Question

What will be the value of the given trigonometric expression?
$\dfrac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }}$

Answer 1

Hint: Rearrange the numerator and denominator terms in such a way that the sum of angles comes equal in magnitude, that is $9\theta ,3\theta $ terms should be together and $5\theta ,7\theta $ should be together. Then apply the trigonometric identity that $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ and$\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$.

Complete step-by-step answer:

Given trigonometric equation is
$\dfrac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }}$
Rearrange its terms we have,
$ \Rightarrow \dfrac{{\sin 9\theta + \sin 3\theta + \sin 7\theta + \sin 5\theta }}{{\cos 9\theta + \cos 3\theta + \cos 7\theta + \cos 5\theta }}$
Now as we know $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ and $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ so use this property in above equation we have,
$ \Rightarrow \dfrac{{2\sin \left( {\dfrac{{9\theta + 3\theta }}{2}} \right)\cos \left( {\dfrac{{9\theta - 3\theta }}{2}} \right) + 2\sin \left( {\dfrac{{7\theta + 5\theta }}{2}} \right)\cos \left( {\dfrac{{7\theta - 5\theta }}{2}} \right)}}{{2\cos \left( {\dfrac{{9\theta + 3\theta }}{2}} \right)\cos \left( {\dfrac{{9\theta - 3\theta }}{2}} \right) + 2\cos \left( {\dfrac{{7\theta + 5\theta }}{2}} \right)\cos \left( {\dfrac{{7\theta - 5\theta }}{2}} \right)}}$
Now simplify the above equation we have,
$ \Rightarrow \dfrac{{2\sin \left( {6\theta } \right)\cos \left( {3\theta } \right) + 2\sin \left( {6\theta } \right)\cos \left( \theta \right)}}{{2\cos \left( {6\theta } \right)\cos \left( {3\theta } \right) + 2\cos \left( {6\theta } \right)\cos \left( \theta \right)}}$
$ \Rightarrow \dfrac{{2\sin \left( {6\theta } \right)\left( {\cos \left( {3\theta } \right) + \cos \left( \theta \right)} \right)}}{{2\cos \left( {6\theta } \right)\left( {\cos \left( {3\theta } \right) + \cos \left( \theta \right)} \right)}}$
Now cancel the common terms we have,
$ \Rightarrow \dfrac{{\sin 6\theta }}{{\cos 6\theta }} = \tan 6\theta $
So this is the required answer.
Hence option (C) is correct.

Note: The trick behind taking $(9\theta ,3\theta ){\text{ and }}\left( {7\theta ,5\theta } \right)$ terms together was to form same terms in both numerator and denominator so that they could be cancelled. It is advisable to remember trigonometric identities as it helps to save a lot of time and proves very useful while dealing with trigonometric problems.