Trig identity

**Stroodle** · February 27th 2010, 04:02 AM

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

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

THanks

**Prove It** · February 27th 2010, 04:10 AM

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

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

THanks

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

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

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

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

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

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

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

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

$= \sin{2\theta}$

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

$= \sin{x}$ .

**Soroban** · February 27th 2010, 06:48 AM

Hello, Stroodle!

Another approach . . .

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

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

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

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

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

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

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

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