The question is as such: If $u = \tan(x/2)$, show that $\cos(x/2) = \frac{1}{\sqrt{1+u^2}}$ and $\sin(x/2) = \frac{u}{\sqrt{1+u^2}}$. I have been mulling over this question for quite some time, but I am not sure how to prove it, as by employing the substitution identity $\tan(x) = \frac{u}{1-u^2}$ I only end up with a more complex equation rather than simplifying anything. By employing the classic LHS and RHS proof, I also get no where because I am unable to reduce the resultant equations, such as
$$\cos(x/2) = \frac{1}{\sqrt{\dfrac{2\tan(x)+2u}{\tan(x)}}}$$
As this is not exactly Weierstrass Substitution, I don't think that method could be applied to formulate a solution. Any pointers or guidance to this question would be helpful.
$\endgroup$ 12 Answers
$\begingroup$Method 1.
$\tan \frac {x}{2} = \frac {\sin\frac x2}{\cos\frac x2}$
$\sin\frac x2 = \alpha u\\\cos \frac x2 = \alpha$
$\sin^2 \frac x2 + \cos^2 \frac x2 = 1\\ \alpha^2u^2 + \alpha^2 = 1\\ \alpha = \frac {1}{\sqrt{u^2 + 1}}$
$\sin\frac x2 = \frac {u}{\sqrt{u^2+1}}\\\cos \frac x2 = \frac {u}{\sqrt{u^2+1}}$
Method 2.
An appeal to geometry
$\sin \frac x2 = \frac {\text{opposite}}{\text{hypotenuse}}\\ \cos \frac x2 = \frac {\text{adjacent}}{\text{hypotenuse}}$
Method 3.
$\tan^2 \frac {x}{2} + 1 = u^2 + 1 = \sec^2 \frac x2 = \frac {1}{\cos^2 \frac x2}$
$\endgroup$ $\begingroup$Recall that $\sec^{2}(.)=1+\tan^{2}(.)$.
Then recall that $\sin(.)=\tan(.)\cos(.)$
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