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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.

Darío A. Gutiérrez
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Nicholas Zhang
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    $90 < X < 360$, and $\cos X > 0$, so $X$ must be in the 4th quadrant. Therefore $\sin X < 0$ and $\tan X < 0$ – Dylan Feb 13 '18 at 15:09
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    You need to think of trig functions as functions instead of triangle ratios, since using triangles limits you to angles between $0$ and $180$ – Dylan Feb 13 '18 at 15:11
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    It may be easier to visualize using unit circle. Draw a unit circle and line $x=2/3$ and find the intersection points. On unit circle, $x$ is cosine and $y$ is sine. – Vasili Feb 13 '18 at 15:15

6 Answers6

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

Jaap Scherphuis
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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 $\frac {-\frac {\sqrt 5}3}{\frac 23}=-\frac {\sqrt 5}2$

Ross Millikan
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Lel $90° < \alpha < 360°$ and $\cos(\alpha) = \frac23$, then $$\arccos\left(\frac23\right) =\alpha = 48.19^°$$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here

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

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

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

Darío A. Gutiérrez
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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.$

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

Narasimham
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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 $$\tan\alpha=-\frac{\sqrt5}{2}.$$

Michael Rozenberg
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