2
$\begingroup$

$\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 $1 + \cos \theta$)

$\endgroup$
5
  • $\begingroup$ $\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
5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks a lot sir. $\endgroup$ – Rider Mar 17 at 9:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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