Double Angle, Half Angle, and Reduction Formulas

Learning Outcomes

• Use double-angle formulas to find exact values.
• Use double-angle formulas to verify identities.
• Use reduction formulas to simplify an expression.
• Use half-angle formulas to find exact values.

Using Double-Angle Formulas to Find Exact Values

In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where $\alpha =\beta$. Deriving the double-angle formula for sine begins with the sum formula,

$\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$

If we let $\alpha =\beta =\theta$, then we have

\begin{align}\sin \left(\theta +\theta \right)&=\sin \theta \cos \theta +\cos \theta \sin \theta \\ \sin \left(2\theta \right)&=2\sin \theta \cos \theta \end{align}

Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, $\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$, and letting $\alpha =\beta =\theta$, we have

\begin{align}\cos \left(\theta +\theta \right)&=\cos \theta \cos \theta -\sin \theta \sin \theta \\ \cos \left(2\theta \right)&={\cos }^{2}\theta -{\sin }^{2}\theta \end{align}

Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:

\begin{align}\cos \left(2\theta \right)&={\cos }^{2}\theta -{\sin }^{2}\theta \\ &=\left(1-{\sin }^{2}\theta \right)-{\sin }^{2}\theta \\ &=1 - 2{\sin }^{2}\theta\end{align}

The second interpretation is:

\begin{align}\cos \left(2\theta \right)&={\cos }^{2}\theta -{\sin }^{2}\theta \\ &={\cos }^{2}\theta -\left(1-{\cos }^{2}\theta \right) \\ &=2{\cos }^{2}\theta -1\end{align}

Similarly, to derive the double-angle formula for tangent, replacing $\alpha =\beta =\theta$ in the sum formula gives

\begin{align}\tan \left(\alpha +\beta \right)&=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ \tan \left(\theta +\theta \right)&=\frac{\tan \theta +\tan \theta }{1-\tan \theta \tan \theta }\\ \tan \left(2\theta \right)&=\frac{2\tan \theta }{1-{\tan }^{2}\theta }\end{align}

A General Note: Double-Angle Formulas

The double-angle formulas are summarized as follows:

\begin{align}\sin \left(2\theta \right)&=2\sin \theta \cos \theta\\\text{ }\\ \cos \left(2\theta \right)&={\cos }^{2}\theta -{\sin }^{2}\theta \\ &=1 - 2{\sin }^{2}\theta \\ &=2{\cos }^{2}\theta -1 \\\text{ }\\ \tan \left(2\theta \right)&=\frac{2\tan \theta }{1-{\tan }^{2}\theta }\end{align}

How To: Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.

1. Draw a triangle to reflect the given information.
2. Determine the correct double-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.

Example 1: Using a Double-Angle Formula to Find the Exact Value Involving Tangent

Given that $\tan \theta =−\frac{3}{4}$ and $\theta$ is in quadrant II, find the following:

1. $\sin \left(2\theta \right)$
2. $\cos \left(2\theta \right)$
3. $\tan \left(2\theta \right)$

Try It

Given $\sin \alpha =\frac{5}{8}$, with $\theta$ in quadrant I, find $\cos \left(2\alpha \right)$.

Example 2: Using the Double-Angle Formula for Cosine without Exact Values

Use the double-angle formula for cosine to write $\cos \left(6x\right)$ in terms of $\cos \left(3x\right)$.

Using Double-Angle Formulas to Verify Identities

Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.

Example 3: Using the Double-Angle Formulas to Establish an Identity

Establish the following identity using double-angle formulas:

$1+\sin \left(2\theta \right)={\left(\sin \theta +\cos \theta \right)}^{2}$

Try It

Establish the identity: ${\cos }^{4}\theta -{\sin }^{4}\theta =\cos \left(2\theta \right)$.

Example 4: Verifying a Double-Angle Identity for Tangent

Verify the identity:

$\tan \left(2\theta \right)=\frac{2}{\cot \theta -\tan \theta }$

Try It

Verify the identity: $\cos \left(2\theta \right)\cos \theta ={\cos }^{3}\theta -\cos \theta {\sin }^{2}\theta$.

Use Reduction Formulas to Simplify an Expression

The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.

We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with $\cos \left(2\theta \right)=1 - 2{\sin }^{2}\theta$. Solve for ${\sin }^{2}\theta :$

$\begin{gathered}\cos \left(2\theta \right)=1 - 2{\sin }^{2}\theta \\ 2{\sin }^{2}\theta =1-\cos \left(2\theta \right) \\ {\sin }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{2} \end{gathered}$

Next, we use the formula $\cos \left(2\theta \right)=2{\cos }^{2}\theta -1$. Solve for ${\cos }^{2}\theta :$

$\begin{gathered}\cos \left(2\theta \right)=2{\cos }^{2}\theta -1 \\ 1+\cos \left(2\theta \right)=2{\cos }^{2}\theta \\ \frac{1+\cos \left(2\theta \right)}{2}={\cos }^{2}\theta \end{gathered}$

The last reduction formula is derived by writing tangent in terms of sine and cosine:

\begin{align}{\tan }^{2}\theta &=\frac{{\sin }^{2}\theta }{{\cos }^{2}\theta } \\ &=\frac{\frac{1-\cos \left(2\theta \right)}{2}}{\frac{1+\cos \left(2\theta \right)}{2}}&& \text{Substitute the reduction formulas.} \\ &=\left(\frac{1-\cos \left(2\theta \right)}{2}\right)\left(\frac{2}{1+\cos \left(2\theta \right)}\right) \\ &=\frac{1-\cos \left(2\theta \right)}{1+\cos \left(2\theta \right)} \end{align}

A General Note: Reduction Formulas

The reduction formulas are summarized as follows:

\begin{align}&{\sin }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{2} \\ &{\cos }^{2}\theta =\frac{1+\cos \left(2\theta \right)}{2} \\ &{\tan }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{1+\cos \left(2\theta \right)} \end{align}

Example 5: Writing an Equivalent Expression Not Containing Powers Greater Than 1

Write an equivalent expression for ${\cos }^{4}x$ that does not involve any powers of sine or cosine greater than 1.

