Question

Prove that: \[\dfrac{{\sin x + \sin 2x + \sin 3x}}{{\cos x + \cos 2x + \cos 3x}} = \tan 2x\]

Answer 1

Hint: In order to prove this, we will use the identities \[\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\] and \[\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\] by combining the first and the last term in the numerator and denominator respectively. After using these identities, we will be getting simplified terms in the numerator and denominator. We will then cancel out some terms from the numerator and denominator and solve further to prove \[LHS = RHS\].

Complete step by step answer:
We need to prove \[\dfrac{{\sin x + \sin 2x + \sin 3x}}{{\cos x + \cos 2x + \cos 3x}} = \tan 2x\]
Let us consider \[LHS\], which is equal to \[\dfrac{{\sin x + \sin 2x + \sin 3x}}{{\cos x + \cos 2x + \cos 3x}}\]. As we know, addition is commutative, we can re-shuffle the second and third term in the numerator as well as the denominator. On doing this, we get
\[LHS \Rightarrow \dfrac{{\sin x + \sin 3x + \sin 2x}}{{\cos x + \cos 3x + \cos 2x}}\]
Now, using the identities \[\left( {\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)} \right)\] on first two terms of the numerator and \[\left( {\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)} \right)\] on first two terms of the denominator, we get
\[LHS \Rightarrow \dfrac{{2\sin \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right) + \sin 2x}}{{2\cos \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right) + \cos 2x}}\]

Now, solving the brackets, we get
\[LHS \Rightarrow \dfrac{{2\sin \left( {\dfrac{{4x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right) + \sin 2x}}{{2\cos \left( {\dfrac{{4x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right) + \cos 2x}}\]
On simplifying, we get
\[LHS \Rightarrow \dfrac{{2\sin \left( {2x} \right)\cos \left( { - x} \right) + \sin 2x}}{{2\cos \left( {2x} \right)\cos \left( { - x} \right) + \cos 2x}}\]
Now, as we know \[\cos \left( { - \theta } \right) = \cos \left( \theta \right)\]. So, using this, we get
\[LHS \Rightarrow \dfrac{{2\sin \left( {2x} \right)\cos \left( x \right) + \sin 2x}}{{2\cos \left( {2x} \right)\cos \left( x \right) + \cos 2x}}\]

Now, taking out \[\sin 2x\] common from the numerator and \[\cos 2x\] common from the denominator, we get
\[LHS \Rightarrow \dfrac{{\sin \left( {2x} \right)\left( {2\cos \left( x \right) + 1} \right)}}{{\cos \left( {2x} \right)\left( {2\cos \left( x \right) + 1} \right)}}\]
Now, cancelling out \[\left( {2\cos \left( x \right) + 1} \right)\] from the numerator and denominator, we get
\[LHS \Rightarrow \dfrac{{\sin \left( {2x} \right)}}{{\cos \left( {2x} \right)}}\]
Now, using \[\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \], we get
\[LHS \Rightarrow \tan 2x\]
Also, we have \[RHS \Rightarrow \tan 2x\].
Hence, we proved \[LHS = RHS = \tan 2x\]

Note: The most important thing we need to remember while solving such types of questions is to analyse which formula is to be applied in order to reach the required form as we can solve trigonometric questions in many ways. Also, we need to be very clear with the formulas as there are just minor differences in the formulae and one wrong term in the formula used will lead to the wrong answer.