1
$\begingroup$

I just encountered the following multiple choice question on a exam. It looks simple but surprisingly I couldn't decipher it! So I decided to mention it here. :)

What is the simplified form of $\dfrac{2 \cos 40^\circ -1}{\sin 50^\circ}$?

  1. $4 \sin 10^\circ$
  2. $4 \cos 10^\circ $
  3. $-4 \sin 10^\circ$
  4. $-4 \cos 10^\circ$

Any hint or help is appreciated. :)

$\endgroup$
2
  • $\begingroup$ The numerator is positive, so this excludes 3 and 4. $\endgroup$
    – egreg
    Commented Mar 9, 2016 at 21:19
  • $\begingroup$ @egreg: Yes, you are right! :) In fact, I am not just looking for the correct answer but a comprehensive solution to obtain the answer. :) $\endgroup$ Commented Mar 9, 2016 at 21:20

3 Answers 3

6
$\begingroup$

Consider $$ 2\sin50^\circ\sin10^\circ=\cos(50^\circ-10^\circ)-\cos(50^\circ+10^\circ) $$

$\endgroup$
1
  • $\begingroup$ Very neatly done (+1) $\endgroup$
    – Nikunj
    Commented Mar 9, 2016 at 21:24
3
$\begingroup$

Since this is from an exam (a badly constructed one, for my taste: what competence is it supposed to test?), I assume it is to be solved under time pressure. Trying out addition theorems takes way too long, and is not sufficient if the solution, like the one just posted by egreg, additionally requires usage of a specific value like cos 60=1/2.

Therefore, it is preferable to proceed from the onset by estimates. To arrive at solution (1), it is sufficient to know that cos 40 and sin 50 are both close to $\sqrt{0.5}\simeq0.7$. Rough estimate then shows that your fraction lays between 0 and 1.

$\endgroup$
1
  • 2
    $\begingroup$ Yes, answers 3 and 4 are automatically excluded for sign reasons, and $4\cos10^\circ$ is almost $4$, far too big. $\endgroup$
    – Lubin
    Commented Mar 9, 2016 at 21:30
1
$\begingroup$

$$2\cdot\dfrac{\cos(50^\circ-x)-\cos(50^\circ+x)}{\sin50^\circ}=2\cdot\dfrac{2\sin50^\circ\sin x}{\sin50^\circ}=4\sin x$$ as $\sin50^\circ\ne0$

Here $x=10^\circ$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .