Question

Prove the following:
$\dfrac{\left( \sin 7x+\sin 5x \right)+\left( \sin 9x+\sin 3x \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)}=\tan 6x$

Answer 1

Hint: In this question, consider the LHS and the formula for $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ and $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$. After that cancel the common terms to prove the desired result, that is to get LHS = RHS.
Complete step-by-step answer:
Here, we have to prove that
$\dfrac{\left( \sin 7x+\sin 5x \right)+\left( \sin 9x+\sin 3x \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)}=\tan 6x$
Let us consider the LHS of the expression given in the question
$E=\dfrac{\left( \sin 7x+\sin 5x \right)+\left( \sin 9x+\sin 3x \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)}$
We know that, $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$. By using this in the above expression, we get,
$E=\dfrac{2\sin \left( \dfrac{7x+5x}{2} \right).\cos \left( \dfrac{7x-5x}{2} \right)+2\sin \left( \dfrac{9x+3x}{2} \right)\cos \left( \dfrac{9x-3x}{2} \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)}$
By simplifying the above expression, we get,
$E=\dfrac{2\sin \left( 6x \right).\cos \left( x \right)+2\sin \left( 6x \right)\cos \left( 3x \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)}$
We know that $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$. By using this in the above expression, we get,
$E=\dfrac{2\sin \left( 6x \right).\cos \left( x \right)+2\sin \left( 6x \right)\cos \left( 3x \right)}{2\cos \left( \dfrac{7x+5x}{2} \right).\cos \left( \dfrac{7x-5x}{2} \right)+2\cos \left( \dfrac{9x+3x}{2} \right).\cos \left( \dfrac{9x-3x}{2} \right)}$
By simplifying the above expression, we get,
$E=\dfrac{2\sin \left( 6x \right).\cos \left( x \right)+2\sin \left( 6x \right)\cos \left( 3x \right)}{2\cos \left( 6x \right).\cos \left( x \right)+2\cos \left( 6x \right).\cos \left( 3x \right)}$
By taking out 2 sin (6x) common from the numerator of the above expression, we get,
$E=\dfrac{\left( 2\sin 6x \right)\left[ \cos x+\cos 3x \right]}{2\cos \left( 6x \right)\cos x+2\cos \left( 6x \right)\cos \left( 3x \right)}$
By taking out 2 cos (6x) common from the denominator of the above expression, we get,
$E=\dfrac{\left( 2\sin 6x \right)\left[ \cos x+\cos 3x \right]}{\left( 2\cos 6x \right)\left[ \cos x+\cos 3x \right]}$
Now, by canceling the like terms of the above expression, we get,
$E=\dfrac{\sin 6x}{\cos 6x}$
We know that, $\dfrac{\sin \theta }{\cos \theta }=\tan \theta$. By using this in the above expression, we get,
$E=\tan 6x$
So, we get, LHS = RHS
Hence proved.
Therefore, we have proved that
$\dfrac{\left( \sin 7x+\sin 5x \right)+\left( \sin 9x+\sin 3x \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)}=\tan 6x$

Note: In these types of questions, students often get confused between the formulas of sin C + sin D or sin C – sin D or cos C + cos D, etc. So, formulas for these expressions must be memorized properly. Also, students must try to club the terms such that we get, $\dfrac{C+D}{2}$ and $\dfrac{C-D}{2}$ as a whole number and not fractional values. In other words, try to take C and D such that (C + D) and (C – D) are even multiples.