Convert a product of trigonometric functions into a sum (or difference) Algebra II

If a and b are two real numbers, then:

\sin\left(a\right)\cos\left(b\right) = \dfrac{1}{2}\sin\left(a+b\right) + \dfrac{1}{2}\sin\left(a - b\right)

Here:

• a = \dfrac{3}{4}x
• b = \dfrac{5\pi}{8}x

Therefore:

\sin\left(\dfrac{3}{4}x\right)\cos\left(\dfrac{5\pi}{8}x\right) = \dfrac{1}{2}\sin\left(\dfrac{3}{4}x+\dfrac{5\pi}{8}x\right) + \dfrac{1}{2}\sin\left(\dfrac{3}{4}x - \dfrac{5\pi}{8}x\right)

\sin\left(\dfrac{3}{4}x\right)\cos\left(\dfrac{5\pi}{8}x\right) = \dfrac{1}{2}\sin\left(\dfrac{6}{8}x+\dfrac{5\pi}{8}x\right) + \dfrac{1}{2}\sin\left(\dfrac{6}{8}x - \dfrac{5\pi}{8}x\right)

Multiplying and dividing the expression by 2 and rearranging gives:

\cos\left(2x\right)\sin\left(4x\right)\cos\left(4x\right)=\dfrac{1}{2}\times\cos\left(2x\right)\left[2\times \sin\left(4x\right)\cos\left(4x\right)\right]

Since 2\sin\left(\theta\right)\cos\left(\theta\right)=\sin\left(2\theta\right) :

\dfrac{1}{2}\times\cos\left(2x\right)\left[2\times \sin\left(4x\right)\cos\left(4x\right)\right]=\dfrac{1}{2}\cos\left(2x\right)\sin\left(2\left(4x\right)\right)=\dfrac{1}{2}\cos\left(2x\right)\sin\left(8x\right)

Applying the product-to-sum identity for trigonometric functions:

\sin\left(a\right)\cos\left(B\right)=\dfrac{1}{2}\left[\sin\left(a+b\right)+\sin\left(a-b\right)\right]

Here:

• a=8x
• b=2x

Therefore:

\dfrac{1}{2}\cos\left(2x\right)\sin\left(8x\right)=\dfrac{1}{4}\left[\sin\left(8x+2x\right)+\sin\left(8x-2x\right)\right]
\dfrac{1}{2}\cos\left(2x\right)\sin\left(8x\right)=\dfrac{1}{4}\left[\sin\left(10x\right)+\sin\left(6x\right)\right]
\dfrac{1}{2}\cos\left(2x\right)\sin\left(8x\right)= \dfrac{1}{4}\sin\left(10x\right)+\dfrac{1}{4}\sin\left(6x\right)

For any two real numbers a and b, we have:

\sin\left(a\right)\sin\left(b\right) = \dfrac{1}{2}\cos\left(a-b\right) - \dfrac{1}{2}\cos\left(a + b\right)

Here:

• a=2x
• b=4x

Therefore:

\sin\left(2x\right)\sin\left(4x\right)

=\dfrac{1}{2}\cos\left(2x-4x\right) - \dfrac{1}{2}\cos\left(2x + 4x\right)
=\dfrac{1}{2}\cos\left(-2x\right) - \dfrac{1}{2}\cos\left(6x\right)

Applying the identity \cos\left(-x\right)=\cos\left(x\right) :

\dfrac{1}{2}\cos\left(2x\right) - \dfrac{1}{2}\cos\left(6x\right)

For any two real numbers a and b, we have:

\sin\left(a\right)\sin\left(b\right) = \dfrac{1}{2}\cos\left(a-b\right) - \dfrac{1}{2}\cos\left(a + b\right)

Here:

• a=3x
• b=x

Therefore:

-\sin\left(3x\right)\sin\left(x\right)
=-\left[\dfrac{1}{2}\cos\left(3x-x\right) - \dfrac{1}{2}\cos\left(3x + x\right)\right]

=-\left[\dfrac{1}{2}\cos\left(2x\right) - \dfrac{1}{2}\cos\left(4x\right)\right]
=\dfrac{1}{2}\cos\left(4x\right) - \dfrac{1}{2}\cos\left(2x\right)

For any two real numbers a and b, we have:

\cos\left(a\right)\cos\left(b\right) = \dfrac{1}{2}\cos\left(a+b\right)+ \dfrac{1}{2}\cos\left(a - b\right):

Here:

• a=\dfrac{x}{2}
• b=\dfrac{3x}{2}

Therefore:

\cos\left(\dfrac{x}{2}\right)\cos\left(\dfrac{3x}{2}\right)

=\dfrac{1}{2}\cos\left(\dfrac{x}{2}+\dfrac{3x}{2}\right) + \dfrac{1}{2}\cos\left(\dfrac{x}{2}-\dfrac{3x}{2}\right)
=\dfrac{1}{2}\cos\left(2x\right) + \dfrac{1}{2}\cos\left(-x\right)

Applying the identity \cos\left(-x\right)=\cos\left(x\right) :

=\dfrac{1}{2}\cos\left(2x\right) - \dfrac{1}{2}\cos\left(x\right)

For any two real numbers a and b, we have:

\sin\left(a\right)\cos\left(b\right) = \dfrac{1}{2}\sin\left(a+b\right) + \dfrac{1}{2}\sin\left(a - b\right)

Here:

• a=\dfrac{x}{2}
• b=\dfrac{x}{3}

Therefore:

\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{3}\right)

=\dfrac{1}{2}\sin\left(\dfrac{x}{2}+\dfrac{x}{3}\right) + \dfrac{1}{2}\sin\left(\dfrac{x}{2}-\dfrac{x}{3}\right)
=\dfrac{1}{2}\sin\left(\dfrac{5x}{6}\right) + \dfrac{1}{2}\sin\left(\dfrac{x}{6}\right)

Since \sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}, the expression can become:

\sin\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{x}{3}\right)

For any two real numbers a and b, we have:

\sin\left(a\right)\cos\left(b\right) = \dfrac{1}{2}\sin\left(a+b\right) +\dfrac{1}{2}\sin\left(a - b\right)

Here:

• a=\dfrac{\pi}{3}
• b=\dfrac{x}{3}

Therefore:

\sin\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{x}{3}\right)

=\dfrac{1}{2}\sin\left(\dfrac{\pi}{3}+\dfrac{x}{3}\right) + \dfrac{1}{2}\sin\left(\dfrac{\pi}{3}-\dfrac{x}{3}\right)
=\dfrac{1}{2}\sin\left(\dfrac{\pi+x}{3}\right) +\dfrac{1}{2}\sin\left(\dfrac{\pi-x}{3}\right)