Homework Year 12 Trig Identities Questions (Check Photos)

(a) See the attached file for part (a). 
(b) To maximize $A(\theta)=100\sin \theta (1+\cos \theta)$, differentiate $A$ to get
\[A'(\theta)=100 \cos \theta (1+\cos \theta)+100\sin \theta (-\sin \theta)\]
\[=100[\cos\theta +\cos^2 \theta -\sin^2 \theta]=100 [\cos \theta +\cos^2\theta- (1-\cos^2 \theta)]\]
\[=100[2\cos^2 \theta +\cos \theta -1]=0\]
\[\Rightarrow     2\cos^2 \theta +\cos \theta -1=0.\]
Using the quadratic formula we get
\[\cos \theta = \frac{-1 \pm\sqrt{1+4\cdot 2}}{2\cdot 2}=\frac{-1\pm 3}{4}=-1, \frac{1}{2}.\]
Hence 
\[ \cos \theta =-1, \text{or}  \cos \theta =\frac{1}{2}\]
\[\rightarrow \theta =\pi  \text{or}  \theta =\frac{\pi}{3}.\]
Clearly $\theta =\pi$ is not addmissible. Therefore  $\theta =\frac{\pi}{3}$ maximizes the cross-section.