Question

prove that -
$\sin(3x) + \sin(2x) - \sin(x) = 4 \sin(x ) \cos( \frac{x}{2} ) \cos( \frac{3x}{2} )$

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Answer 1

Let x = 2y. Then we want to show

sin(6y) + sin(4y) - sin(2y) = 4 sin(2y) cos(y) cos(3y)

Recall the angle sum identities,

sin(x ± y) = sin(x) cos(y) ± cos(x) sin(y)

cos(x ± y) = cos(x) cos(y) ∓ sin(x) sin(y)

which lets us write

sin(6y) = sin(4y + 2y) = sin(4y) cos(2y) + cos(4y) sin(2y)

sin(4y) = sin(2y + 2y) = 2 sin(2y) cos(2y)

cos(4y) = cos(2y + 2y) = cos²(2y) - sin²(2y)

Ultimately, we use these identities to rewrite the left side as

sin(6y) + sin(4y) - sin(2y)

= 2 (sin(4y) cos(2y) + cos(4y) sin(2y)) + 2 sin(2y) cos(2y) - sin(2y)

= 2 sin(2y) cos²(2y) + (cos²(2y) - sin²(2y)) sin(2y) + 2 sin(2y) cos(2y) - sin(2y)

Notice the underlined common factor of sin(2y). If we remove this from both sides of the identity we want to prove, then it remains to show that

2 cos²(2y) + (cos²(2y) - sin²(2y)) + 2 cos(2y) - 1 = 4 cos(y) cos(3y)

or

3 cos²(2y) - sin²(2y) + 2 cos(2y) - 1 = 4 cos(y) cos(3y)

Recall the Pythagorean identity,

cos²(x) + sin²(x) = 1

which lets us write

3 cos²(2y) - sin²(2y) + 2 cos(2y) - 1

= 3 cos²(2y) - (1 - cos²(2y)) + 2 cos(2y) - 1

= 4 cos²(2y) + 2 cos(2y) - 2

= 2 (2 cos²(2y) + cos(2y) - 1)

= 2 (2 cos(2y) - 1) (cos(2y) + 1)

Recall the half-angle identity for cosine,

cos²(x/2) = 1/2 (1 + cos(x))

which means

cos(2y) + 1 = 2 • 1/2 (1 + cos(2y)) = 2 cos²(y)

and so

2 (2 cos(2y) - 1) (cos(2y) + 1)

= 4 (2 cos(2y) - 1) cos²(y)

= 4 cos(y) (2 cos(2y) - 1) cos(y)

Now notice the underlined factor of 4 cos(y), which also appears in the right side of the identity we want to prove. Eliminate this term and all that's left is to show that

(2 cos(2y) - 1) cos(y) = cos(3y)

which follows from a combination of the identities I mentioned above:

(2 cos(2y) - 1) cos(y)

= 2 cos(2y) cos(y) - cos(y)

= 2 • 1/2 (cos(2y + y) + cos(2y - y)) - cos(y)

= (cos(3y) + cos(y)) - cos(y)

= cos(3y)

as required.

Answer 2

Trigonometric identities involve equations that use the trigonometric functions that are true for all variables in the equation

The identity is given as:

$\sin(3x) + \sin(2x) - \sin(x) = 4\sin(x)\cos(\frac x2)\cos(\frac{3x}{2})$

Let x = 2y.

