1. ## Trig identity

Hey there,

I have this problem that I've been able to solve, but only after what seems like an excessive amount of working. Was wondering if anyone knows a quicker way to go about it?

Given that $\displaystyle t=tan\left (\frac{x}{2}\right )$ show that $\displaystyle sin(x)=\frac{2t}{1+t^2}$

I've done this by first showing that $\displaystyle cos(x)=\frac{1-t^2}{1+t^2}$, but I'm sure there must be a shorter way, or something simple that I'm missing?

THanks

2. Originally Posted by Stroodle
Hey there,

I have this problem that I've been able to solve, but only after what seems like an excessive amount of working. Was wondering if anyone knows a quicker way to go about it?

Given that $\displaystyle t=tan\left (\frac{x}{2}\right )$ show that $\displaystyle sin(x)=\frac{2t}{1+t^2}$

I've done this by first showing that $\displaystyle cos(x)=\frac{1-t^2}{1+t^2}$, but I'm sure there must be a shorter way, or something simple that I'm missing?

THanks
$\displaystyle \frac{2t}{1 + t^2} = \frac{2\tan{\frac{x}{2}}}{1 + \tan^2{\frac{x}{2}}}$

$\displaystyle = \frac{\frac{2\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}} }{1 + \frac{\sin^2{\frac{x}{2}}}{\cos^2{\frac{x}{2}}}}$

$\displaystyle = \frac{\frac{2\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}} }{\frac{\cos^2{\frac{x}{2}} + \sin^2{\frac{x}{2}}}{\cos^2{\frac{x}{2}}}}$

$\displaystyle = \frac{\frac{2\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}} }{\frac{1}{\cos^2{\frac{x}{2}}}}$

$\displaystyle = \frac{2\sin{\frac{x}{2}}\cos^2{\frac{x}{2}}}{\cos{ \frac{x}{2}}}$

$\displaystyle = 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}$

Now let $\displaystyle \theta = \frac{x}{2}$, so that this becomes

$\displaystyle 2\sin{\theta}\cos{\theta}$

$\displaystyle = \sin{2\theta}$

$\displaystyle = \sin{\frac{2x}{2}}$

$\displaystyle = \sin{x}$.

3. Hello, Stroodle!

Another approach . . .

$\displaystyle \text{Given that: }\,t\:=\:\tan\tfrac{x}{2},\, \text{ show that: }\:\sin x \:=\:\frac{2t}{1+t^2}$

We have: .$\displaystyle \tan\tfrac{x}{2} \:=\:\frac{t}{1} \:=\:\frac{opp}{adj}$

Angle $\displaystyle \tfrac{x}{2}$ is in a right triangle with: .$\displaystyle opp = t,\;adj = 1$

Pythagorus says: .$\displaystyle hyp \,=\,\sqrt{1+t^2}$

Hence: .$\displaystyle \sin\tfrac{x}{2} \:=\:\frac{t}{\sqrt{1+t^2}},\;\;\cos\tfrac{x}{2} \:=\:\frac{1}{\sqrt{1+t^2}}$ .[1]

$\displaystyle \text{We know that: }\;\sin x \;=\;2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$ .[2]

Substitute [1] into [2]: .$\displaystyle \sin x \;=\;2\left(\frac{t}{\sqrt{1+t^2}}\right)\left(\fr ac{1}{\sqrt{1+t^2}}\right) \;=\;\frac{2t}{1+t^2}$