Question

if t = tan(x/2), -π<x<π, show that
cos x = (1-t^2)/(1+t^2) and sin x = (2t)/(1+t^2)

Answer 1

Answer on Question #53985 – Math – Calculus

Question

If $t = \tan \left( \frac{x}{2} \right)$,

then

$$\cos x = \frac{1 - t^2}{1 + t^2}, \quad \sin x = \frac{2t}{1 + t^2}.$$

Solution

We have

$$\frac{1 - t^2}{1 + t^2} = \frac{1 - \left(\tan \left(\frac{x}{2}\right)\right)^2}{1 + \left(\tan \left(\frac{x}{2}\right)\right)^2} = \frac{1 - \left(\frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}\right)^2}{1 + \left(\frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}\right)^2} = \frac{\left(\cos \left(\frac{x}{2}\right)\right)^2 - \left(\sin \left(\frac{x}{2}\right)\right)^2}{\left(\cos \left(\frac{x}{2}\right)\right)^2} = \frac{\left(\cos \left(\frac{x}{2}\right)\right)^2 + \left(\sin \left(\frac{x}{2}\right)\right)^2}{\left(\cos \left(\frac{x}{2}\right)\right)^2}$$ $$= \frac{\cos \left(2 \cdot \frac{x}{2}\right)}{\left(\cos \left(\frac{x}{2}\right)\right)^2} \cdot \frac{\left(\cos \left(\frac{x}{2}\right)\right)^2}{\left(\cos \left(\frac{x}{2}\right)\right)^2 + \left(\sin \left(\frac{x}{2}\right)\right)^2} = \frac{\cos x}{1} = \cos x;$$ $$\frac{2t}{1 + t^2} = \frac{2 \tan \left(\frac{x}{2}\right)}{1 + \left(\tan \left(\frac{x}{2}\right)\right)^2} = \frac{\frac{2 \sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}}{1 + \left(\frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}\right)^2} = \frac{2 \sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)} \cdot \frac{\left(\cos \left(\frac{x}{2}\right)\right)^2}{\left(\cos \left(\frac{x}{2}\right)\right)^2 + \left(\sin \left(\frac{x}{2}\right)\right)^2} = \frac{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{1} = \sin \left(2 \cdot \frac{x}{2}\right) = \sin x.$$

So

$$\cos x = \frac{1 - t^2}{1 + t^2}, \quad \sin x = \frac{2t}{1 + t^2},$$

where $t = \tan \left( \frac{x}{2} \right)$.

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