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The problem is as follows:

Find the value of B:

$$B=\frac{\sin40^{\circ}-\sqrt{3}\cos40^{\circ}}{\sin10^{\circ}\cos10^{\circ}}$$

I tried to use product to sum and sum to product identities but neither the sum or the difference of $40$ and $10$ seem to produce an "important angle" let's say $30^{\circ}$, $60^{\circ}$, $45^{\circ}$, $37^{\circ}$, $53^{\circ}$.

What should I do?.

Community
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Chris Steinbeck Bell
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  • What is $\cos(x+30^\circ)$? – Angina Seng Oct 19 '17 at 06:09
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    The coefficients of the trig terms in the numerator are $\sqrt{3}$ and ... somewhat sneakily ... $1$. It would be awfully convenient if each of those were divided by $2$. – Blue Oct 19 '17 at 06:11
  • See also: https://math.stackexchange.com/questions/172471/solving-e-frac1-sin10-circ-frac-sqrt3-cos10-circ and https://math.stackexchange.com/questions/10661/find-the-value-of-displaystyle-sqrt3-cdot-cot-20-circ-4-cdot-cos – lab bhattacharjee Oct 19 '17 at 10:31
  • @labbhattacharjee Thanks for those links, they've helped me a lot. – Chris Steinbeck Bell Oct 19 '17 at 13:43

2 Answers2

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$\displaystyle B = \frac{\sin(40)-\sqrt{3}\cos(40)}{\sin(10)\cos(10)}$

$\displaystyle = \frac{2(\frac{1}{2}\sin(40)-\frac{\sqrt{3}}{2}\cos(40))}{\sin(10)\cos(10)}$

$\displaystyle = \frac{2(\sin(30)\sin(40)-\cos(30)\cos(40))}{\sin(10)\cos(10)}$

$\displaystyle = \frac{-2\cos(70)}{\sin(10)\cos(10)}$

$\displaystyle = \frac{-4\cos(70)}{\sin(20)}$

Now, remember that $\cos(90-x)=\sin(x)$?

Yeah me neither lol.

$=\boxed{-4}$

Saketh Malyala
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  • Nice and neat way to solve it. I must admit that I had identified the $\sqrt(3)$ but missed to divide it by $2$. The other way to solve it could be by using the sum of angles in sines which would result into $\sin(-20^{\circ})=-\sin(20^{\circ})$ thus avoiding to recall the fact from the formula of symmetry. Although one way or another you have to remember the other. – Chris Steinbeck Bell Oct 19 '17 at 13:48
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One has $$B = 2\frac{\sin 40^\circ\sin 30^\circ - \cos 40^\circ\cos 30^\circ}{\sin 10^\circ\cos 10^\circ} = 2\frac{-\cos (40^\circ + 30^\circ)}{\sin 10^\circ \cos 10^\circ} = 2\frac{-\sin 20^\circ}{\sin 10^\circ\cos 10^\circ}=-4.$$

GAVD
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