Question

The value of $\dfrac{\sin 3\theta +\sin 5\theta +\sin 7\theta +\sin 9\theta }{\cos 3\theta +\cos 5\theta +\cos 7\theta +\cos 9\theta }$ is equal to
(a) $\tan 3\theta$
(b) $\cot 3\theta$
(c) $\tan 6\theta$
(d) $\cot 6\theta$

Answer 1

We should rearrange the terms and group them such that $3\theta$ term comes together with $9\theta$ , and the $5\theta$ term comes together with $7\theta$ . The advantage of this grouping is that we can get our result in a fewer number of steps. We must then use the formula $\sin C+\sin D=\dfrac{1}{2}\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ for addition of sine, and the formula $\cos C+\cos D=\dfrac{1}{2}\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ for addition of cosines. On further simplifying, we can easily get our answer.

Complete step by step solution:
Let us assume a variable $x$ such that $x=\dfrac{\sin 3\theta +\sin 5\theta +\sin 7\theta +\sin 9\theta }{\cos 3\theta +\cos 5\theta +\cos 7\theta +\cos 9\theta }$ .
We can rearrange and group the terms in numerator and denominator and write it as follows,
$x=\dfrac{\left( \sin 3\theta +\sin 9\theta \right)+\left( \sin 5\theta +\sin 7\theta \right)}{\left( \cos 3\theta +\cos 9\theta \right)+\left( \cos 5\theta +\cos 7\theta \right)}...\left( i \right)$
We know about the identity for the addition of sine, which says
$\sin C+\sin D=\dfrac{1}{2}\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
and for the addition of cosines, we have
$\cos C+\cos D=\dfrac{1}{2}\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
We should use these two identities in equation (i) to get,
x=\dfrac{\left\\{ \dfrac{1}{2}\sin \left( \dfrac{3\theta +9\theta }{2} \right)\cos \left( \dfrac{3\theta -9\theta }{2} \right) \right\\}+\left\\{ \dfrac{1}{2}\sin \left( \dfrac{5\theta +7\theta }{2} \right)\cos \left( \dfrac{5\theta -7\theta }{2} \right) \right\\}}{\left\\{ \dfrac{1}{2}\cos \left( \dfrac{3\theta +9\theta }{2} \right)\cos \left( \dfrac{3\theta -9\theta }{2} \right) \right\\}+\left\\{ \dfrac{1}{2}\cos \left( \dfrac{5\theta +7\theta }{2} \right)\cos \left( \dfrac{5\theta -7\theta }{2} \right) \right\\}}
Cancelling $\dfrac{1}{2}$ from each term and evaluating the equation further, we get the following simplified version,
$x=\dfrac{\sin 6\theta \cos \left( -3\theta \right)+\sin 6\theta \cos \left( -\theta \right)}{\cos 6\theta \cos \left( -3\theta \right)+\cos 6\theta \cos \left( -\theta \right)}$
We all know that for any angle,$\cos \left( -\Phi \right)=\cos \Phi$ .
We can use the above point to further simplify our equation,
$x=\dfrac{\sin 6\theta \cos 3\theta +\sin 6\theta \cos \theta }{\cos 6\theta \cos 3\theta +\cos 6\theta \cos \theta }$
Now, taking $\sin 6\theta$ as common from the numerator, and $\cos 6\theta$ as common from the denominator, we get the following
$x=\dfrac{\sin 6\theta \left( \cos 3\theta +\cos \theta \right)}{\cos 6\theta \left( \cos 3\theta +\cos \theta \right)}$
We can now cancel the term $\left( \cos 3\theta +\cos \theta \right)$ from numerator and denominator to get
$x=\dfrac{\sin 6\theta }{\cos 6\theta }$
We know that $\dfrac{\sin \Phi }{\cos \Phi }=\tan \Phi$ .
Using the above identity, we can write
$x=\tan 6\theta$

So, the correct answer is “Option c”.

Note: Some students, in eagerness, may consider the variable $\theta$ (theta) as 0 (zero). We should clearly note the difference between the two. We can solve this problem directly without rearranging the terms in the first step. But then we have to apply the formula for addition of cosines one more time, which will be a lengthy process.