Double-Angle Formulae

In the previous concept, we used addition and subtraction formulae for trigonometric functions. Now, we take another look at those same formulae. The double-angle formulae are a special case of the sum formulae, where $\alpha = \beta$. In other words, they allow us to find the trigonometric function of an angle that is double a special angle. Formulae can be derived to find the sine, cosine, and tangent in such cases, and these formulae are useful in simplifying trigonometric expressions. 

Deriving the double-angle formula for sine begins with the sum formula that was introduced previously: $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$.

If we let $\alpha = \beta = \theta$, then we have:

$\displaystyle{ \begin{aligned} \sin(\theta + \theta) &= \sin \theta \cos \theta + \cos \theta \sin \theta \\ \sin(2\theta) &= 2\sin \theta \cos \theta \end{aligned} }$

The double-angle formula for cosine can be derived similarly, and is:

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

Notice that we can apply the Pythagorean identities to get two more variations of the cosine formula:

$\displaystyle{ \begin{aligned} \cos{\left(2\theta \right)} &= \cos^2 \theta - \sin^2 \theta \\ &= \left(1- \sin^2 \theta \right) - \sin^2 \theta \\ &= 1- 2\sin^2 \theta \end{aligned} }$

$\displaystyle{ \begin{aligned} \cos{\left(2\theta\right)} &= \cos^2 \theta - \sin^2 \theta \\ &= \cos^2 \theta - \left(1- \cos^2 \theta \right) \\ &= 2 \cos^2 \theta -1 \end{aligned} }$

Similarly, to derive the double-angle formula for tangent, replacing $\alpha = \beta = \theta$ in the sum formula gives

$\displaystyle{ \begin{aligned} \tan{\left(\alpha + \beta \right)} &= \frac{ \tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\ \tan{\left(\theta + \theta \right)} &= \frac{ \tan \theta + \tan \theta}{1 - \tan \theta \tan \theta} \\ \tan{\left(2\theta \right)} &= \frac{ 2\tan \theta }{1 - \tan^2 \theta} \end{aligned} }$

The double-angle formulae are summarized as follows:

• $\sin{\left(2\theta\right)} = 2\sin \theta \cos \theta $
• $\cos{\left(2\theta\right)} = \cos^2 \theta - \sin^2 \theta$
• $\cos{\left(2\theta\right)}= 1- 2\sin^2 \theta$
• $\cos{\left(2\theta\right)} = 2 \cos^2 \theta -1$
• $\displaystyle{\tan{\left(2\theta\right)} = \frac{ 2\tan \theta }{1 - \tan^2 \theta} }$

Example

Find $\sin(60^{\circ})$ using the function $\sin(30^{\circ})$.

Notice that $30^{\circ}$ is a special angle, and $60^{\circ}$ is twice its value. We will apply the double-angle formula for sine: $\sin(2\theta) = 2\sin \theta \cos \theta $.

In this case, we let $\theta = 30^{\circ}$ because we want to solve $\sin(60^{\circ}) = \sin(2\cdot30^{\circ})$. Substitute $\theta = 30^{\circ}$ into the formula:

$\displaystyle{ \begin{aligned} \sin{\left(60^{\circ}\right)} &= \sin{\left(30^{\circ} + 30^{\circ}\right)} \\ &= 2\sin (30^{\circ}) \cos (30^{\circ}) \end{aligned} }$

From the unit circle, we can identify that $\displaystyle{\sin (30^{\circ}) = \frac{1}{2}}$ and $\displaystyle{\cos (30^{\circ}) = \frac{\sqrt{3}}{2}}$. Substitute these values into the formula:

$\displaystyle{ \sin{\left(60^{\circ}\right)} = 2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) }$

Simplify:

$\displaystyle{ \begin{aligned} \sin{\left(60^{\circ}\right)} &= 2\left(\frac{\sqrt{3}}{4}\right) \\ &= \frac{\sqrt{3}}{2} \end{aligned} }$

Half-Angle Formulae

The half-angle formulae can be derived from the double-angle formulae. They are useful for finding the trigonometric function of an angle $\theta$ which is half of a special angle $\alpha$ (in other words, $\displaystyle{\theta = \frac{\alpha}{2}}$). The half-angle formulae are as follows:

• $\displaystyle{ \sin{\left(\frac{\alpha}{2}\right)} = \pm \sqrt{\frac{1- \cos \alpha}{2}} }$
• $\displaystyle{ \cos{\left(\frac{\alpha}{2}\right)} = \pm \sqrt{\frac{1+ \cos \alpha}{2}} }$
• $\displaystyle{ \tan{\left(\frac{\alpha}{2}\right)} = \pm \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} }$
• $\displaystyle{ \tan{\left(\frac{\alpha}{2}\right)} = \frac{\sin \alpha}{1 + \cos \alpha} }$
• $\displaystyle{ \tan{\left(\frac{\alpha}{2}\right)} = \frac{1 - \cos \alpha}{\sin \alpha} }$

Although some of the formulas have a $\pm$ sign, only one sign is applied. The sign that is applied in each case depends on the magnitude of the angle. Recall that different signs are applied to trigonometric functions that fall in each of the four quadrants (according to the mnemonic rule "A Smart Trig Class").

Example

Find $\sin(15^{\circ})$ using a half-angle formula.

Recall that $30^{\circ}$ is a special angle, and note that $\displaystyle{15^{\circ} = \frac{30^{\circ}}{2}}$. We can thus apply the half-angle formula with $\alpha = 30^{\circ}$:

$\displaystyle{ \begin{aligned} \sin{\left(15^{\circ}\right)} &= \sin{\left(\frac{30^{\circ}}{2}\right)} \\ &= \pm \sqrt{\frac{1- \cos(30^{\circ})}{2}} \end{aligned} }$

Substitute $\displaystyle{\cos{\left(30^{\circ}\right)} = \frac{\sqrt{3}}{2} }$, which is provided in the unit circle, and simplify:

$\displaystyle{ \begin{aligned} \sin{\left(15^{\circ}\right)} &= \sqrt{\frac{1- \frac{\sqrt{3}}{2}}{2}} \\ &= \sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}} \\ &= \sqrt{\frac{2-\sqrt{3}}{4}} \\ &= \frac{\sqrt{2-\sqrt{3}}}{4} \\ \end{aligned} }$

Notice that we used only the positive root because $15^{\circ}$$$ falls in the first quadrant and $\sin(15^{\circ})$ is therefore positive.