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

Applying the identity \sin\left(-x\right)=-\sin\left(x\right) we get:

\sin\left(\dfrac{\pi}{3}\right)-\sin\left(\dfrac{\pi}{6}\right)

Since for any really numbers a and b
\sin\left(a\right) -\sin\left(b\right) = 2\sin\left(\dfrac{a - b}{2}\right)\cos\left(\dfrac{a + b}{2}\right)

And here:

• \dfrac{\pi}{3}
• b=\dfrac{\pi}{6}

The expression can be converted to :

2\sin\left(\dfrac{\dfrac{\pi}{3}-\dfrac{\pi}{6}}{2}\right)\cos\left(\dfrac{\dfrac{\pi}{3}+\dfrac{\pi}{6}}{2}\right)=2\sin\left(\dfrac{\pi}{12}\right)\cos\left(\dfrac{\pi}{4}\right)

Applying the identity \cos\left(x\right)=\sin\left(\dfrac{\pi}{2}-x\right) we get:

\sin\left(\dfrac{\pi}{3}\right)+\cos\left(\dfrac{\pi}{3}\right)=\sin\left(\dfrac{\pi}{3}\right)+\sin\left(\dfrac{\pi}{2}-\dfrac{\pi}{3}\right)=\sin\left(\dfrac{\pi}{3}\right)+\sin\left(\dfrac{\pi}{6}\right)

Since for any really numbers a and b
\sin\left(a\right) + \sin\left(b\right) = 2\sin\left(\dfrac{a + b}{2}\right)\cos\left(\dfrac{a - b}{2}\right)

And here:

• a=\dfrac{\pi}{3}
• b=\dfrac{\pi}{6}

The expression can be converted to :

2\sin\left(\dfrac{\dfrac{\pi}{3}+\dfrac{\pi}{6}}{2}\right)\cos\left(\dfrac{\dfrac{\pi}{3}-\dfrac{\pi}{6}}{2}\right)=2\sin\left(\dfrac{\pi}{4}\right)\cos\left(\dfrac{\pi}{12}\right)

Since for any really numbers a and b

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

And here:

• a=\dfrac{\pi}{2}
• b=\dfrac{3\pi}{2}

The expression can be converted to:

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

Applying the identity:

\cos\left(-x\right)=\cos\left(x\right)

We get:

Applying the identity \sin\left(2x\right)=2\sin\left(x\right)\cos\left(x\right) we get:

2\sin^{2}\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{\pi}{6}\right)+\sin\left(\dfrac{2\pi}{3}\right)\sin\left(\dfrac{\pi}{3}\right)

=2\sin^{2}\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{\pi}{6}\right)+2\sin\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{\pi}{3}\right)\cdot\sin\left(\dfrac{\pi}{3}\right)

=2\sin^{2}\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{\pi}{6}\right)+2\sin^{2}\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{\pi}{3}\right)

=2\sin^{2}\left(\dfrac{\pi}{3}\right)\left[\cos\left(\dfrac{\pi}{6}\right)+\cos\left(\dfrac{\pi}{3}\right)\right]

Since for any really numbers a and b
\cos\left(a\right)+\cos\left(b\right)=2\cos\left(\dfrac{a+b}{2}\right)\cos\left(\dfrac{a-b}{2}\right)

And here:

• a=\dfrac{\pi}{6}
• b=\dfrac{\pi}{3}

The expression can be converted to :

2\sin^{2}\left(\dfrac{\pi}{3}\right)\cdot2\cos\left(\dfrac{\dfrac{\pi}{6}+\dfrac{\pi}{3}}{2}\right)\cos\left(\dfrac{\dfrac{\pi}{6}-\dfrac{\pi}{3}}{2}\right)=4\sin^{2}\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{\pi}{4}\right)\cos\left(-\dfrac{\pi}{12}\right)

Since \cos\left(-x\right)=\cos\left(x\right) :

Applying the identity 2\cos^{2}\left(x\right)-1=\cos\left(2x\right) and rearranging the expression we get:

2\cos^{2}\left(\dfrac{\pi}{9}\right)+2\cos^{2}\left(\dfrac{\pi}{18}\right)-2

=2\cos^{2}\left(\dfrac{\pi}{9}\right)-1+2\cos^{2}\left(\dfrac{\pi}{18}\right)-1

=\cos\left(2\cdot\dfrac{\pi}{9}\right)+\cos\left(2\cdot\dfrac{\pi}{18}\right)

=\cos\left(\dfrac{2\pi}{9}\right)+\cos\left(\dfrac{\pi}{9}\right)

Since for any really numbers a and b
\cos\left(a\right)-\cos\left(b\right)=-2\sin\left(\dfrac{a+b}{2}\right)\sin\left(\dfrac{a-b}{2}\right)

And here:

• a=\dfrac{2\pi}{9}
• b=\dfrac{\pi}{9}

The expression can be converted to :

2\cos\left(\dfrac{\dfrac{2\pi}{9}+\dfrac{\pi}{9}}{2}\right)\cos\left(\dfrac{\dfrac{2\pi}{9}-\dfrac{\pi}{9}}{2}\right)=2\cos\left(\dfrac{\pi}{6}\right)\cos\left(\dfrac{\pi}{18}\right)

Since for any really numbers a and b

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

And here:

• a=\dfrac{2\pi}{5}
• b=\dfrac{\pi}{7}

The expression can be converted to :

\cos\left(\dfrac{2\pi}{5}\right)-\cos\left(\dfrac{\pi}{7}\right)=-2\sin\left(\dfrac{\dfrac{2\pi}{5}+\dfrac{\pi}{7}}{2}\right)\sin\left(\dfrac{\dfrac{2\pi}{5}-\dfrac{\pi}{7}}{2}\right)=-2\sin\left(\dfrac{19\pi}{70}\right)\sin\left(\dfrac{9\pi}{70}\right)

Simplifying the expression :

\sin\left(\dfrac{\pi}{5}+\dfrac{\pi}{4}\right)+\sin\left(\dfrac{\pi}{10}-\dfrac{\pi}{20}\right)=\sin\left(\dfrac{9\pi}{20}\right)+\sin\left(\dfrac{\pi}{20}\right)

Since for any really numbers a and b
\sin\left(a\right) + \sin\left(b\right) = 2\sin\left(\dfrac{a + b}{2}\right)\cos\left(\dfrac{a - b}{2}\right)

And here:

• a=\dfrac{9\pi}{20}
• b=\dfrac{\pi}{20}

The expression can be converted to :

2\sin\left(\dfrac{\dfrac{9\pi}{20}+\dfrac{\pi}{20}}{2}\right)\cos\left(\dfrac{\dfrac{9\pi}{20}-\dfrac{\pi}{20}}{2}\right)=2\sin\left(\dfrac{\pi}{4}\right)\cos\left(\dfrac{\pi}{5}\right)