Hi, I was solving an integral and got this substitution

t= tg(x/2) then sin(x) = (2t)/(1+t^2) and cos(x) = (1-t^2)/(1+t^2)

How did I get sin(x) = (2t)/(1+t^2)?

I know that t = sinx/(cosx+1) and sinx = [(cosx+1)*t]

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- Jun 18th 2013, 12:27 PM #1

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- Jun 18th 2013, 01:54 PM #2

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## Re: How to get sin(x)?

Unfortunately, you start off by saying you "

**know**" something that isn't true!

$\displaystyle tan(x)= \frac{sin(x)}{cos(x)}$ so $\displaystyle tan(x(/2)= \frac{sin(x/2)}{cos(x/2)}= t$. We have the trig identities $\displaystyle sin(x/2)= \sqrt{(1/2)(1- cos(x)}$ and $\displaystyle cos(x/2)= \sqrt{(1/2)(1+ cos(x})}$ so $\displaystyle t= \sqrt{\frac{1- cos(x)}{1+ cos(x)}}$, $\displaystyle t^2= \frac{1- cos(x)}{1+ cos(x)}$, $\displaystyle t^2(1+ cos(x))= 1- cos(x)$, $\displaystyle (t^2+ 1)cos(x)= 1- t^2$ so that $\displaystyle cos(x)= \frac{1- t^2}{1+ t^2}$.

- Jun 18th 2013, 03:53 PM #3

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## Re: How to get sin(x)?

An alternative route through this is to start with the identity

$\displaystyle \tan(2A)= \frac{2\tan(A)}{1-\tan^{2}(A)}$

from which

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

Let $\displaystyle t=\tan(x/2), $ and we have $\displaystyle \tan x = \frac{2t}{1-t^{2}}.$

Now draw a right-angled triangle with $\displaystyle \text{ 'base angle' } x,\text{ 'opposite' }2t\text{ and 'adjacent' }1-t^{2}.$ Use Pythagoras to calculate the hypotenuse $\displaystyle \text{ }1+t^{2}\text{ },$ and so read off the expressions for $\displaystyle \sin x \text{ and } \cos x.$

- Jun 27th 2013, 11:42 AM #4

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