Integration with trigonomitry

Question

Hi I was trying to find the area between the following curves (below) however I am unsure how to continue from the trigonometry which gets presented:

Curves:

$$y = 2\sin(x)\\y = \cos\left(\frac{x}{2}\right)$$

I have found that the curves meet at $x=\pi$ and $x=2 \arcsin(\frac14)$. I am supposed to find the area being $\frac94$ however this is what I got, I don't know how to simplify: $$A = 2\cos \left( 2\arcsin\left(\frac14\right)\right) + 2\sin \left( 2\arcsin\left(\frac14\right)\right)$$ I tried letting $z = \arcsin(\frac14)$ meaning I get: $$A = 2\cos(2z) + \frac12$$ which then can simplify to: $$A = 2(1-\sin^2(z))+\frac12$$ but then I have no idea. Thanks and good luck!

Answer 1

Are you requested to find the red-coloured area?

Then it's simply $$A=\int_{2\arcsin\left(\frac14\right)} ^{\pi} 2\sin(x)-\int_{2\arcsin\left(\frac14\right)} ^{\pi} \cos\left(\frac{x}{2}\right)\\=\left[-2\cos(\pi)+2\cos\left(2\arcsin\left(\frac14\right)\right)\right]-\left[2\sin(\frac{\pi}{2})-2\sin\left(\frac{2\arcsin\left(\frac14\right)}{2}\right)\right]\\=2-2+2\sin\left(\frac{2\arcsin\left(\frac14\right)}{2}\right)+2\cos\left(2\arcsin\left(\frac14\right)\right)\\=\frac12+\frac74\\=\frac94=2.25$$

Answer 2

$$\sin\left(\arcsin\left(\frac14\right)\right) = \frac14$$ This is all you need to do from where you have reached.