Question

Prove the following:
$\dfrac{{\sin x + \sin 3x}}{{\cos x + \cos 3x}} = \tan 2x$

Answer 1

We can take the LHS of the given equation. Then we can simplify its numerator using the trigonometric identities $\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ . We can simplify the denominator using the identity $\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ . On doing further calculations, we will obtain the RHS of the equation. We can say the given equation is true when LHS=RHS

Complete step-by-step answer:
We need to prove that $\dfrac{{\sin x + \sin 3x}}{{\cos x + \cos 3x}} = \tan 2x$
Let us look at the LHS,
$LHS = \dfrac{{\sin x + \sin 3x}}{{\cos x + \cos 3x}}$ … (1)
We can consider the numerator of the LHS
We know that $\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
We can substitute the values,
$\Rightarrow \sin \left( x \right) + \sin \left( {3x} \right) = 2\sin \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right)$
On simplification, we get,
$\Rightarrow \sin \left( x \right) + \sin \left( {3x} \right) = 2\sin \left( {2x} \right)\cos \left( { - x} \right)$
We know that $\cos \left( { - x} \right) = \cos \left( x \right)$ . So, we get,
$\Rightarrow \sin \left( x \right) + \sin \left( {3x} \right) = 2\sin \left( {2x} \right)\cos \left( x \right)$ … (2)
We can consider the denominator of the LHS
We know that $\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
We can substitute the values,
$\Rightarrow \cos \left( x \right) + \cos \left( {3x} \right) = 2\cos \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right)$
On simplification, we get,
$\Rightarrow \cos \left( x \right) + \cos \left( {3x} \right) = 2\cos \left( {2x} \right)\cos \left( { - x} \right)$
We know that $\cos \left( { - x} \right) = \cos \left( x \right)$ . So, we get,
$\Rightarrow \cos \left( x \right) + \cos \left( {3x} \right) = 2\cos \left( {2x} \right)\cos \left( x \right)$ … (3)
We can substitute (3) and (2) in (1)
$\Rightarrow LHS = \dfrac{{2\sin 2x\cos x}}{{2\cos 2x\cos x}}$
On further simplification, we get,
$\Rightarrow LHS = \dfrac{{\sin 2x}}{{\cos 2x}}$
We know that $\tan A = \dfrac{{\sin A}}{{\cos A}}$ . So, we get,
$\Rightarrow LHS = \tan 2x$
RHS is also equal to $\tan 2x$ . So, we can write,
LHS=RHS.
Hence the equation is proved.

Note: We must be familiar with the following trigonometric identities used in this problem.
$\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\cos \left( A \right) - \cos \left( B \right) = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\sin \left( { - x} \right) = - \sin \left( x \right)$
$\cos \left( { - x} \right) = \cos \left( x \right)$
We must know the values of trigonometric functions at common angles. Adding $\pi$ or multiples of $\pi$ with the angle retains the ratio and adding $\dfrac{\pi }{2}$ or odd multiples of $\dfrac{\pi }{2}$ will change the ratio. While converting the angles we must take care of the sign of the ratio in its respective quadrant. In the 1st quadrant all the trigonometric ratios are positive. In the 2nd quadrant only sine and sec are positive. In the third quadrant, only tan and cot are positive and in the fourth quadrant, only cos and sec are positive. The angle measured in the counterclockwise direction is taken as positive and angle measured in the clockwise direction is taken as negative.