Double Angle Identities

The trigonometric double angle formulas give a relationship between the basic trigonometric functions applied to twice an angle in terms of trigonometric functions of the angle itself.

Tips for remembering the following formulas:

We can substitute the values \((2x)\) into the sum formulas for \(\sin\) and \(\cos.\) Using the 45-45-90 and 30-60-90 degree triangles, we can easily see the relationships between \(\sin x\) and \(\cos x\) by the lengths they represent. The several \(\cos 2x\) definitions can be derived by using the Pythagorean theorem and \(\tan x = \frac{\sin x}{\cos x}.\)

Double Angle Formulas

\[\begin{align} \sin 2x &= 2\sin x\cos x \\\\ \cos 2x &= \cos^2 x - \sin^2 x \\ &= 2\cos^2 x - 1 \\ &= 1 - 2\sin^2 x \\\\ \tan 2x &= \frac{2\tan x}{1 - \tan^2 x} \end{align} \]

From these formulas, we also have the following identities:

\[\begin{align} \sin^2 x&=\frac{1}{2}(1-\cos 2x)\\\\ \cos^2 x&=\frac{1}{2}(1+\cos 2x)\\\\ \sin x\cos x&=\frac{1}{2}\sin 2x\\\\ \tan^2 x&=\frac{1-\cos 2x}{1+\cos 2x}. \end{align} \]

We have the following formulas:

\[\begin{align} \sinh 2x &=2\sinh x\cosh x\\\\ \cosh 2x &=\cosh^2 x+\sinh^2 x\\ &=2\cosh^2 x-1\\ &=2\sinh^2 x+1\\\\ \tanh 2x &=\frac{2\tanh x}{1+\tanh^2 x}. \end{align} \]

If \(\sin\alpha=\frac{1}{2}\) and \(0<\alpha<\frac{\pi}{2},\) what is the value of \(\cos2\alpha?\)

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Given \(\sin\alpha=\frac{1}{2}\) and \(0<\alpha<\frac{\pi}{2},\) we have \[\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\left(\frac{1}{2}\right)^2}=\frac{\sqrt{3}}{2}.\] This implies \[\cos2\alpha=\cos^2\alpha-\sin^2\alpha=\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{1}{2}\right)^2=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}. \ _\square\]

Use the double angle identity to solve the following problems:

If \(\sin\theta=\frac{5}{13}\) and \(0<\theta<\frac{\pi}{2},\) what is the value of \(\sin2\theta?\)

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Given \(\sin\theta=\frac{5}{13}\) and \(0<\theta<\frac{\pi}{2},\) we have \[\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\left(\frac{5}{13}\right)^2}=\frac{12}{13}.\] This implies \[\sin2\theta=2\sin\theta\cos\theta =2\cdot \frac{5}{13} \cdot \frac{12}{13}=\frac{120}{169}. \ _\square\]

If \(\sin\alpha=\frac{3}{5}\) and \(\frac{\pi}{2}<\alpha<\pi,\) what is the value of \(\tan2\alpha?\)

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Given \(\sin\alpha=\frac{3}{5}\) and \(\frac{\pi}{2}<\alpha<\pi,\) we have \[\cos\alpha=-\sqrt{1-\sin^2\alpha}=-\sqrt{1-\left(\frac{3}{5}\right)^2}=-\frac{4}{5}.\] This implies \(\displaystyle{\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=-\frac{3}{4}}.\) Hence, it follows that \[\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{2\cdot\left(-\frac{3}{4}\right)}{1-\left(-\frac{3}{4}\right)^2}=-\frac{24}{7}. \ _\square\]

What is \(\theta\) \((0<\theta<\pi)\) that satisfies the following:

\[\frac{\tan2\theta+\sec2\theta-1}{\tan2\theta+\sec2\theta+1}=-1?\]

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Observe that the given equation can be rewritten as follows:

\[\begin{align} \frac{\tan2\theta+\sec2\theta-1}{\tan2\theta+\sec2\theta+1} &=\frac{\frac{\sin2\theta}{\cos2\theta}+\frac{1}{\cos2\theta}-1}{\frac{\sin2\theta}{\cos2\theta}+\frac{1}{\cos2\theta}+1}\\\\ &=\frac{\sin2\theta+1-\cos2\theta}{\sin2\theta+1+\cos2\theta}\\\\ &=\frac{2\sin\theta\cos\theta+2\sin^2\theta}{2\sin\theta\cos\theta+2\cos^2\theta}\\\\ &=\frac{\sin\theta(\cos\theta+\sin\theta)}{\cos\theta(\sin\theta+\cos\theta)}\\\\ &=\tan\theta=-1. \end{align}\]

Then, since \(0<\theta<\pi,\) \(\theta=\frac{3\pi}{4}. \) \(_\square\)

If \(\csc\alpha=\frac{2\sqrt{3}}{3}\) for \(0<\alpha<\frac{\pi}{2},\) what is \(\sin2\alpha+\cos2\alpha?\)

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Since \(\sin\alpha=\frac{1}{\csc\alpha}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\) and \(0<\alpha<\frac{\pi}{2},\) \(\alpha=\frac{\pi}{3}\) and \(\cos\alpha=\frac{1}{2}.\) Hence,

\[\begin{align} \sin2\alpha+\cos2\alpha &=2\sin\alpha\cos\alpha+\big( \cos^2\alpha - \sin^2\alpha \big) \\ &=2\cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}+ \left(\frac{1}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2 \\ &=\frac{\sqrt{3}}{2}+\frac{1}{4}-\frac{3}{4} \\ &=\frac{\sqrt{3}-1}{2}. \ _\square \end{align}\]

What is \(X\) in the following identity: \(\displaystyle{1+\cot^2\theta=\frac{X}{\sin^2 2\theta}}?\)

\[\begin{array} (a) \cos^2\theta &&&(b)\ 4\cos^2\theta &&&(c) \sin^2\theta &&&(d)\ 4\sin^2\theta \end{array}\]

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Since \(1+\cot^2\theta=\csc^2\theta\), we have

\[\begin{align} \csc^2\theta&=\frac{X}{\sin^22\theta}\\ \left(\frac{\sin2\theta}{\sin\theta}\right)^2&=X. \end{align}\]

Using the identity \(\sin2\theta=2\sin\theta\cos\theta\) gives

\[X=\left(\frac{\sin2\theta}{\sin\theta}\right)^2=\left(\frac{2\sin\theta\cos\theta}{\sin\theta}\right)^2=4\cos^2\theta.\]

Therefore, the answer is \((b).\) \(_\square\)

Use the double angle formulas to prove the identity

\[ \csc 2\theta - \cot 2\theta = \tan \theta.\]

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We have

\[\begin{align} \csc 2\theta - \cot 2\theta &= \dfrac{1}{\sin 2\theta} - \dfrac{\cos 2\theta}{\sin 2\theta} \\ \\ &= \dfrac{1-\cos 2\theta}{\sin 2\theta}. \end{align}\]

Using the identities \(1-\cos 2\theta = 2 \sin^2 \theta\) and \(\sin 2\theta = 2\sin \theta \cos \theta\) gives

\[\csc 2\theta - \cot 2\theta = \dfrac{1-\cos 2\theta}{\sin 2\theta} = \dfrac{2 \sin^2 \theta}{2 \sin \theta \cos \theta} = \dfrac{\sin \theta}{\cos \theta} = \tan \theta. \ _\square \]