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3.10: Double Angle Identities

Difficulty Level: At Grade Created by: CK-12
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Finding the values for trig functions is pretty familiar to you by now. The trig functions of some particular angles may even seem obvious, since you've worked with them so many times. In some cases, you might be able to use this knowledge to your benefit to make calculating the values of some trig equations easier. For example, if someone asked you to evaluate

\begin{align*}\cos 120^\circ\end{align*}

without consulting a table of trig values, could you do it?

You might notice right away that this is equal to four times \begin{align*}30^\circ\end{align*}. Can this help you? Read this Concept, and at its conclusion you'll know how to use certain formulas to simplify multiples of familiar angles to solve problems.

Watch This

James Sousa: Double Angle Identities

Guidance

Here we'll start with the sum and difference formulas for sine, cosine, and tangent. We can use these identities to help derive a new formula for when we are given a trig function that has twice a given angle as the argument. For example, \begin{align*}\sin (2 \theta)\end{align*}. This way, if we are given \begin{align*}\theta\end{align*} and are asked to find \begin{align*}\sin (2 \theta)\end{align*}, we can use our new double angle identity to help simplify the problem. Let's start with the derivation of the double angle identities.

One of the formulas for calculating the sum of two angles is:

\begin{align*}\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\end{align*}

If \begin{align*}\alpha\end{align*} and \begin{align*}\beta\end{align*} are both the same angle in the above formula, then

\begin{align*}\sin (\alpha + \alpha) & = \sin \alpha \cos \alpha + \cos \alpha \sin \alpha \\ \sin 2 \alpha & = 2 \sin \alpha \cos \alpha\end{align*}

This is the double angle formula for the sine function. The same procedure can be used in the sum formula for cosine, start with the sum angle formula:

\begin{align*}\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\end{align*}

If \begin{align*}\alpha\end{align*} and \begin{align*}\beta\end{align*} are both the same angle in the above formula, then

\begin{align*}\cos (\alpha + \alpha )& = \cos \alpha \cos \alpha - \sin \alpha \sin \alpha \\ \cos 2 \alpha & = \cos^2 \alpha - \sin^2 \alpha\end{align*}

This is one of the double angle formulas for the cosine function. Two more formulas can be derived by using the Pythagorean Identity, \begin{align*}\sin^2 \alpha + \cos^2 \alpha = 1\end{align*}.

\begin{align*}\sin^2 \alpha = 1 - \cos^2 \alpha\end{align*} and likewise \begin{align*}\cos^2 \alpha = 1 - \sin^2 \alpha\end{align*}

\begin{align*}\text{Using}\ \sin^2 \alpha & = 1 - \cos^2 \alpha: && \text{Using}\ \cos^2 \alpha = 1 - \sin^2 \alpha: \\ \cos 2 \alpha & = \cos^2 \alpha - \sin^2 \alpha && \qquad \ \ \cos 2 \alpha = \cos^2 \alpha - \sin^2 \alpha \\ & = \cos^2 \alpha - (1 - \cos^2 \alpha) && \qquad \qquad \quad \ = (1 - \sin^2 \alpha) - \sin^2 \alpha \\ & = \cos^2 \alpha - 1 + \cos^2 \alpha && \qquad \qquad \quad \ = 1 - \sin^2 \alpha - \sin^2 \alpha \\ & = 2 \cos^2 \alpha - 1 && \qquad \qquad \quad \ = 1 - 2 \sin^2 \alpha\end{align*}

Therefore, the double angle formulas for \begin{align*}\cos 2 \alpha\end{align*} are:

\begin{align*}\cos 2 \alpha & = \cos^2 \alpha - \sin^2 \alpha \\ \cos 2 \alpha & = 2 \cos^2 \alpha -1 \\ \cos 2 \alpha & = 1 - 2 \sin^2 \alpha\end{align*}

Finally, we can calculate the double angle formula for tangent, using the tangent sum formula:

\begin{align*}\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}\end{align*}

If \begin{align*}\alpha\end{align*} and \begin{align*}\beta\end{align*} are both the same angle in the above formula, then

\begin{align*}\tan (\alpha + \alpha) & = \frac{\tan \alpha + \tan \alpha}{1 - \tan \alpha \tan \alpha} \\ \tan 2 \alpha & = \frac{2 \tan \alpha}{1 - \tan^2 \alpha}\end{align*}

We can use these formulas to help simplify calculations of trig functions of certain arguments.

