Question

Find the value of $ \dfrac{{\sin a\sin 3a + \sin 3a\sin 7a + \sin 5a\sin 15a}}{{\sin a\cos 3a + \sin 3a\cos 7a + \sin 5a\cos 15a}} $
a) $ \sin 11a $
b) $ \cot 11a $
c) $ \cos 11a $
d) $ \tan 11a $

Answer 1

Hint: We will be using the formulas of conversion of multiplied trigonometric function to added trigonometric function and then simplify each term. The formula or identities that we are going to use is listed here. After simplification we will try to make our simplified expression match to any of the given options and conclude our answer.
Formula used:
From trigonometry we have the following identities:
1) $ \sin A\sin B = \dfrac{1}{2}\cos (A - B) - \dfrac{1}{2}\cos (A + B) $
2) $ \cos A\sin B = \dfrac{1}{2}\sin (A + B) - \dfrac{1}{2}\sin (A - B) $
3) $ \sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) $
4) $ \cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) $
5) $ \dfrac{{\sin A}}{{\cos A}} = \tan A $

Complete step by step solution:
The given trigonometric expression is:
 $ \dfrac{{\sin a\sin 3a + \sin 3a\sin 7a + \sin 5a\sin 15a}}{{\sin a\cos 3a + \sin 3a\cos 7a + \sin 5a\cos 15a}} $
 $ = \dfrac{{(\sin a\sin 3a) + (\sin 3a\sin 7a) + (\sin 5a\sin 15a)}}{{(\cos 3a\sin a) + (\cos 7a\sin 3a) + (\cos 15a\sin 5a)}} $
Applying the formula (1) we have the following:
\[ = \dfrac{{\dfrac{1}{2}\cos (3a - a) - \dfrac{1}{2}\cos (3a + a) + \dfrac{1}{2}\cos (7a - 3a) - \dfrac{1}{2}\cos (7a + 3a) + \dfrac{1}{2}\cos (15a - 5a) - \dfrac{1}{2}\cos (15a + 10a)}}{{\dfrac{1}{2}\sin (3a + a) - \dfrac{1}{2}\sin (3a - a) + \dfrac{1}{2}\sin (7a + 3a) - \dfrac{1}{2}\sin (7a - 3a) + \dfrac{1}{2}\sin (15a + 5a) - \dfrac{1}{2}\sin (15a - 5a)}}\]
\[ = \dfrac{{\dfrac{1}{2}\cos (2a) - \dfrac{1}{2}\cos (4a) + \dfrac{1}{2}\cos (4a) - \dfrac{1}{2}\cos (10a) + \dfrac{1}{2}\cos (10a) - \dfrac{1}{2}\cos (20a)}}{{\dfrac{1}{2}\sin (4a) - \dfrac{1}{2}\sin (2a) + \dfrac{1}{2}\sin (10a) - \dfrac{1}{2}\sin (4a) + \dfrac{1}{2}\sin (20a) - \dfrac{1}{2}\sin (10a)}}\]
Taking common factor $ \dfrac{1}{2} $ out we get:
\[ = \dfrac{{\dfrac{1}{2}\left( {\cos (2a) - \cos (4a) + \cos (4a) - \cos (10a) + \cos (10a) - \cos (20a)} \right)}}{{\dfrac{1}{2}\left( {\sin (4a) - \sin (2a) + \sin (10a) - \sin (4a) + \sin (20a) - \sin (10a)} \right)}}\]
Cancelling $ \dfrac{1}{2} $ we get:
\[ = \dfrac{{\cos (2a) - \cos (4a) + \cos (4a) - \cos (10a) + \cos (10a) - \cos (20a)}}{{\sin (4a) - \sin (2a) + \sin (10a) - \sin (4a) + \sin (20a) - \sin (10a)}}\]
On simplification of the above expression we get:
\[\dfrac{{\cos (2a) - \cos (20a)}}{{ - \sin (2a) + \sin (20a)}}\]
\[ = \dfrac{{\cos (2a) - \cos (20a)}}{{\sin (20a) - \sin (2a)}}\]
Applying formula (3) and (4) to the above expression we get:
\[\Rightarrow \dfrac{{2\sin \left( {\dfrac{{20a + 2a}}{2}} \right)\sin \left( {\dfrac{{20a - 2a}}{2}} \right)}}{{2\cos \left( {\dfrac{{20a + 2a}}{2}} \right)\sin \left( {\dfrac{{20a - 2a}}{2}} \right)}}\]
\[ = \dfrac{{\sin \left( {\dfrac{{20a + 2a}}{2}} \right)}}{{\cos \left( {\dfrac{{20a + 2a}}{2}} \right)}}\]
The angles in the sine and cosine function is same so we can write this as tan function:
\[ = \tan \left( {\dfrac{{20a + 2a}}{2}} \right) = \tan \left( {\dfrac{{22a}}{2}} \right) = \tan 11a\]
So, the correct answer is “Option D”.

Note: The most crucial part of the problem is the use of formulas. Always remember the basic trigonometric identities to solve the problem. It is impossible to solve the problem without knowing the formulas. Since, the simplification requires plenty of cancellations and patience, try to do all the algebraic manipulations in a stable mind and use neat and clean paper. There are other alternative ways of doing the same problem by using the identities of sine and cosine functions with use of tangent function. The problem there is, it will result in complicated angles and the number of functions we will get will be more. Moreover, it will make your copy look ugly and may result in something unexpected.