Proving the Half-Angle Formula for Tangent Using Trigonometric Identities

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The discussion focuses on proving the half-angle formula for tangent, specifically the equation tan(x/2) = (1 - cos(x))/sin(x). Participants suggest using the Weierstrass substitution or double angle identities to simplify the expression. There is a request for step-by-step guidance to understand how to apply these methods effectively. The conversation emphasizes the importance of showing previous attempts to facilitate better assistance. Overall, the thread highlights the need for clarity in problem-solving approaches in trigonometry.
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Homework Statement


tanx/2 = (1-cosx)/sinx


Homework Equations





The Attempt at a Solution


This is where i got on the right side
i don't know where to finish...(1-cosx)/sinx
 
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Well, are you familiar with Weierstrass substitution? They are also known as t-formula:
http://pear.math.pitt.edu/Calculus2/week3/3_2li5.html

Another method would be to use the Double angle identities:
\cos (2\theta) = 1- 2\sin^2 \theta and \sin (2\theta) = 2 \sin \theta \cos \theta.

Into those, let \theta = x/2, and put those back into what you have and simplify.
 
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I found that a lot of people have been telling me that but i don't understand how you use that to make the sides equal is there a way you can show me step by step
 
No really I can't ! You didn't even tell me which of the two methods I posted you want help with. Show us what you have tried to do yourself, so we can help you get over what you're stuck with.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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