Example 6: Using the Power-Reducing Formulas to Prove an Identity

Use the power-reducing formulas to prove

${\sin }^{3}\left(2x\right)=\left[\frac{1}{2}\sin \left(2x\right)\right]\left[1-\cos \left(4x\right)\right]$

Try It

Use the power-reducing formulas to prove that $10{\cos }^{4}x=\frac{15}{4}+5\cos \left(2x\right)+\frac{5}{4}\cos \left(4x\right)$.

Using Half-Angle Formulas to Find Exact Values

The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace $\theta$ with $\frac{\alpha }{2}$, the half-angle formula for sine is found by simplifying the equation and solving for $\sin \left(\frac{\alpha }{2}\right)$. Note that the half-angle formulas are preceded by a $\pm$ sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which $\frac{\alpha }{2}$ terminates.

The half-angle formula for sine is derived as follows:

\begin{align}{\sin }^{2}\theta &=\frac{1-\cos \left(2\theta \right)}{2}\\ {\sin }^{2}\left(\frac{\alpha }{2}\right)&=\frac{1-\left(\cos 2\cdot \frac{\alpha }{2}\right)}{2} \\ &=\frac{1-\cos \alpha }{2} \\ \sin \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1-\cos \alpha }{2}} \end{align}

To derive the half-angle formula for cosine, we have

\begin{align}{\cos }^{2}\theta &=\frac{1+\cos \left(2\theta \right)}{2}\\ {\cos }^{2}\left(\frac{\alpha }{2}\right)&=\frac{1+\cos \left(2\cdot \frac{\alpha }{2}\right)}{2} \\ &=\frac{1+\cos \alpha }{2} \\ \cos \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1+\cos \alpha }{2}} \end{align}

For the tangent identity, we have

\begin{align}{\tan }^{2}\theta &=\frac{1-\cos \left(2\theta \right)}{1+\cos \left(2\theta \right)} \\ {\tan }^{2}\left(\frac{\alpha }{2}\right)&=\frac{1-\cos \left(2\cdot \frac{\alpha }{2}\right)}{1+\cos \left(2\cdot \frac{\alpha }{2}\right)} \\ &=\frac{1-\cos \alpha }{1+\cos \alpha }\hfill \\ \tan \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }} \end{align}

A General Note: Half-Angle Formulas

The half-angle formulas are as follows:

\begin{align}\sin \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1-\cos \alpha }{2}} \\ \text{ } \\ \cos \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1+\cos \alpha }{2}} \\ \text{ } \\ \tan \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }} \\ &=\frac{\sin \alpha }{1+\cos \alpha } \\ &=\frac{1-\cos \alpha }{\sin \alpha }\end{align}

Example 7: Using a Half-Angle Formula to Find the Exact Value of a Sine Function

Find $\sin \left({15}^{\circ }\right)$ using a half-angle formula.

How To: Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.

1. Draw a triangle to represent the given information.
2. Determine the correct half-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.

Example 8: Finding Exact Values Using Half-Angle Identities

Given that $\tan \alpha =\frac{8}{15}$ and $\alpha$ lies in quadrant III, find the exact value of the following:

1. $\sin \left(\frac{\alpha }{2}\right)$
2. $\cos \left(\frac{\alpha }{2}\right)$
3. $\tan \left(\frac{\alpha }{2}\right)$

Try It

Given that $\sin \alpha =-\frac{4}{5}$ and $\alpha$ lies in quadrant IV, find the exact value of $\cos \left(\frac{\alpha }{2}\right)$.

Example 9: Finding the Measurement of a Half Angle

Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of $\theta$ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If $\tan \theta =\frac{5}{3}$ for higher-level competition, what is the measurement of the angle for novice competition?

Key Equations

 Double-angle formulas \begin{align}\sin \left(2\theta \right)&=2\sin \theta \cos \theta\\\text{ }\\ \cos \left(2\theta \right)&={\cos }^{2}\theta -{\sin }^{2}\theta \\ &=1 - 2{\sin }^{2}\theta \\ &=2{\cos }^{2}\theta -1 \\\text{ }\\ \tan \left(2\theta \right)&=\frac{2\tan \theta }{1-{\tan }^{2}\theta }\end{align} Reduction formulas \begin{align}&{\sin }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{2} \\ &{\cos }^{2}\theta =\frac{1+\cos \left(2\theta \right)}{2} \\ &{\tan }^{2}\theta =\frac{1-\cos \left(2\theta \right)}{1+\cos \left(2\theta \right)} \end{align} Half-angle formulas \begin{align}\sin \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1-\cos \alpha }{2}} \\ \text{ } \\ \cos \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1+\cos \alpha }{2}} \\ \text{ } \\ \tan \left(\frac{\alpha }{2}\right)&=\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }} \\ &=\frac{\sin \alpha }{1+\cos \alpha } \\ &=\frac{1-\cos \alpha }{\sin \alpha }\end{align}

Key Concepts

• Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent.
• Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term.
• Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not.

Glossary

double-angle formulas
identities derived from the sum formulas for sine, cosine, and tangent in which the angles are equal
half-angle formulas
identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions
reduction formulas
identities derived from the double-angle formulas and used to reduce the power of a trigonometric function