- Thread starter
- #1
Hi suvadip,Can it be proved?
\(\displaystyle \left(\frac{-2\sin A}{1-\cos A}\right)\cos\left(\frac{A}{2}\right)\tan^{-1}\left[\cos \left(\frac{A}{2}\right)\right]=\frac{\pi^2-4A^2}{8}\)
Please look at the attached sheet. How to reach the final answerHi suvadip,
This trigonometric identity is not valid. For example substituting \(A=\dfrac{\pi}{2}\) we get zero in the right hand side whereas \(-\dfrac{1}{\sqrt{2}}\tan^{-1}\dfrac{1}{\sqrt{2}}\neq 0\) on the right hand side.
Your attachment is not very clear and I cannot read it. If you can post a more clear attachment or write down the equations using LaTeX I would perhaps be able to help.Please look at the attached sheet. How to reach the final answer
Your attachment is not very clear and I cannot read it. If you can post a more clear attachment or write down the equations using LaTeX I would perhaps be able to help.
How do you get this? After differentiating wrt A, I get,I have to prove
\(\displaystyle \int_0^{\pi/2}\frac{log(1+cosA cosx)}{cosx}dx=\frac{\pi^2-4A^2}{8}\)
I have arrived at
\(\displaystyle \frac{d}{dA}(I)=-sinA \int_0^1\frac{2}{1+cosA +(1-cos A)z^2}dz\)
You are correct so far (having differentiated under the integral sign and then made the substitution $z = \tan(x/2)$). The next step is to write this as $$\frac{d}{dA}(I)=\frac{-2\sin A}{1-\cos A} \int_0^1\frac{1}{z^2 +\frac{1+\cos A}{1-\cos A}}dz.$$ Now use the fact that \(\displaystyle \frac{1+\cos A}{1-\cos A} = \cot^2(A/2)\) to get $$\frac{d}{dA}(I)= \frac{-2\sin A}{1-\cos A} \int_0^1\frac{1}{z^2 +\cot^2(A/2)}dz = \frac{-2\sin A}{1-\cos A} \Bigl[\tan(A/2)\arctan\bigl(z\tan(A/2)\bigr)\Bigr]_0^1 = \frac{-2\sin A}{1-\cos A}\frac A2\tan(A/2)$$ (provided that $|A| < \pi$, so that $\arctan\bigl(\tan(A/2)\bigr) = A/2$). Using half-angle formulas again, you can write this as $$\frac{\frac{-2}{1+\tan^2(A/2)}}{1- \frac{1-\tan^2(A/2)}{1+\tan^2(A/2)}}A\tan(A/2) = \frac{-A\tan(A/2)}{\tan^2(A/2)} = -A\cot(A/2).$$ If that last expression was just $A$ (without the $\cot(A/2)$) then you could integrate it to get $-A^2/2$ which, together with the initial value $0$ when $A = \pi/2$, would give the formula that you are looking for.I have to prove
\(\displaystyle \int_0^{\pi/2}\frac{\log(1+\cos A \cos x)}{\cos x}dx=\frac{\pi^2-4A^2}{8}\)
I have arrived at
\(\displaystyle \frac{d}{dA}(I)=-\sin A \int_0^1\frac{2}{1+\cos A +(1-\cos A)z^2}dz\) where I is the given integral. Am I correct so far and how to proceed at the answer from there? I have used the Leibnitz's rule for differentiation under the sign of integration.
I'm very willing to believe that you're right, but I'll leave it to the OP to sort out the details.Hi Opalg!
I am not sure but I seem to be getting the final answer. If the integral I reached is evaluated further, I get
$$\frac{dI}{dA}=2\left(\tan^{-1}\left(\frac{\cos A}{\sin A}\right)-\tan^{-1}\left(\frac{1+\cos A}{\sin A}\right)\right)$$
Using the formula for $\tan^{-1}a-\tan^{-1}b$, I simplified it to
$$\frac{dI}{dA}=-A$$
Please check if I did anything wrong, thank you.