Example A

If \begin{align*}\sin a = \frac{5}{13}\end{align*} and \begin{align*}a\end{align*} is in Quadrant II, find \begin{align*}\sin 2a\end{align*}, \begin{align*}\cos 2a\end{align*}, and \begin{align*}\tan 2a\end{align*}.

Solution: To use \begin{align*}\sin 2a = 2 \sin a \cos a\end{align*}, the value of \begin{align*}\cos a\end{align*} must be found first.

\begin{align*}& =\cos^2 a + \sin^2 a = 1\\ & = \cos^2 a + \left (\frac{5}{13} \right )^2 = 1\\ & = \cos^2 a + \frac{25}{169} = 1\\ & = \cos^2 a = \frac{144}{169}, \cos a = \pm \frac{12}{13}\end{align*}.

However since \begin{align*}a\end{align*} is in Quadrant II, \begin{align*}\cos a\end{align*} is negative or \begin{align*}\cos a = - \frac{12}{13}\end{align*}.

\begin{align*}\sin 2a = 2 \sin a \cos a = 2 \left (\frac{5}{13} \right ) \times \left (- \frac{12}{13} \right ) = \sin 2a = - \frac{120}{169}\end{align*}

For \begin{align*}\cos 2a\end{align*}, use \begin{align*}\cos(2a) = \cos^2 a - \sin^2 a\end{align*}

\begin{align*}\cos(2a) & = \left (- \frac{12}{13} \right )^2 - \left (\frac{5}{13} \right )^2 \ \text{or}\ \frac{144-25}{169} \\ \cos (2a) & = \frac{119}{169}\end{align*}

For \begin{align*}\tan 2a\end{align*}, use \begin{align*}\tan 2a = \frac{2 \tan a}{1 - \tan^2 a}\end{align*}. From above, \begin{align*}\tan a = \frac{\frac{5}{13}}{-\frac{12}{13}} = - \frac{5}{12}\end{align*}.

\begin{align*}\tan(2a) = \frac{2 \cdot \frac{-5}{12}}{1 - \left (\frac{-5}{12} \right )^2} = \frac{\frac{-5}{6}}{1 - \frac{25}{144}} = \frac{\frac{-5}{6}}{\frac{119}{144}} = - \frac{5}{6} \cdot \frac{144}{119} = - \frac{120}{119}\end{align*}

Example B

Find \begin{align*}\cos 4 \theta\end{align*}.

Solution: Think of \begin{align*}\cos 4 \theta\end{align*} as \begin{align*}\cos (2 \theta + 2 \theta)\end{align*}.

\begin{align*}\cos 4 \theta = \cos (2 \theta + 2 \theta) = \cos 2 \theta \cos 2 \theta - \sin 2 \theta \sin 2 \theta = \cos^2 2 \theta - \sin^2 2 \theta\end{align*}

Now, use the double angle formulas for both sine and cosine. For cosine, you can pick which formula you would like to use. In general, because we are proving a cosine identity, stay with cosine.

\begin{align*}& = (2 \cos^2 \theta - 1)^2 - (2 \sin \theta \cos \theta)^2 \\ & = 4 \cos^4 \theta - 4 \cos^2 \theta + 1 - 4 \sin^2 \theta \cos^2 \theta \\ & = 4 \cos^4 \theta - 4 \cos^2 \theta + 1 - 4 (1 - \cos^2 \theta) \cos^2 \theta \\ & = 4 \cos^4 \theta - 4 \cos^2 \theta + 1 - 4 \cos^2 \theta + 4 \cos^4 \theta \\ & = 8 \cos^4 \theta - 8 \cos^2 \theta + 1\end{align*}

