What is the Triganomic Identity for Simplifying Trig Expressions?

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The discussion focuses on simplifying the trigonometric expression \(\frac{1-\cos\theta}{\sin\theta} = \tan \frac{a}{2}\), with the assumption that \(a\) refers to \(\theta\). Key identities are introduced, including \(\cos^2{u} = \frac{1}{2}(1+ \cos{2u})\) and \(\sin^2{u} = \frac{1}{2}(1 - \cos{2u})\), to facilitate simplification. The transformation of \(\tan^2{u}\) is also mentioned, leading to a more manageable form of the expression. Additionally, the relationship \(\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\) is highlighted, along with the identity \(1-\cos\theta=2\sin^{2}\frac{\theta}{2}\). This discussion provides a methodical approach to simplifying trigonometric expressions using established identities.
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Nothing special, I'm just not clicking on this question

\frac {1-\cos\theta}{\sin\theta} = \tan \frac {a}{2}

cheers
 
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I'm going to assume by a you mean \theta.

In that case, use

\cos^2{u} = \frac{1}{2}(1+ \cos{2u})

and

\sin^2{u} = \frac{1}{2}(1 - \cos{2u})

\Longrightarrow \tan^2{u} = \frac{1 - \cos{2u}}{1 + \cos{2u}}

to get rid of the annoying \frac{\theta}{2}, and then multiply by an appropriate version of 1.
 
\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}

1-\cos\theta=2\sin^{2}\frac{\theta}{2}


Daniel.
 
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