So, we have:

$\sin(6y) + \sin(4y) - \sin(2y) = 4\sin(2y)\cos(y)\cos(3y)$

Expand

$\sin(4y + 2y) + \sin(2y + 2y) - \sin(2y) = 4\sin(2y)\cos(y)\cos(3y)$

Expand the identities using the angle sum identities,

$\sin(4y)\cos(2y) + \cos(4y)\sin(2y) + 2\sin(2y)\cos(2y) - \sin(2y) = 4\sin(2y)\cos(y)\cos(3y)$

Expand cos(4y) and sin(4y)

$2\sin(2y)\cos^2(2y) + [\cos^2(2y) - \sin^2(2y)]\sin(2y) + 2\sin(2y)\cos(2y) - \sin(2y) = 4\sin(2y)\cos(y)\cos(3y)$

Factor out sin(2y)

$\sin(2y)[2\cos^2(2y) + \cos^2(2y) - \sin^2(2y) + 2\cos(2y) - 1] = 4\sin(2y)\cos(y)\cos(3y)$

Evaluate the like terms

$\sin(2y)[3\cos^2(2y) - \sin^2(2y) + 2\cos(2y) - 1] = 4\sin(2y)\cos(y)\cos(3y)$

Substitute

$\sin^2(2y) = 1 - \cos^2(2y)$

$\sin(2y)[3\cos^2(2y) - 1 + \cos^2(2y) + 2\cos(2y) - 1] = 4\sin(2y)\cos(y)\cos(3y)$

Evaluate the like terms

$\sin(2y)[4\cos^2(2y) - 1 + 2\cos(2y) - 1] = 4\sin(2y)\cos(y)\cos(3y)$

$\sin(2y)[4\cos^2(2y) + 2\cos(2y) - 2] = 4\sin(2y)\cos(y)\cos(3y)$

Factor out 2

$2\sin(2y)[2\cos^2(2y) + \cos(2y) - 1] = 4\sin(2y)\cos(y)\cos(3y)$

Expand

$2\sin(2y)[2\cos^2(2y) +2\cos(2y) - \cos(2y) - 1] = 4\sin(2y)\cos(y)\cos(3y)$

Factorize

$2\sin(2y)[2\cos(2y)( \cos(2y) + 1) -1( \cos(2y) + 1)] = 4\sin(2y)\cos(y)\cos(3y)$

Factor out cos(2y) + 1

$2\sin(2y)[( 2\cos(2y) - 1)( \cos(2y) + 1)] = 4\sin(2y)\cos(y)\cos(3y)$

By half identity, we have:

$\cos\²(\frac x2) = \frac 12 (1 + \cos(x))$

Multiply both sides by 2

$2\cos\²(\frac x2) = (1 + \cos(x))$

Replace x with 2y

$2\cos\²(y) = 1 + \cos(2y)$

So, we have:

$2\sin(2y)[( 2\cos(2y) - 1)( 2\cos^2(y))] = 4\sin(2y)\cos(y)\cos(3y)$

$4\sin(2y)(2 \cos(2y) - 1)(\cos^2(y)) = 4\sin(2y)\cos(y)\cos(3y)$

Factor out cos(y)

$4\sin(2y)(\cos(y)(2 \cos(2y) - 1)(\cos(y)) = 4\sin(2y)\cos(y)\cos(3y)$

Expand

$4\sin(2y)(\cos(y)(2 \cos(2y)\cos(y) - \cos(y)) = 4\sin(2y)\cos(y)\cos(3y)$

Using the half identity, we have:

$4\sin(2y)(\cos(y)(2 * \frac 12 *\cos(2y + y) + \cos(2y - y) - \cos(y)) = 4\sin(2y)\cos(y)\cos(3y)$

$4\sin(2y)(\cos(y)(\cos(3y) + \cos(y) - \cos(y)) = 4\sin(2y)\cos(y)\cos(3y)$

$4\sin(2y)\cos(y)\cos(3y) = 4\sin(2y)\cos(y)\cos(3y)$

Recall that:

x = 2y

So, we have:

$4\sin(x)\cos(\frac x2)\cos(\frac {3x}2) = 4\sin(x)\cos(\frac x2)\cos(\frac {3x}2)$

Both sides of the equations are the same.

Hence, the identity $\sin(3x) + \sin(2x) - \sin(x) = 4\sin(x)\cos(\frac x2)\cos(\frac{3x}{2})$ has been proved

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