Example C

Solve the trigonometric equation \begin{align*}\sin 2x = \sin x\end{align*} such that \begin{align*}(- \pi \le x < \pi)\end{align*}

Solution: Using the sine double angle formula:

\begin{align*}& \qquad \qquad \quad \ \ \sin 2x = \sin x \\ & \qquad \quad 2 \sin x \cos x \ = \sin x \\ & \ 2 \sin x \cos x - \sin x = 0 \\ & \ \ \ \sin x (2 \cos x -1) = 0 \\ & \quad \Bigg\downarrow \qquad \ \ \searrow \\ & \qquad \qquad \qquad 2 \cos x - 1 = 0 \\ & \qquad \qquad \qquad \quad \ \ 2 \cos x = 1 \\ & \sin x = 0 \\ & \quad \ \ x = 0, - \pi \qquad \ \cos x = \frac{1}{2} \\ & \qquad \qquad \qquad \qquad \qquad x = \frac{\pi}{3}, - \frac{\pi}{3}\end{align*}

Vocabulary

Double Angle Identity: A double angle identity relates the a trigonometric function of two times an argument to a set of trigonometric functions, each containing the original argument.

Guided Practice

1. If \begin{align*}\sin x = \frac{4}{5}\end{align*} and \begin{align*}x\end{align*} is in Quad II, find the exact values of \begin{align*}\cos 2x, \sin 2x\end{align*} and \begin{align*}\tan 2x\end{align*}

2. Find the exact value of \begin{align*}\cos^2 15^\circ - \sin^2 15^\circ\end{align*}

3. Verify the identity: \begin{align*}\cos 3 \theta = 4 \cos^3\theta - 3 \cos \theta\end{align*}

Solutions:

1. If \begin{align*}\sin x = \frac{4}{5}\end{align*} and in Quadrant II, then cosine and tangent are negative. Also, by the Pythagorean Theorem, the third side is \begin{align*}3(b = \sqrt{5^2 - 4^2})\end{align*}. So, \begin{align*}\cos x = - \frac{3}{5}\end{align*} and \begin{align*}\tan x = - \frac{4}{3}\end{align*}. Using this, we can find \begin{align*}\sin 2x, \cos 2x\end{align*}, and \begin{align*}\tan 2x\end{align*}.

\begin{align*}& && \ \cos 2x = 1 - \sin^2 x && \ \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \\ & && \qquad \quad = 1 -2 \cdot \left (\frac{4}{5} \right)^2 && \qquad \quad \ = \frac{2 \cdot - \frac{4}{3}}{1 - \left (- \frac{4}{3} \right )^2} \\ & \sin 2x = 2 \sin x \cos x && \qquad \quad = 1 - 2 \cdot \frac{16}{25} && \qquad \quad \ = \frac{-\frac{8}{3}}{1 - \frac{16}{9}} = - \frac{8}{3} \div - \frac{7}{9} \\ & \qquad \ \ = 2 \cdot \frac{4}{5} \cdot - \frac{3}{5} && \qquad \quad = 1 - \frac{32}{25} && \qquad \quad \ = - \frac{8}{3} \cdot - \frac{9}{7} \\ & \qquad \ \ = - \frac{24}{25} && \qquad \quad =- \frac{7}{25} && \qquad \quad \ = \frac{24}{7}\end{align*}

2. This is one of the forms for \begin{align*}\cos 2x\end{align*}.

\begin{align*}\cos^2 15^\circ - \sin^2 15^\circ & = \cos (15^\circ \cdot 2) \\ & = \cos 30^ \circ \\ & = \frac{\sqrt{3}}{2}\end{align*}

