1. ## Explain working

Seek explanation of getting the second line, $= 8(\pi+ \frac{\pi}{2}) - (0)$.

$A=8 \left [\theta + 2sin\theta + \frac{\theta}{2} + \frac{sin2\theta}{4} \right ]^\pi_0$

$= 8(\pi+ \frac{\pi}{2}) - (0)$

$=8 \pi + 4 \pi = 12\pi$ Ans

I know that $sin2 \theta = 2sin \theta cos \theta$

so that,

$2sin \theta + \frac{sin2 \theta}{4} = 2sin\theta + \frac{2sin \theta cos \theta}{4}$

$=2sin \theta \left ( 1 + \frac {cos \theta}{4} \right )$

$............................$

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2. Originally Posted by ashura
Seek explanation of getting the second line, $= 8(\pi+ \frac{\pi}{2}) - (0)$.

$A=8 \left [\theta + 2sin\theta + \frac{\theta}{2} + \frac{sin2\theta}{4} \right ]^\pi_0$

$= 8(\pi+ \frac{\pi}{2}) - (0)$

$=8 \pi + 4 \pi = 12\pi$ Ans

I know that $sin2 \theta = 2sin \theta cos \theta$

so that,

$2sin \theta + \frac{sin2 \theta}{4} = 2sin\theta + \frac{2sin \theta cos \theta}{4}$

$=2sin \theta \left ( 1 + \frac {cos \theta}{4} \right )$
Hello, Ashura,

you only have to remember that $\sin(0)=\sin(\pi)=\sin(2\cdot\pi)=...=0$

thus when you plug in pi all summands containing the sine function will become zero and when you plug in zero even the theta-summands will become zero too.

EB

3. Thanks, I understand the 0 result when 0 is pluged in for theta. I'm not seeing how the $\left (\pi + \frac {\pi}{2} \right )$ result is reached when pi is plugged in for theta. That's what I was trying to work out in the last 2 lines of my post.

4. Originally Posted by ashura
$8 \left [\theta + 2sin\theta + \frac{\theta}{2} + \frac{sin2\theta}{4} \right ]$ for $\theta = \pi$
$8 \left [\pi + 2sin\pi + \frac{\pi}{2} + \frac{sin2\pi}{4} \right ]$

As Earboth mentioned:
$sin\pi = 0$
$sin2\pi = 0$

So:
$8 \left [\pi + 0 + \frac{\pi}{2} + \frac{0}{4} \right ]$

$8 \left [\pi + \frac{\pi}{2} \right ]$

-Dan

5. Thanks, didn't realize that $sin \, \pi$ was really 0.