Prove tan(θ / 2) = sin θ / (1 + cos θ) for θ in quadrant 1. I've looked at http://openstudy.com/study#/updates/4f78f650e4b0ddcbb89e7fe4 and http://openstudy.com/study#/updates/511ffe90e4b03d9dd0c5162b but I still can't figure it out!

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|dw:1440229429187:dw| I can get to here but I don't know how to prove it from this point on.
Multiply numerator and denominator with sqrt( 1+cos theta). |dw:1440231020172:dw| simply this to get answer Source: http://www.mathskey.com/question2answer/
i haven't looked at the other thread but this seems the most obvious way to take it \(\large tan\frac{θ }{ 2} = \frac{sin θ}{ 1 + cos θ} \) so you are proving \(\large \frac{sin\frac{θ }{ 2}}{cos\frac{θ }{ 2}} = \frac{2sin \frac{θ }{ 2} cos \frac{θ }{ 2}}{ 1 + (1 - 2 sin^2 \frac{θ }{ 2})} \) now, simplify the RHS

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as for the significance of Q1 |dw:1440235915730:dw|
How do I simplify the right hand side? Sorry, I'm having a lot of trouble with this.
|dw:1440237322888:dw|
\[(2-2\sin^2 \theta)^2\] Is this right?
i will latex it so you see it clearly however, before doing so, i assume you know the double angle formula. that is how you break this down in the first place https://gyazo.com/943d7627c9c451b6b7653f69176505d6 yes? then, for the *denominator* \(1 + (1-2 sin^2 \frac{\theta}{2}) = \) \(1 + 1-2 sin^2 \frac{\theta}{2} = \) \(2-2 sin^2 \frac{\theta}{2} = \) \(2(1- sin^2 \frac{\theta}{2}) = !!!\)
cos(theta)+1 or \[2 \cos^2(\theta/2)\] ?
https://gyazo.com/5d5b168435f599a323eefa614185c8ed \(\checkmark\) now go back and compare numerator and denominator
https://gyazo.com/af818b18771bbb89f7bff94ece2513fe
\[(2\sin (\theta/2) \cos (\theta/2)) /2\cos^2(θ/2)\] \[(\sin \theta/2)/(\cos \theta/2)\] \[=\tan (\theta/2)\] Is this right?
\(\huge \checkmark\)
Thank you so much! I finally feel like I get it now!
cool!

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