Question

How can I demonstrate: $\dfrac{(\sin x+\sin 3x+\sin 5x)}{(\cos x+\cos 3x+\cos 5x)}=\tan 3x$?

Answer 1

Hint: We will first take the LHS of the given equation. We will prove the given trigonometric equation using the trigonometric formulae. We will first consider \[\sin x\] and \[\sin 5x\]in the numerator, and use the formula \[\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\]. Then, in the denominator, we will consider \[\cos x\] and \[\cos 5x\], then we will use the formula\[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\]. On solving further, we get the RHS. Hence, proved.

Complete step by step solution:
According to the given question, we have to prove the given equation. For this, we will make use of the various trigonometric formulae we learnt.
We will take the LHS in the given equation and make it equal to RHS.
The given equation is:
$\dfrac{(\sin x+\sin 3x+\sin 5x)}{(\cos x+\cos 3x+\cos 5x)}=\tan 3x$-----(1)
Taking LHS, we have,
\[\dfrac{(\sin x+\sin 3x+\sin 5x)}{(\cos x+\cos 3x+\cos 5x)}\]
We can see in RHS that the angle that tangent function has, is \[3x\].
So, in the numerator of the LHS, we will first consider the terms \[\sin x\] and \[\sin 5x\]. We will use the formula, that is, \[\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\], we get,
\[\Rightarrow \dfrac{(\sin 5x+\sin x+\sin 3x)}{(\cos x+\cos 3x+\cos 5x)}\]
Applying the formula, we get,
\[\Rightarrow \dfrac{(2\sin \left( \dfrac{5x+x}{2} \right)\cos \left( \dfrac{5x-x}{2} \right)+\sin 3x)}{(\cos x+\cos 3x+\cos 5x)}\]
\[\Rightarrow \dfrac{(2\sin \left( \dfrac{6x}{2} \right)\cos \left( \dfrac{4x}{2} \right)+\sin 3x)}{(\cos x+\cos 3x+\cos 5x)}\]
\[\Rightarrow \dfrac{(2\sin \left( 3x \right)\cos \left( 2x \right)+\sin 3x)}{(\cos x+\cos 3x+\cos 5x)}\]
Now, in the denominator, we will consider the terms \[\cos x\] and \[\cos 5x\], and using the formula \[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\], we get,
\[\Rightarrow \dfrac{2\sin 3x\cos 2x+\sin 3x}{(\cos 5x+\cos x)+\cos 3x}\]
Applying the formula, we get,
\[\Rightarrow \dfrac{2\sin 3x\cos 2x+\sin 3x}{2\cos \left( \dfrac{5x+x}{2} \right)\cos \left( \dfrac{5x-x}{2} \right)+\cos 3x}\]
\[\Rightarrow \dfrac{2\sin 3x\cos 2x+\sin 3x}{2\cos \left( \dfrac{6x}{2} \right)\cos \left( \dfrac{4x}{2} \right)+\cos 3x}\]
Solving further, we get,
\[\Rightarrow \dfrac{2\sin 3x\cos 2x+\sin 3x}{2\cos 3x\cos 2x+\cos 3x}\]
We can see common terms both in the numerator as well as in the denominator, so taking the common terms out, we get,
\[\Rightarrow \dfrac{\sin 3x(2\cos 2x+1)}{\cos 3x(2\cos 2x+1)}\]
Cancelling the common terms, we get the expression as,
\[\Rightarrow \dfrac{\sin 3x}{\cos 3x}\]
We know that the ratio of sine and cosine function gives us tangent function. So, we get,
\[\Rightarrow \tan 3x\]
Since, LHS is equal to RHS,
Therefore, we have proved the given $\dfrac{(\sin x+\sin 3x+\sin 5x)}{(\cos x+\cos 3x+\cos 5x)}=\tan 3x$.

Note: The formula used in solving the question should be used keeping in mind the pre-requisites as well as the outcome it would give. We deliberately chose sine and cosine functions with angles \[x\] and \[5x\], so that when the appropriate formula is applied we will function with angle \[3x\]. \[3x\] is required in the LHS expression because we can see \[3x\] in the RHS.