Question

Prove that \[\dfrac{\left( \sin 7x+\sin 5x \right)+\left( \sin 9x+\sin 3x \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)}=\tan 6x\]

Answer 1

Hint: First take the LHS. Solve the numerator and denominator separately by using basic trigonometric formula that is \[\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)\], \[\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)\]. Then put these values back in the LHS and simplify it to get \[\tan 6x\] in LHS. Thus you need to prove that LHS = RHS.

Complete step-by-step answer:

We have been given a trigonometric expression, where we need to prove that the LHS is equal to\[\tan 6x\].
\[\dfrac{\left( \sin 7x+\sin 5x \right)+\left( \sin 9x+\sin 3x \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)}=\tan 6x\]
Now let us consider the LHS of the given expression.
\[\dfrac{\left( \sin 7x+\sin 5x \right)+\left( \sin 9x+\sin 3x \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)}=LHS\]\[\to (1)\]
Now let us consider the numerator of the LHS.
\[\left( \sin 7x+\sin 5x \right)+\left( \sin 9x+\sin 3x \right)\]\[\to (2)\]
We know the basic trigonometric formula,
\[\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)\]
Let us apply this formula in (1)
\[2\sin \left( \dfrac{7x+5x}{2} \right)\cos \left( \dfrac{7x-5x}{2} \right)+2\sin \left( \dfrac{9x+3x}{2} \right)\cos \left( \dfrac{9x-3x}{2} \right)\]
Let us simplify the above expression,
\[\begin{align}
& =2\sin \left( \dfrac{12x}{2} \right)\cos \left( \dfrac{2x}{2} \right)+2\sin \left( \dfrac{12x}{2} \right)\cos \left( \dfrac{6x}{2} \right) \\
& \\
& =2\sin 6x\cos x+2\sin 6x\cos 3x \\
& \\
& =2\sin 6x\left( \cos x+\cos 3x \right)\text{ }\to (3) \\
\end{align}\]
Now, let us solve the denominator.
\[\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)\]\[\to (4)\]
We know the trigonometric formula,
\[\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)\]
Let’s apply this formula in equation (3)
\[2\cos \left( \dfrac{7x+5x}{2} \right)\cos \left( \dfrac{7x-5x}{2} \right)+2\cos \left( \dfrac{9x+3x}{2} \right)\cos \left( \dfrac{9x-3x}{2} \right)\]
Let us simplify the above expression,
\[\begin{align}
& =2\cos \left( \dfrac{12x}{2} \right)\cos \left( \dfrac{2x}{2} \right)+2\cos \left( \dfrac{12x}{2} \right)\cos \left( \dfrac{6x}{2} \right) \\
& \\
& =2\cos 6x\cos x+2\cos 6x\cos 3x \\
& \\
& =2\cos 6x\left( \cos x+\cos 3x \right)\text{ }\to (5) \\
\end{align}\]
Thus, we have found the numerator and denominator of the LHS in equation (1)
\[LHS=\dfrac{\left( \sin 7x+\sin 5x \right)+\left( \sin 9x+\sin 3x \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)}\]
Then substitute the value in equation (3) and (5)
\[LHS=\dfrac{2\sin 6x\left( \cos x+\cos 3x \right)}{2\cos 6x\left( \cos x+\cos 3x \right)}\]
Cancel out \[2\left( \cos x+\cos 3x \right)\]from the numerator and denominator.
Thus we get,
\[LHS=\dfrac{\sin 6x}{\cos 6x}=\tan 6x\].
We know that\[\dfrac{\sin x}{\cos x}=\tan x\].
Thus \[LHS=RHS=\tan 6x\]
Hence we 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: We can solve this in a few steps without taking the numerator and denominator separately. Just apply the basic trigonometric formula and simplify it. We have separately solved the numerator and denominator to get the clarity in the solution.