3. Step 1: Use the cosine sum formula

\begin{align*}\cos 3 \theta & = 4 \cos^3 \theta - 3 \cos \theta \\ \cos (2 \theta + \theta) & = \cos 2 \theta \cos \theta - \sin 2 \theta \sin \theta\end{align*}

Step 2: Use double angle formulas for \begin{align*}\cos 2\theta\end{align*} and \begin{align*}\sin 2\theta\end{align*}

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

Step 3: Distribute and simplify.

\begin{align*}& = 2 \cos^3 \theta - \cos \theta - 2 \sin^2 \theta \cos \theta \\ & = - \cos \theta (-2 \cos^2 \theta + 2 \sin^2 \theta + 1) \\ & = - \cos \theta [- 2 \cos^2 \theta + 2 (1 - \cos^2 \theta) + 1] && \rightarrow \text{Substitute}\ 1 - \cos^2 \theta \ \text{for}\ \sin^2 \theta \\ & = - \cos \theta [- 2 \cos^2 \theta + 2 - 2 \cos^2 \theta + 1] \\ & = - \cos \theta (-4 \cos^2 \theta + 3) \\ & = 4 \cos^3 \theta - 3 \cos \theta\end{align*}

Concept Problem Solution

Since the problem wants you to find:

\begin{align*}\cos 120^\circ\end{align*}

You can simplify this into a familiar angle:

\begin{align*}\cos (2 \times 60^\circ)\end{align*}

And then apply the double angle identity:

\begin{align*}\cos (2 \times 60^\circ) &= 2 \cos^2 60^\circ - 1 \\ &= (2)(\cos 60^\circ)(\cos 60^\circ) - 1 \\ &= (2)(\frac{1}{2})(\frac{1}{2}) - 1 \\ &= -\frac{1}{2}\end{align*}

Practice

Simplify each expression so that it is in terms of \begin{align*}\sin(x)\end{align*} and \begin{align*}\cos(x)\end{align*}.

  1. \begin{align*}\sin2x+\cos x\end{align*}
  2. \begin{align*}\sin2x+\cos2x\end{align*}
  3. \begin{align*}\sin3x+\cos2x\end{align*}
  4. \begin{align*}\sin2x+\cos3x\end{align*}

Solve each equation on the interval \begin{align*}[0,2\pi)\end{align*}.

  1. \begin{align*}\sin(2x)=2\sin(x)\end{align*}
  2. \begin{align*}\cos(2x)=\sin(x)\end{align*}
  3. \begin{align*}\sin(2x)-\tan(x)=0\end{align*}
  4. \begin{align*}\cos^2(x)+\cos(x)=\cos(2x)\end{align*}
  5. \begin{align*}\cos(2x)=\cos(x)\end{align*}

Simplify each expression so that only one calculation would be needed in order to evaluate.

  1. \begin{align*}2\cos^2(15^\circ)-1\end{align*}
  2. \begin{align*}2\sin(25^\circ)\cos(25^\circ)\end{align*}
  3. \begin{align*}1-2\sin^2(35^\circ)\end{align*}
  4. \begin{align*}\cos^2(60^\circ)-\sin^2(60^\circ)\end{align*}
  5. \begin{align*}2\sin(125^\circ)\cos(125^\circ)\end{align*}
  6. \begin{align*}1-2\sin^2(32^\circ)\end{align*}

Vocabulary

Double Angle Identity

Double Angle Identity

A double angle identity relates the trigonometric function of two times an argument to a set of trigonometric functions, each containing the original argument.
Half Angle Identity

Half Angle Identity

A half angle identity relates a trigonometric function of one half of an argument to a set of trigonometric functions, each containing the original argument.
identity

identity

An identity is a mathematical sentence involving the symbol “=” that is always true for variables within the domains of the expressions on either side.
power reducing identity

power reducing identity

A power reducing identity relates the power of a trigonometric function containing a given argument to a set of trigonometric functions, each containing the original argument.

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Sep 26, 2012
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