Double Angle, Half Angle, and Reduction Formulas: Learn It 1

Lumen Learning, OpenStax

Double Angle, Half Angle, and Reduction Formulas: Learn It 1

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 [latex]\alpha =\beta[/latex]. Deriving the double-angle formula for sine begins with the sum formula,

[latex]\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta[/latex]

If we let [latex]\alpha =\beta =\theta[/latex], then we have

[latex]\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}[/latex]

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

[latex]\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}[/latex]

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

[latex]\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}[/latex]

The second interpretation is:

[latex]\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}[/latex]

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

[latex]\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}[/latex]

double angle formulas

The double-angle formulas are summarized as follows:

[latex]\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}[/latex]

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.

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

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

[latex]\sin \left(2\theta \right)[/latex]
[latex]\cos \left(2\theta \right)[/latex]
[latex]\tan \left(2\theta \right)[/latex]

If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given [latex]\tan \theta =-\frac{3}{4}[/latex], such that [latex]\theta[/latex] is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because [latex]\theta[/latex] is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:

[latex]\begin{align}{\left(-4\right)}^{2}+{\left(3\right)}^{2}&={c}^{2}\\ 16+9&={c}^{2}\\ 25&={c}^{2}\\ c&=5\end{align}[/latex]

Now we can draw a triangle:

Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-4,0), and (-4,3). The angle at the origin is theta. The angle formed by the side (-4,3) to (-4,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 5.

Let’s begin by writing the double-angle formula for sine.
[latex]\sin \left(2\theta \right)=2\sin \theta \cos \theta[/latex]

We see that we to need to find [latex]\sin \theta[/latex] and [latex]\cos \theta[/latex]. Based on Figure 2, we see that the hypotenuse equals 5, so [latex]\sin \theta =\frac{3}{5}[/latex], and [latex]\cos \theta =−\frac{4}{5}[/latex]. Substitute these values into the equation, and simplify.
Thus,

[latex]\begin{align}\sin \left(2\theta \right)&=2\left(\frac{3}{5}\right)\left(−\frac{4}{5}\right) \\ &=−\frac{24}{25} \end{align}[/latex]
Write the double-angle formula for cosine.
[latex]\cos \left(2\theta \right)={\cos }^{2}\theta -{\sin }^{2}\theta[/latex]

Again, substitute the values of the sine and cosine into the equation, and simplify.

[latex]\begin{align}\cos \left(2\theta \right)&={\left(−\frac{4}{5}\right)}^{2}−{\left(\frac{3}{5}\right)}^{2} \\ &=\frac{16}{25}−\frac{9}{25} \\ &=\frac{7}{25}\end{align}[/latex]
Write the double-angle formula for tangent.
[latex]\tan \left(2\theta \right)=\frac{2\tan \theta }{1−{\tan }^{2}\theta }[/latex]

In this formula, we need the tangent, which we were given as [latex]\tan \theta =−\frac{3}{4}[/latex]. Substitute this value into the equation, and simplify.

[latex]\begin{align}\tan \left(2\theta \right)&=\frac{2\left(-\frac{3}{4}\right)}{1-{\left(-\frac{3}{4}\right)}^{2}} \\ &=\frac{-\frac{3}{2}}{1-\frac{9}{16}} \\ &=-\frac{3}{2}\left(\frac{16}{7}\right) \\ &=-\frac{24}{7} \end{align}[/latex]

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

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