Author Topic: Trig identity question (Read 905 times) Tweet Share

Stroodle · « **on:** February 27, 2010, 11:01:43 pm »

Hey, I have this problem that I've been able to solve, but only after a crap load of working. Was wondering if anyone knows a quicker way:

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

QuantumJG · « **Reply #1 on:** February 28, 2010, 12:13:11 am »

Quote from: Stroodle on February 27, 2010, 11:01:43 pm

Hey, I have this problem that I've been able to solve, but only after a crap load of working. Was wondering if anyone knows a quicker way:

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

At the moment your method sounds like the best.

So tan( $\dfrac{x}{2}$ ) = t

tan(x) = $\dfrac{sin(x)}{cos(x)}$ = $\dfrac{2t}{1-t^{2}} \rightarrow sin(x) = cos(x) \dfrac{2t}{1-t^{2}}$

Now,

$cos(x) = 1 - 2sin^{2}(\dfrac{x}{2})$

And,

$sin^{2}(\dfrac{x}{2}) = t^{2}cos^{2}(\dfrac{x}{2})$

$\rightarrow sin^{2}(\dfrac{x}{2}) = t(1 - sin^{2}(\dfrac{x}{2}))$

$\rightarrow sin^{2}(\dfrac{x}{2}) = \dfrac{t^(2)}{1+t^(2)}$

$\therefore cos(x) = \dfrac{1 + t^{2}}{1 + t^{2}} - \dfrac{2t^{2}}{1 + t^{2}}$

$\therefore cos(x) = \dfrac{1 - t^{2}}{1 + t^{2}}$

$\therefore sin(x) = \dfrac{1 - t^{2}}{1 + t^{2}} \times \dfrac{2t}{1-t^{2}}$

This is basically your method, right?

Stroodle · « **Reply #2 on:** February 28, 2010, 12:23:55 am »

Awesome, also just found another simple way if anyones interested:

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

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

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

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

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

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

Then letting $\theta = \frac{x}{2}$ , this becomes

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

$= \sin{2\theta}$

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

$= \sin{x}$

QuantumJG · « **Reply #3 on:** February 28, 2010, 09:08:39 am »

Quote from: Stroodle on February 28, 2010, 12:23:55 am

Awesome, also just found another simple way if anyones interested:

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

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

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

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

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

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

Then letting $\theta = \frac{x}{2}$ , this becomes

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

$= \sin{2\theta}$

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

$= \sin{x}$

This is a much simpler solution.

Mao · « **Reply #4 on:** February 28, 2010, 11:07:43 am »

Given $t = \tan(x/2)$ , by the double angle formula for tan, $\tan(x) = \frac{2t}{1-t^2}$

Interpreting this in a geometric sense, let the O = 2t, A = 1-t^2, computing pythagoras:

$H^2 = O^2 + A^2 = 4t^2 + 1 - 2t^2 + t^4 = t^4 + 2t^2 + 1 = (t^2 + 1)^2$

$\implies H = t^2 + 1$

$\sin(x) = \frac{O}{H} = \frac{2t}{t^2 + 1}$

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