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


$\sqrt{4A^2\cos^2 \dfrac {\theta} {2}}$ (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 $1 + \cos \theta$)

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

  • $\begingroup$ Thanks a lot sir. $\endgroup$ – Rider Mar 17 at 9:16

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