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If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as

a) $\frac{1+t^2}{1-t^2}$

b) $\frac{2t}{1+t^2}$

c) $\frac{1-t^2}{1+t^2}$

d) $\frac{2t}{1-t^2}$

Attempt: I tried using the half angle formula but it just leaves me with an expression in terms of $\tan{x}$'s and I don't know how to go to $t$, let alone express $\cos{x}$ in $t$.

Parseval
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4 Answers4

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Note that $$\sin^{2}\left(\frac{x}{2}\right) = \frac{1-\cos x}{2},\quad \cos^{2}\left(\frac{x}{2}\right) = \frac{1+\cos x}{2},$$

so $$\tan^{2}\left(\frac{x}{2}\right) =\frac{1-\cos x}{1+\cos x}.$$

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

we get

\begin{align*} t^{2} &= \frac{1-\cos x}{1+\cos x}\\ \implies t^{2} + t^{2}\cos x &= 1 - \cos x\\ \implies (1+t^{2})\cos x &= 1-t^{2}\\ \implies \cos x &= \frac{1-t^{2}}{1+t^{2}} \end{align*}

*Edited based on Arnaldo's suggestion.

DMcMor
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  • why didn't you go straight from $\tan^{2}\left(\frac{x}{2}\right) =\frac{1-\cos x}{1+\cos x}\to t^2 =\frac{1-\cos x}{1+\cos x}$ and just isolated $\cos x$? – Arnaldo May 09 '17 at 19:30
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    @Arnaldo because when I worked this out for something else recently I started with $\cos x$ and manipulated it until I got something in terms of $\tan\left(\frac{x}{2}\right)$ so I just copy pasted here, but yes that is a much cleaner argument. – DMcMor May 09 '17 at 20:19
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Following your idea you can do:

$$\tan x=\frac{2\tan (x/2)}{1-\tan^2(x/2)}=\frac{2t}{1-t^2}$$

so,

$$\frac{\sin x}{\cos x}=\frac{2t}{1-t^2}\to \frac{\sin^2 x}{\cos^2 x}=\frac{4t^2}{(1-t^2)^2}$$

using proportion properties we get:

$$\frac{\cos^2 x}{\sin^2 x+\cos^2 x}=\frac{(1-t^2)^2}{4t^2+(1-t^2)^2}\to \cos^2x=\frac{(1-t^2)^2}{(1+t^2)^2}\to \cos x=\frac{1-t^2}{1+t^2}$$

Arnaldo
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Since you are given four choices, all you have to do is eliminate the wrong ones. If $x=0$, then $t=0$ and $\cos(x)=1$. The formulas in b) and d) give $0$ which is wrong. If $x=\pi/2$, then $t=1$ and the formula in a) gives division by $0$. The only choice left is c).

Somos
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  • Wonderful! This is what I needed, a quick way of eliminating the wrong ones instead of spending 30 minutes deriving the correct answer. – Parseval May 09 '17 at 20:30
  • @Parseval: If that were what you wanted, you could say that and avoid people spent their time trying to write a more constructive answer. – Arnaldo May 09 '17 at 21:27
  • Sorry, but the wording of the problem as multiple choice strongly hinted that this was a test question with no proof being needed. Some people saw this right away. Others didn't. – Somos May 09 '17 at 21:46
  • @Arnaldo, I think many people actually enjoy writing down a more constructive answer, myself included. Also, this can benefit future students and other people that want to know how to constructively prove that c) is the correct answer. – Parseval May 10 '17 at 05:09
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In my opinion the simplest solution is as follows:
$\tan(\frac{x}{2})=t\implies\cos(\frac{x}{2})=\frac{1}{\sqrt{t^2+1}}$;
this can be seen by constructing a triangle with opposite side of angle $\frac{x}{2}$ equal to $t$, and adjacent side equal to $1$.

Now, it is known $\cos(x)=2\cos^2(\frac{x}{2})-1$.
$\cos(\frac{x}{2})=\frac{1}{\sqrt{t^2+1}}\implies\cos^2(\frac{x}{2})=\frac{1}{{t^2+1}}$.
Thus $\cos(x)=\frac{2}{{t^2+1}}-1=\frac{1-t^2}{1+t^2}$.

gfppoy
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