<meta http-equiv="refresh" content="1; url=/nojavascript/"> Double Angle Identities ( Read ) | Trigonometry | CK-12 Foundation
Dismiss
Skip Navigation

Double Angle Identities

%
Best Score
Practice Double Angle Identities
Practice
Best Score
%
Practice Now
Double Angle Identities
 0  0  0

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

\cos 120^\circ

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

You might notice right away that this is equal to four times 30^\circ . 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, \sin (2 \theta) . This way, if we are given \theta and are asked to find \sin (2 \theta) , 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:

\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta

If \alpha and \beta are both the same angle in the above formula, then

\sin (\alpha + \alpha) & = \sin \alpha \cos \alpha + \cos \alpha \sin \alpha \\\sin 2 \alpha & = 2 \sin \alpha \cos \alpha

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:

\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta

If \alpha and \beta are both the same angle in the above formula, then

\cos (\alpha + \alpha )& = \cos \alpha \cos \alpha - \sin \alpha \sin \alpha \\\cos 2 \alpha & = \cos^2 \alpha - \sin^2 \alpha

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

\sin^2 \alpha = 1 - \cos^2 \alpha and likewise \cos^2 \alpha = 1 - \sin^2 \alpha

\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

Therefore, the double angle formulas for \cos 2 \alpha are:

\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

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

\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}

If \alpha and \beta are both the same angle in the above formula, then

\tan (\alpha + \alpha) & = \frac{\tan \alpha + \tan \alpha}{1 - \tan \alpha \tan \alpha} \\\tan 2 \alpha & = \frac{2 \tan \alpha}{1 - \tan^2 \alpha}

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

Example A

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

Solution: To use \sin 2a = 2 \sin a \cos a , the value of \cos a must be found first.

& =\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} .

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

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

For \cos 2a , use \cos(2a) = \cos^2 a - \sin^2 a

\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}

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

\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}

Example B

Find \cos 4 \theta .

Solution: Think of \cos 4 \theta as \cos (2 \theta + 2 \theta) .

\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

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.

& = (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

Example C

Solve the trigonometric equation \sin 2x = \sin x such that (- \pi \le x < \pi)

Solution: Using the sine double angle formula:

& \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}

Vocabulary

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

Guided Practice

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

2. Find the exact value of \cos^2 15^\circ - \sin^2 15^\circ

3. Verify the identity: \cos 3 \theta = 4 \cos^3\theta - 3 \cos \theta

Solutions:

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

& && \ \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}

2. This is one of the forms for \cos 2x .

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

3. Step 1: Use the cosine sum formula

\cos 3 \theta & = 4 \cos^3 \theta - 3 \cos \theta \\\cos (2 \theta + \theta) & = \cos 2 \theta \cos \theta - \sin 2 \theta \sin \theta

Step 2: Use double angle formulas for \cos 2\theta and \sin 2\theta

= (2 \cos^2 \theta - 1) \cos \theta - (2 \sin \theta \cos \theta) \sin \theta

Step 3: Distribute and simplify.

& = 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

Concept Problem Solution

Since the problem wants you to find:

\cos 120^\circ

You can simplify this into a familiar angle:

\cos (2 \times 60^\circ)

And then apply the double angle identity:

\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}

Practice

Simplify each expression so that it is in terms of \sin(x) and \cos(x) .

  1. \sin2x+\cos x
  2. \sin2x+\cos2x
  3. \sin3x+\cos2x
  4. \sin2x+\cos3x

Solve each equation on the interval [0,2\pi) .

  1. \sin(2x)=2\sin(x)
  2. \cos(2x)=\sin(x)
  3. \sin(2x)-\tan(x)=0
  4. \cos^2(x)+\cos(x)=\cos(2x)
  5. \cos(2x)=\cos(x)

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

  1. 2\cos^2(15^\circ)-1
  2. 2\sin(25^\circ)\cos(25^\circ)
  3. 1-2\sin^2(35^\circ)
  4. \cos^2(60^\circ)-\sin^2(60^\circ)
  5. 2\sin(125^\circ)\cos(125^\circ)
  6. 1-2\sin^2(32^\circ)

Image Attributions

Reviews

Email Verified
Well done! You've successfully verified the email address .
OK
Please wait...
Please wait...
ShareThis Copy and Paste

Original text