# Can $\frac{\sqrt{3}}{\sin20^{\circ}}-\frac{1}{\cos20^{\circ}}$ have two values?

I would like to confirm a solution. The question goes as:

Show that $$\frac{\sqrt{3}}{\sin20^{\circ}}-\frac{1}{\cos20^{\circ}}=4$$

Firstly I combined the two terms to form something like: $$\dfrac{\sqrt{3}\cos20^{\circ} - \sin20^{\circ}}{\sin20^{\circ}\cos20^{\circ}}.$$

Clearly, the numerator is of the form $$a\cos\theta+b\sin\theta$$, and can thus be expressed as $$R\sin(\theta+\alpha)$$, where $$R=\sqrt{a^2+b^2}$$, and $$\alpha=\arctan(\dfrac{a}{b})$$. Following this method, I obtain something like this: $$\dfrac{2\sin(20^{\circ}-60^{\circ})}{\sin20^{\circ}\cos20^{\circ}},$$ which results in the expression to be equal to $$-4$$. But, we took $$R$$ to be the principal square root, which was $$2$$. If we take it to be the negative root, $$-2$$, it follows that the expression is equal to $$4$$. So, can this expression have two values?

Edit: I would also like to mention how I computed the numerator:

Numerator:

$$\sqrt{3}\cos20^{\circ} - \sin20^{\circ}$$.

Now, this can be represented as $$R\sin(\theta+\alpha)$$.
Computation of $$R$$: $$R=\sqrt{a^2+b^2}\implies R=\sqrt{(3+1)}=2$$
Computation of $$\alpha$$: $$\tan\alpha=\dfrac{a}{b}\implies \tan\alpha=\dfrac{\sqrt3}{-1}\implies \alpha=-60^{\circ}$$, because $$\tan(-x)=-\tan x$$, and in this case $$\tan\alpha=-\sqrt3$$, and $$-\tan(60^{\circ})=-\sqrt3$$, so $$\alpha=-60^{\circ}$$.

Hence, we conclude that numerator:$$2\sin(20+(-60))=2\sin(-40)=-2\sin(40^{\circ})$$

• The numerator should actually be $\dfrac{2\sin(60^\circ-20^\circ)}{\sin 20^\circ\cos 20^\circ}$. Looks like you made a mistake while applying the formula you mentioned. See it rather simply as follows: $$\frac{\sqrt 3\cos 20^\circ - \sin 20^\circ}{\sin 20^\circ\cos 20^\circ}=\frac{2(\frac{\sqrt 3}2\cos 20^\circ - \frac 12\sin 20^\circ)}{\sin 20^\circ\cos 20^\circ}=\frac{2(\sin 60^\circ\cos 20^\circ - \cos 60^\circ\sin 20^\circ)}{\sin 20^\circ\cos 20^\circ}=\frac{2\sin(60^\circ - 20^\circ)}{\sin 20^\circ \cos 20^\circ}$$ Jul 16, 2016 at 9:09
• Jul 16, 2016 at 9:11
• To all users, I added my computation of the numerator. Jul 16, 2016 at 9:32

The expression would have only one value, and that is 4. I'll first tell you the reason for this, then your mistake in calculating the numerator, which led you to the wrong conclusion.

1) So the reason why the expression would have one specific value is because the values of $\sin20^\circ$ and $\cos20^\circ$, on which the expression depends, have just one specific well-defined values of their own. Just by knowing this, you could have concluded that your calculation or reasoning has gone wrong somewhere.

2) The mistake which led you to the wrong conclusion is in your calculation of the numerator. Your numerator originally was -

$\sqrt{3}\cos20^{\circ} - \sin20^{\circ}$, which can be written as

$2(\frac{\sqrt{3}}{2}\cos20^{\circ} - \frac{1}{2}\sin20^{\circ})$ ...... (1)

And you were trying to express it as -

$R\sin(\theta+\alpha)$, which can be expanded to express it as follows -

$R(\sin\theta\cos\alpha + \cos\theta\sin\alpha)$ ..... (2)

If you compare (1) with (2), you can see that the way you calculated your value of $R=2$ was correct. Also, on comparison, the value of $\alpha = 20^\circ$. But here's where you got your calculation wrong -

On comparison, the value of $\theta$ must be such that

$\sin\theta = \frac{\sqrt{3}}{2}$ and $\cos\theta = -\frac{1}{2}$

Since the sine is positive while the cosine is negative, $\theta$ must lie in the second quadrant, and the value would therefore be $\theta = 180^\circ - 60^\circ = 120^\circ$ (and not the value you found out which was $-60^\circ$, which would, in fact, lie in the fourth quadrant). Now the numerator will correctly be -

$2\sin{(120^\circ + 20^\circ)} = 2\sin{140^\circ} = 2\sin{40^\circ}$. Now you can put this correct numerator back in your final reduced expression and get the only value possible for the given expression, and that is 4.

When $R=-2,-2\sin\theta=\sqrt3\iff\sin\theta=?,-2\cos\theta=1\iff\cos\theta=?$

Observe that $\theta$ lies in the third quadrant $\theta=(2n+1)180^\circ+60^\circ$ where $n$ is any integer

$$\sin\{20^\circ-(2n+1)180^\circ+60^\circ)\}=-\sin220^\circ$$

and $$\sin220^\circ=\sin(180+40)^\circ=-\sin40^\circ$$