Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

$$\sinθ = 2\sin^2θ+\cos^2θ-1$$

Am I on the right track?

share|cite|improve this question
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
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 $\sin^2 \theta + \cos^2 \theta = 1.\;$

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

share|cite|improve this answer
Clean and clear +1 – Amzoti Aug 3 '13 at 1:09

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?
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.