# Prove the following: \[\sin 2x + 2\sin 4x + \sin 6x = 4{\text{co}}{{\text{s}}^2}x\sin 4x\].

Answer

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Hint: Use 2x = x + x to express 2sin(4x) as sin(4x) + sin(4x) and then simplify the expression using the sum of two sine terms formula. Then you can take the common terms and again apply the sum of two sine terms formula to obtain the result.

Complete step-by-step answer:

We start solving by assigning the left-hand side of the equation to LHS.

\[LHS = \sin 2x + 2\sin 4x + \sin 6x...........(1)\]

We know that 2x can be written as x + x. Using this trick in equation (1), we express 2sin(4x) as sin(4x) + sin(4x).

\[LHS = \sin 2x + \sin 4x + \sin 4x + \sin 6x.........(2)\]

Group first two terms and the next two terms in equation (2).

\[LHS = (\sin 2x + \sin 4x) + (\sin 4x + \sin 6x)........(3)\]

We know the formula for adding two sine terms, which is given as follows:

\[\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)...........(4)\]

Using formula in equation (4) in equation (3) for the terms in the bracket, we get:

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

Simplifying the above equation, we get:

\[LHS = 2\sin \left( {\dfrac{{6x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right) + 2\sin \left( {\dfrac{{10x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right)\]

Simplifying further, we get:

\[LHS = 2\sin \left( {3x} \right)\cos \left( { - x} \right) + 2\sin \left( {5x} \right)\cos \left( { - x} \right)\]

We know that cos(-x) = cos(x), then we get:

\[LHS = 2\sin 3x\cos x + 2\sin 5x\cos x.........(5)\]

We can take 2cos(x) as a common term in equation (5) and simplify further.

\[LHS = 2\cos x\left( {\sin 3x + \sin 5x} \right)\]

Again, applying equation (4) in this equation, we get:

\[LHS = 2\cos x\left( {2\sin \left( {\dfrac{{3x + 5x}}{2}} \right)\cos \left( {\dfrac{{3x - 5x}}{2}} \right)} \right)\]

Simplifying further, we get:

\[LHS = 2\cos x\left( {2\sin \left( {\dfrac{{8x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right)} \right)\]

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

Multiplying terms outside with terms inside the bracket, we get:

\[LHS = 4\cos x\sin 4x\cos \left( { - x} \right)\]

We know that cos(-x) = cos(x).

\[LHS = 4\cos x\sin 4x\cos x\]

Multiplying the cos(x), we get:

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

\[LHS = RHS\]

Hence, the left-hand side of the equation is equal to the right-hand side which completes the proof.

Note: You can also combine sin(2x) and sin(6x) and apply the sum of sine terms formula to get 2sin(4x) cos(2x). Then you can take 2sin(4x) as a common term and use the cos(2x) formula in terms of cos(x) to get the final expression.

Complete step-by-step answer:

We start solving by assigning the left-hand side of the equation to LHS.

\[LHS = \sin 2x + 2\sin 4x + \sin 6x...........(1)\]

We know that 2x can be written as x + x. Using this trick in equation (1), we express 2sin(4x) as sin(4x) + sin(4x).

\[LHS = \sin 2x + \sin 4x + \sin 4x + \sin 6x.........(2)\]

Group first two terms and the next two terms in equation (2).

\[LHS = (\sin 2x + \sin 4x) + (\sin 4x + \sin 6x)........(3)\]

We know the formula for adding two sine terms, which is given as follows:

\[\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)...........(4)\]

Using formula in equation (4) in equation (3) for the terms in the bracket, we get:

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

Simplifying the above equation, we get:

\[LHS = 2\sin \left( {\dfrac{{6x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right) + 2\sin \left( {\dfrac{{10x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right)\]

Simplifying further, we get:

\[LHS = 2\sin \left( {3x} \right)\cos \left( { - x} \right) + 2\sin \left( {5x} \right)\cos \left( { - x} \right)\]

We know that cos(-x) = cos(x), then we get:

\[LHS = 2\sin 3x\cos x + 2\sin 5x\cos x.........(5)\]

We can take 2cos(x) as a common term in equation (5) and simplify further.

\[LHS = 2\cos x\left( {\sin 3x + \sin 5x} \right)\]

Again, applying equation (4) in this equation, we get:

\[LHS = 2\cos x\left( {2\sin \left( {\dfrac{{3x + 5x}}{2}} \right)\cos \left( {\dfrac{{3x - 5x}}{2}} \right)} \right)\]

Simplifying further, we get:

\[LHS = 2\cos x\left( {2\sin \left( {\dfrac{{8x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right)} \right)\]

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

Multiplying terms outside with terms inside the bracket, we get:

\[LHS = 4\cos x\sin 4x\cos \left( { - x} \right)\]

We know that cos(-x) = cos(x).

\[LHS = 4\cos x\sin 4x\cos x\]

Multiplying the cos(x), we get:

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

\[LHS = RHS\]

Hence, the left-hand side of the equation is equal to the right-hand side which completes the proof.

Note: You can also combine sin(2x) and sin(6x) and apply the sum of sine terms formula to get 2sin(4x) cos(2x). Then you can take 2sin(4x) as a common term and use the cos(2x) formula in terms of cos(x) to get the final expression.

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