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Sine, Cosine and Tangent of a Half Angle

Mathematics Grade 10 Difficulty 2/5

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Sine, Cosine, and Tangent of a Half Angle

🎯 Big Idea

Half-angle formulas allow us to find the exact values of trigonometric functions for angles that are half of a known angle. For example, if we know the trig values for $60°$, we can find the exact values for $30°$. More powerfully, we can find exact values for angles like $22.5°$ (half of $45°$) or $15°$ (half of $30°$) that aren't standard angles on the unit circle.

📐 Deriving the Half-Angle Formulas

We start from the double-angle formula for cosine:

$$\cos(2\alpha) = 1 - 2\sin^2(\alpha)$$ $$\cos(2\alpha) = 2\cos^2(\alpha) - 1$$

Now, let $\theta = 2\alpha$, which means $\alpha = \frac{\theta}{2}$. Substituting:

For Sine

Starting with: $\cos\theta = 1 - 2\sin^2\!\left(\frac{\theta}{2}\right)$

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

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

$$\sin\!\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos\theta}{2}}$$

For Cosine

Starting with: $\cos\theta = 2\cos^2\!\left(\frac{\theta}{2}\right) - 1$

$2\cos^2\!\left(\frac{\theta}{2}\right) = 1 + \cos\theta$

$\cos^2\!\left(\frac{\theta}{2}\right) = \frac{1 + \cos\theta}{2}$

$$\cos\!\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos\theta}{2}}$$

For Tangent

Since $\tan\!\left(\frac{\theta}{2}\right) = \frac{\sin(\theta/2)}{\cos(\theta/2)}$, we get:

$$\tan\!\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}$$

There are also two very useful sign-free alternative forms:

$$\tan\!\left(\frac{\theta}{2}\right) = \frac{\sin\theta}{1 + \cos\theta} = \frac{1 - \cos\theta}{\sin\theta}$$

These forms are derived by multiplying numerator and denominator by appropriate conjugates and using the identity $\sin^2\theta + \cos^2\theta = 1$.

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