Question

Prove the following:
\[\sin 2x + 2\sin 4x + \sin 6x = 4{\cos ^2}x\sin 4x\]

Answer 1

Hint: We can take the LHS of the given equation and add the 1st and last term. Then we can simplify it 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 take the common terms outside and then use the identity $\cos 2x = {\cos ^2}x - 1$. 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 \[\sin 2x + 2\sin 4x + \sin 6x = 4{\cos ^2}x\sin 4x\]

Let us look at the LHS,

$LHS = \sin 2x + 2\sin 4x + \sin 6x$

We can add the 1st and 3rd terms of the LHS.

$ \Rightarrow LHS = 2\sin 4x + \left( {\sin 2x + \sin 6x} \right)$

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,

 \[ \Rightarrow \sin \left( {2x} \right) + \sin \left( {6x} \right) = 2\sin \left( {\dfrac{{2x + 6x}}{2}} \right)\cos \left( {\dfrac{{2x - 6x}}{2}} \right)\]

On simplification, we get,

\[ \Rightarrow \sin \left( {2x} \right) + \sin \left( {6x} \right) = 2\sin \left( {4x} \right)\cos \left( { - 2x} \right)\]

We know that \[\cos \left( { - x} \right) = \cos \left( x \right)\]

\[ \Rightarrow \sin \left( {2x} \right) + \sin \left( {6x} \right) = 2\sin 4x\cos 2x\]

We can substitute it in the LHS.

\[ \Rightarrow LHS = 2\sin 4x + 2\sin 4x\cos 2x\]

We can take the common term \[2\sin 4x\].

 \[ \Rightarrow LHS = 2\sin 4x\left( {1 + \cos 2x} \right)\]

We know that $\cos 2x = 2{\cos ^2}x - 1$

\[ \Rightarrow LHS = 2\sin 4x\left( {1 + 2{{\cos }^2}x - 1} \right)\]

On simplification, we get,

\[ \Rightarrow LHS = 4{\cos ^2}x\sin 4x\]

RHS is also equal to \[2{\cos ^2}x\sin 4x\]. So, we can write,

$LHS = RHS$.

Hence the equation is proved.

Note: We must be familiar to 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 following figure gives us an idea about the signs of different trigonometric functions. The angle measured in the counter clockwise direction is taken as positive and angle measured in the clockwise direction is taken as negative.