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$$\sin\dfrac \theta2 = \sin^2θ+\cos^2θ-1$$

$$\sinθ = 2\sin^2θ+\cos^2θ-1$$

Am I on the right track?

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Convert to what? The Dark Side? –  Pedro Tamaroff Aug 2 '13 at 19:56
    
It doesn't look like the right track. In your first equation, the right-hand side is $0$. A clearer description of what is asked for might be helpful. Perhaps the question asks you to express $\sin(\theta/2)$ in terms of trigonometric functions of $\theta$. –  André Nicolas Aug 2 '13 at 20:00
    
That was the question word to word. –  Little Jon Aug 2 '13 at 20:34
    
Do you want to express $\sin {\frac{\theta}2}$ in terms of something else? - the question is confusing, and the assertions you have put do not help without an explanation of where they come from and what they are supposed to demonstrate. –  Mark Bennet Aug 2 '13 at 21:12
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2 Answers

up vote 2 down vote accepted

In the event that you are being asked to solve for $\theta$ given the equation $$\sin\left(\frac \theta2\right) = \sin^2 \theta + \cos^2 \theta - 1,$$

...note that by the Pythagorean Theorem, we know that $\sin^2 \theta + \cos^2 \theta = 1.\;$

$$\begin{align}\sin\left(\frac \theta2\right) & = \underbrace{(\sin^2 \theta + \cos^2\theta)}_{= 1} - 1 \\ \\ & = 1 - 1 \\ \\& = 0\end{align}$$ $$\implies \frac{\theta}{2} = \sin^{-1}(0) \implies \theta = 2 \sin^{-1}(0)$$

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Clean and clear +1 –  Amzoti Aug 3 '13 at 1:09
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I can't begin to imagine how you got those identities... $\cos^2 + \sin^2 = 1$ so the RHS of the first is zero, for instance.

Here's a hint: start with the formula

$$\cos (2\psi) = \cos^2(\psi) - \sin^2(\psi).$$

  1. Can you use any trig identities to write the right-hand side only in terms of sines of $\psi$?
  2. Now set $\psi = \theta/2$. Can you rearrange terms to get a useful formula?
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