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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.

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  • $\begingroup$ Note you can use something like $y=x/2$ to change your question to $u = \tan\left(y\right)$ is given and to show $\cos\left(y\right) = \frac{1}{\sqrt{1 + u^2}}$ and $\sin\left(y\right) = \frac{u}{\sqrt{1 + u^2}}$. In other words, since $x$ is only ever used as $x/2$, this doesn't really have anything to do with "half angle trigonometric identities", as your question title states. $\endgroup$ – John Omielan Feb 7 at 23:43
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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

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$\sin \frac x2 = \frac {\text{opposite}}{\text{hypotenuse}}\\ \cos \frac x2 = \frac {\text{adjacent}}{\text{hypotenuse}}$

Method 3.

$\tan^2 \frac {x}{2} + 1 = u^2 + 1 = \sec^2 \frac x2 = \frac {1}{\cos^2 \frac x2}$

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Recall that $\sec^{2}(.)=1+\tan^{2}(.)$.

Then recall that $\sin(.)=\tan(.)\cos(.)$

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