Question

The expansion $\dfrac{{\sin \dfrac{{7\pi }}{{24}} + \sin \dfrac{{5\pi }}{{24}} + \sin \dfrac{{9\pi }}{{24}} + \sin \dfrac{{3\pi }}{{24}}}}{{\cos \dfrac{{7\pi }}{{24}} + \cos \dfrac{{5\pi }}{{24}} + \cos \dfrac{{9\pi }}{{24}} + \cos \dfrac{{3\pi }}{{24}}}}$
A. 1
B. $2 - \sqrt 3 $
C. $\sqrt 2 - 1$
D. $\dfrac{1}{{\sqrt 3 }}$

Answer 1

Hint: We will first simplify the given expression. Apply the formula$\sin A + \sin B = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$ in the numerator and $\cos A + \cos B = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$ in the denominator. Take the common terms out and then cancel the terms which are common in numerator and denominator. Substitute the known values to get the required answer.

Complete step-by-step answer:
We have to find the value of $\dfrac{{\sin \dfrac{{7\pi }}{{24}} + \sin \dfrac{{5\pi }}{{24}} + \sin \dfrac{{9\pi }}{{24}} + \sin \dfrac{{3\pi }}{{24}}}}{{\cos \dfrac{{7\pi }}{{24}} + \cos \dfrac{{5\pi }}{{24}} + \cos \dfrac{{9\pi }}{{24}} + \cos \dfrac{{3\pi }}{{24}}}}$
We will apply the formula $\sin A + \sin B = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( ={\dfrac{{x - y}}{2}} \right)$ and $\cos A + \cos B = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$
=$\dfrac{{2\sin \left( {\dfrac{{7\pi }}{{24}} + \dfrac{{5\pi }}{{24}}} \right)\cos \left( {\dfrac{{7\pi }}{{24}} - \dfrac{{5\pi }}{{24}}} \right) + 2\sin \left( {\dfrac{{9\pi }}{{24}} + \dfrac{{3\pi }}{{24}}} \right)\cos \left( {\dfrac{{9\pi }}{{24}} - \dfrac{{3\pi }}{{24}}} \right)}}{{2\cos \left( {\dfrac{{7\pi }}{{24}} + \dfrac{{5\pi }}{{24}}} \right)\cos \left( {\dfrac{{7\pi }}{{24}} - \dfrac{{5\pi }}{{24}}} \right) + 2\cos \left( {\dfrac{{9\pi }}{{24}} + \dfrac{{3\pi }}{{24}}} \right)\cos \left( {\dfrac{{9\pi }}{{24}} - \dfrac{{3\pi }}{{24}}} \right)}}$
We will simplify the brackets.
=$\dfrac{{2\sin \left( {\dfrac{\pi }{2}} \right)\cos \left( {\dfrac{\pi }{{12}}} \right) + 2\sin \left( {\dfrac{\pi }{2}} \right)\cos \left( {\dfrac{\pi }{4}} \right)}}{{2\cos \left( {\dfrac{\pi }{2}} \right)\cos \left( {\dfrac{\pi }{{12}}} \right) + 2\cos \left( {\dfrac{\pi }{2}} \right)\cos \left( {\dfrac{\pi }{4}} \right)}}$
We will now take $2\sin \left( {\dfrac{\pi }{2}} \right)$ common from numerator and similarly, we can take $2\cos \left( {\dfrac{\pi }{2}} \right)$ common from denominator.
$
= \dfrac{{2\sin \left( {\dfrac{\pi }{2}} \right)\left( {\cos \left( {\dfrac{\pi }{{12}}} \right) + \cos \left( {\dfrac{\pi }{4}} \right)} \right)}}{{2\cos \left( {\dfrac{\pi }{2}} \right)\left( {\cos \left( {\dfrac{\pi }{{12}}} \right) + \cos \left( {\dfrac{\pi }{4}} \right)} \right)}} \\
   = \dfrac{{\sin \left( {\dfrac{\pi }{2}} \right)}}{{\cos \left( {\dfrac{\pi }{2}} \right)}} \\
$
Now, we know that $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $
Therefore, the above expression can be simplified as,
$\dfrac{{\sin \left( {\dfrac{\pi }{2}} \right)}}{{\cos \left( {\dfrac{\pi }{2}} \right)}} = \tan \dfrac{\pi }{2}$
And the value of $\tan \dfrac{\pi }{2}$ is not defined.
Hence, the value of the expression is not defined.

Note: The best way to solve these types of questions is to simplify the given expression using the trigonometric identities. Here, while applying the formula of sum of trigonometric ratios, we have taken the term together such that their sum is equal. For example, $\left( {\dfrac{{7\pi }}{{24}} + \dfrac{{5\pi }}{{24}}} \right) = \left( {\dfrac{{9\pi }}{{24}} + \dfrac{{3\pi }}{{24}}} \right) = \dfrac{\pi }{2}$
This helps to reduce calculations. Also, do not try to find the value of each term as it will be very time consuming.