9.3: Double-Angle Identities

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Page ID: 203495

Roy Simpson
Cosumnes River College

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The Double Angle Identities

Theorem: Double-Angle Identities

\[ \begin{array}{rcl}
\sin\left( 2\theta \right) & = & 2 \sin\left( \theta \right) \cos\left( \theta \right) \\[6pt] \cos\left( 2\theta \right) & = & \cos^2\left( \theta \right) - \sin^2\left( \theta \right) \\[6pt] & = & 2\cos^2\left( \theta \right) - 1 \\[6pt] & = & 1 - 2 \sin^2\left( \theta \right) \\[6pt] \tan\left( 2\theta \right) & = & \dfrac{2 \tan\left( \theta \right)}{1 - \tan^2\left( \theta \right)} \\[6pt] \end{array} \nonumber \]

Proof: Deriving the Double-Angle Identity for sine begins with the Sum Identity,\[\sin(\alpha+\beta)=\sin \left(\alpha\right) \cos \left(\beta\right)+\cos \left(\alpha\right) \sin \left(\beta\right). \nonumber \]If we let \(\alpha=\beta=\theta\), then we have\[\begin{array}{rrcl}
& \sin(\theta+\theta) & = & \sin \left(\theta\right) \cos \left(\theta\right)+\cos \left(\theta\right) \sin \left(\theta\right) \\[6pt] \implies & \sin(2\theta) & = & 2\sin \left(\theta\right) \cos \left(\theta\right) \\[6pt] \end{array} \nonumber \]Deriving the Double-Angle Identity for cosine gives us three options. First, starting from the Sum Identity,\[\cos(\alpha+\beta)=\cos \left(\alpha\right) \cos \left(\beta\right) - \sin \left(\alpha\right) \sin \left(\beta\right),\nonumber \]and letting \(\alpha=\beta=\theta\), we have\[\begin{array}{rrcl}
& \cos(\theta+\theta) & = & \cos \left(\theta\right) \cos \left(\theta\right)-\sin \left(\theta\right) \sin \left(\theta\right) \\[6pt] \implies & \cos(2\theta) & = & \cos^2 \left(\theta\right) - \sin^2 \left(\theta\right) \\[6pt] \end{array} \nonumber \]Using the Pythagorean Identities, we can expand this Double-Angle Identity for cosine and get two more variations. The first variation is\[\begin{array}{rcl}
\cos(2\theta) & = & \cos^2 \left(\theta\right) - \sin^2 \left(\theta\right) \\[6pt] & = & \left(1-\sin^2 \left(\theta\right)\right)-\sin^2 \left(\theta\right) \\[6pt] & = & 1 - 2 \sin^2\left( \theta \right) \\[6pt] \end{array}\nonumber \]The second variation is\[\begin{array}{rcl}
\cos(2\theta) & = & \cos^2 \left(\theta\right)-\sin^2 \left(\theta\right) \\[6pt] & = & \cos^2 \left(\theta\right)-\left(1-\cos^2 \left(\theta\right)\right) \\[6pt] & = & 2 \cos^2 \left(\theta\right)-1 \\[6pt] \end{array} \nonumber \]Similarly, to derive the Double-Angle Identity for the tangent, replacing \(\alpha=\beta=\theta\) in the sum formula gives\[\begin{array}{rrcl}
& \tan(\alpha+\beta) & = & \dfrac{\tan \left(\alpha\right)+\tan \left(\beta\right)}{1-\tan \left(\alpha\right) \tan \left(\beta\right)} \\[6pt] \implies & \tan(\theta+\theta) & = & \dfrac{\tan \left(\theta\right)+\tan \left(\theta\right)}{1-\tan \left(\theta\right) \tan \left(\theta\right)} \\[6pt] \implies & \tan(2\theta) & = & \dfrac{2\tan \left(\theta\right)}{1-\tan^2 \left(\theta\right)} \\[6pt] \end{array} \nonumber \]

Caution: Don't Factor Out of Functions!

\[ \text{trig}(2 \theta ) \neq 2 \text{trig}(\theta) \nonumber \]

Finding Exact Values of Trigonometric Functions Involving Double Angles

Example \( \PageIndex{ 1 } \): Using double angles with triangles

Let's consider a right triangle with an interior unkown angle of \(\theta\), where we are given two sides. Find the values of \( \sin(2 \theta) \), \( \cos(2\theta)\), and \(\tan(2\theta)\).

Example \(\PageIndex{2}\): A popular style of problem revisited.

If \(\alpha\) is a Quadrant III angle with \(\sin(\alpha) = -\frac{12}{13}\), and \(\beta\) is a Quadrant IV angle with \(\tan(\beta) = -5\), find the following

\(\sin(2 \alpha)\)
\(\cos(2 \alpha)\)
\(\cos(2 \beta)\)

Revisiting Proofs of Identities

Example \(\PageIndex{3}\): We can now do even more proofs!

Prove each identity.

\(\cos^4 \left(\theta\right)-\sin^4 \left(\theta\right) =\cos\left(2\theta\right)\)
\(\cos(2\theta)\cos \theta=\cos^3 \left(\theta\right)−\cos \left(\theta\right) \sin^2 \left(\theta\right)\)

Revisiting Graphs of Trigonometric Functions

Example \(\PageIndex{4}\): We can graph more interesting functions!

Graph the function.\[ g(x) = \dfrac{6\tan\left( \frac{\pi}{4}x \right)}{1 - \tan^2\left( \frac{\pi}{4} x \right)} \nonumber \]

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