90^{\circ}<\alpha<360^{\circ},$ we obtain $270^{\circ}<\alpha<360^{\circ},$ which gives $\tan\alpha<0$.

In another hand, since $$1+\tan^2\alpha=\frac{1}{\cos^2\alpha},$$ we obtain $$|\tan\alpha|=\sqrt{\frac{9}{4}-1}$$ or $$|\tan\alpha|=\frac{\sqrt5}{2},$$ which gives $$\tan\alpha=-\frac{\sqrt5}{2}.$$

Answer 6

Lel $90° < \alpha < 360°$ and $\cos(\alpha) = \frac23$, then $$\arccos\left(\frac23\right) =\alpha = 48.19^°$$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$

Because $\alpha$ has to be bigger than $90^°$, we take the other angle $\beta$ of equal $\cos(\alpha)$ but negative $\sin(\alpha)$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$

So by means of the following identity: $$\sin \left(\arccos \left(x\right)\right)=\sqrt{1-x^2}$$ We get

$$-\sin(\alpha) = -\sin\left(\arccos \left(\frac32\right)\right)=-\sqrt{1-\left(\frac32\right)^2} = -\frac{\sqrt{5}}{3}$$