###### Supplemental Videos

The main topics of this section are also presented in the following videos:

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The main topics of this section are also presented in the following videos:

In the last section we developed a few important trigonometric identities and some examples. In this section, we will continue working with a few special trigonometric identities and do some more examples. All of these examples can be derived from basic trigonometric principles. However, in this book we will focus not on deriving the identities but rather in using the identities to solve problems. For this reason we do not include derivations of the identities but instead refer you to other texts on trigonometry for such derivations.

What happens to the cosine or sine of an angle if we cut the angle in half? You might expect the cosine or sine to be cut in half, but this is not the case. Instead, we have the Half-Angle Identities.

\begin{align*}
\cos\left(\frac{\theta}{2}\right) \amp = \pm \sqrt{\frac{1+\cos(\theta)}{2}} \\
\\
\sin\left(\frac{\theta}{2}\right) \amp = \pm \sqrt{\frac{1-\cos(\theta)}{2}}
\end{align*}

There are examples that we can solve using the Half-Angle Identities.

Find the exact value for \(\cos(15^{\circ})\text{.}\)

Solution

We first note that \(15^{\circ}\) is half of \(30^{\circ}\text{.}\) Therefore,

\begin{equation*}
\cos(15^{\circ})=\cos\left(\frac{30^{\circ}}{2}\right) = \pm \sqrt{\frac{1+\cos(30^{\circ})}{2}}.
\end{equation*}

We can evaluate the cosine. Since 15 degrees is in the first quadrant, we need the positive result.

\begin{equation*}
\cos(15^{\circ}) = \sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}
\end{equation*}

The sum and difference of angles identities are often used to rewrite expressions in other forms, or to rewrite an angle in terms of simpler angles.

\begin{align*}
\sin(\theta + \phi) \amp = \sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta) \\
\\
\sin(\theta - \phi) \amp = \sin(\theta)\cos(\phi) - \sin(\phi)\cos(\theta) \\
\\
\cos(\theta + \phi) \amp = \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi) \\
\\
\cos(\theta - \phi) \amp = \cos(\theta)\cos(\phi) + \sin(\theta)\sin(\phi)
\end{align*}

Find the exact value of \(\cos(75^{\circ})\)

Solution

Since we know that \(75^{\circ}=30^{\circ}+45^{\circ}\) we can evaluate \(\cos(75^{\circ})\) using an angle sum identity.

\begin{align*}
\cos(30^{\circ} + 45^{\circ}) \amp = \cos(30^{\circ})\cos(45^{\circ}) - \sin(30^{\circ})\sin(45^{\circ}) \\
\amp = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}-\frac{1}{2}\cdot\frac{\sqrt{2}}{2} \\
\amp = \frac{\sqrt{6}-\sqrt{2}}{4}
\end{align*}

The Angle Sum and Difference Identities can also be used to solve more complicated trigonometric equations.

Solve \(\sin(x)\sin(2x)+\cos(x)\cos(2x)=\frac{\sqrt{3}}{2}\)

Solution

We first recognize that the left hand side of the equation is result of the difference of angles identity for cosine. Therefore,

\begin{align*}
\sin(x)\sin(2x)+\cos(x)\cos(2x)\amp=\frac{\sqrt{3}}{2}
\end{align*}

is equivalent to the equation

\begin{align*}
\cos(x-2x)\amp=\frac{\sqrt{3}}{2}
\end{align*}

Thus we have

\begin{align*}
\cos(x-2x)\amp=\frac{\sqrt{3}}{2} \\
\cos(-x)\amp=\frac{\sqrt{3}}{2} \\
\cos(x)\amp=\frac{\sqrt{3}}{2} \\
x\amp=\frac{\pi}{6}+2 \pi k, \frac{11 \pi}{6}+2\pi k
\end{align*}

The above examples demonstrate just a little bit of the power of the Half-Angle and Angle Sum and Difference Identities. If you go on to take a Calculus course you may also see these identities come up again as you calculate derivatives.