Sin/Cos Identity

**DjamaalRobinson** · September 1st 2008, 02:47 PM

How would i prove this identity
(cos2θ)/(1+sin2θ) = (cotθ-1)/(cotθ+1)

**Soroban** · September 1st 2008, 03:10 PM

Hello, DjamaalRobinson!

Prove: . $\frac{\cos2\theta}{1+\sin2\theta} \;=\;\frac{\cot\theta - 1}{\cot\theta + 1}$

The right side is: . $\frac{\dfrac{\cos\theta}{\sin\theta} - 1}{\dfrac{\cos\theta}{\sin\theta} + 1}$

Multiply by $\frac{\sin\theta}{\sin\theta}\!:\;\;\frac{\sin\the ta\left(\dfrac{\cos\theta}{\sin\theta} - 1\right)}{\sin\theta\left(\dfrac{\cos\theta}{\sin\ theta} + 1\right)} \;=\;\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}$

Multiply by $\frac{\cos\theta + \sin\theta}{\cos\theta + \sin\theta}\!:$ . $\frac{\cos\theta+\sin\theta}{\cos\theta+\sin\theta } \cdot\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}$ . $=\;\frac{\cos^2\!\theta - \sin^2\!\theta}{\cos^2\!\theta + 2\sin\theta\cos\theta + \sin^2\!\theta}$

We have: . $\frac{\overbrace{\cos^2\!\theta - \sin^2\!\theta}^{\text{This is }\cos2\theta}}{\underbrace{\sin^2\!\theta + \cos^2\!\theta}_{\text{This is 1}} + \underbrace{2\sin\theta\cos\theta}_{\text{This is }\sin2\theta}} \;=\;\frac{\cos2\theta}{1 + \sin2\theta}$ . . . ta-DAA!

**Krizalid** · September 1st 2008, 04:41 PM

$\begin{aligned}<br /> \frac{\cos 2\theta }{1+\sin 2\theta }&=\frac{(\cos \theta +\sin \theta )(\cos \theta -\sin \theta )}{1+2\sin \theta \cos \theta } \\ <br /> & =\frac{(\cos \theta +\sin \theta )(\cos \theta -\sin \theta )}{(\cos \theta +\sin \theta )^{2}} \\ <br /> & =\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=\frac{\cot \theta -1}{\cot \theta +1}.\quad\blacksquare<br /> \end{aligned}$

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