Double Angle and Half Angle Formulas of Trigonometric Functions
Overview
Back in the day, when the owners of sushi restaurants were high school students, there used to be formulas like angle addition, double angle, and sum-difference formulas in the curriculum, but nowadays, it’s understood they are not. All the following formulas can be derived from the sum formulas, so it’s better to learn the derivation process and derive them as needed rather than memorizing them all.
$$ \begin{align*} \sin ( \theta_{1} \pm \theta_{2}) &= \sin \theta_{1} \cos \theta_{2} \pm \sin \theta_{2} \cos \theta_{2} \\ \cos ( \theta_{1} \pm \theta_{2}) &= \cos \theta_{1} \cos\theta_{2} \mp \sin\theta_{1} \sin\theta_{2} \\ \tan ( \theta_{1} \pm \theta_{2}) &= \dfrac{\tan\theta_{1} \pm \tan\theta_{2}}{1 \mp \tan\theta_{1}\tan\theta_{2}} \end{align*} $$
Double Angle Formula
$$ \begin{align*} \sin 2\theta &=2\sin\theta\cos\theta \\ \cos 2\theta &=\cos^{2}\theta-\sin^{2}\theta=2\cos^{2}\theta-1=1-2\sin^{2}\theta \\ \tan 2\theta &=\dfrac{2\tan\theta}{1-\tan^{2}\theta} \end{align*} $$
Proof
The double angle formula is used to eliminate cosine when multiplying sine and cosine. Or, when terms related to angles are divided between $\theta$ and $2\theta$, it is used to adjust to $\theta$. It can be derived assuming $\theta_{1} = \theta_{2}=\theta$ from the sum formulas.
$\sin$
$$ \begin{cases} \sin(\theta+\theta)=\sin(\theta+\theta)=\sin 2\theta \\ \sin(\theta+\theta) = \sin \theta \cos \theta + \sin \theta \cos \theta = 2 \sin \theta \cos \theta \end{cases} $$
$$ \implies \sin 2\theta =2\sin\theta\cos\theta $$
$\cos$
$$ \begin{cases} \cos(\theta+\theta)=\cos(\theta+\theta)=\cos 2\theta \\ \cos(\theta+\theta)=\cos \theta \cos\theta - \sin\theta \sin\theta=\cos^{2}\theta-\sin^{2}\theta=2\cos^{2}\theta-1=1-2\sin^{2}\theta \end{cases} $$
$$ \implies \cos 2\theta=\cos^{2}\theta-\sin^{2}\theta=2\cos^{2}\theta-1=1-2\sin^{2}\theta $$
$\tan$
$$ \tan 2\theta =\dfrac{\sin 2\theta}{\cos 2\theta}=\dfrac{2\sin\theta\cos\theta}{\cos^{2}\theta-\sin^{2}\theta} $$
Dividing numerator and denominator by $\cos^{2}\theta$ results in the following.
$$ \tan 2\theta =\dfrac{2\tan\theta}{1-\tan^{2}\theta} $$
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Half-Angle Formula
$$ \begin{align*} \sin^{2} \dfrac{\theta}{2}&=\dfrac{1}{2}(1-\cos\theta) \\ \cos^{2} \dfrac{\theta}{2}&=\dfrac{1}{2}(\cos\theta+1) \\ \tan^{2} \dfrac{\theta}{2}&=\dfrac{1-\cos\theta}{1+\cos\theta} \end{align*} $$
Proof
The half-angle formula is useful in several calculations, such as reducing the order when integrating trigonometric functions. It can be derived using the cosine double angle formula.
$\sin$
$$ \begin{align*} &&\cos 2\theta &=1-2\sin^{2}\theta \\ \implies && 2\sin^{2}\theta&=1-\cos2\theta \\ \implies && \sin^{2}\theta&=\dfrac{1}{2}(1-\cos 2\theta) \end{align*} $$
Here, substituting $\theta$ with $\dfrac{\theta}{2}$ we obtain the following.
$$ \sin^{2} \dfrac{\theta}{2}=\dfrac{1}{2}(1-\cos\theta) $$
$\cos$
$$ \begin{align*} &&\cos 2\theta &=2\cos^{2}\theta-1 \\ \implies && 2\cos^{2}\theta&=\cos 2\theta+1 \\ \implies && \cos^{2}\theta&=\dfrac{1}{2}(\cos 2\theta+1) \end{align*} $$
Here, substituting $\theta$ with $\dfrac{\theta}{2}$ we obtain the following.
$$ \cos^{2} \dfrac{\theta}{2}=\dfrac{1}{2}(\cos\theta+1) $$
$\tan$
$$ \tan^{2} \dfrac{\theta}{2}=\dfrac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\dfrac{\frac{1}{2}(1-\cos\theta)}{\frac{1}{2}(\cos\theta+1)}=\dfrac{1-\cos\theta}{1+\cos\theta} $$
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