# How to solve this step to get this

$$\sqrt{2A^2(1+\cos \theta)}$$

to

$$\sqrt{4A^2\cos^2 \dfrac {\theta} {2}}$$ (Here, divided by $$2$$ is only under $$\theta$$)

I have solved till : $$\dfrac {2A\sqrt{1+\cos\theta}}{2}$$ (divided by $$2$$ is whole under $$1 + \cos \theta$$)

• $\cos(\theta)=\cos (2\theta/2)=2\cos^2(\theta/2)-1\implies 2A^2(1+\cos(\theta))=4A\cos^2(\theta/2)$ – Koro Mar 15 at 10:16
• @Koro Thanks. How did you solve the 3rd step ? – Rider Mar 15 at 10:18
• Your "till" result is wrong. – user65203 Mar 15 at 10:52
• @Koro: I think you're missing a factor of $A$ when you went to the third step (it should be $4A^2)$. – bjcolby15 Mar 15 at 11:21
• @bjcolby15: Ooops, I missed that. You are right. It should be $4A^2$ instead of $4A$ in last step. – Koro Mar 15 at 11:29

Hint: Using the trigonometric identity (a variation of the half-angle formula) $$\cos^2 \dfrac {\theta}{2} = \dfrac {1 + \cos \theta}{2}$$ rewrite this expression in terms of $$1 + \cos \theta$$ only, and the answer will be immediate.