Given that $\alpha$ is an acute angle and $\cos\alpha=\frac{1+\sqrt{5}}{4}$, then $\sin\frac{\alpha}{2}=$
A. $\frac{3-\sqrt{5}}{8}$
B. $\frac{-1+\sqrt{5}}{8}$
C. $\frac{3-\sqrt{5}}{4}$
D. $\frac{-1+\sqrt{5}}{4}$

Let $a _ { 0 } = \frac { 1 } { 2 }$ and $a _ { n }$ be defined inductively by
$$a _ { n } = \sqrt { \frac { 1 + a _ { n - 1 } } { 2 } } , n \geq 1 .$$
(a) Show that for $n = 0,1,2 , \ldots$,
$$a _ { n } = \cos \theta _ { n } \text { for some } 0 < \theta _ { n } < \frac { \pi } { 2 } ,$$
and determine $\theta _ { n }$.
(b) Using (a) or otherwise, calculate
$$\lim _ { n \rightarrow \infty } 4 ^ { n } \left( 1 - a _ { n } \right)$$

Let $\alpha$ and $\beta$ be nonzero real numbers such that $2 ( \cos \beta - \cos \alpha ) + \cos \alpha \cos \beta = 1$. Then which of the following is/are true?
[A] $\tan \left( \frac { \alpha } { 2 } \right) + \sqrt { 3 } \tan \left( \frac { \beta } { 2 } \right) = 0$
[B] $\sqrt { 3 } \tan \left( \frac { \alpha } { 2 } \right) + \tan \left( \frac { \beta } { 2 } \right) = 0$
[C] $\tan \left( \frac { \alpha } { 2 } \right) - \sqrt { 3 } \tan \left( \frac { \beta } { 2 } \right) = 0$
[D] $\sqrt { 3 } \tan \left( \frac { \alpha } { 2 } \right) - \tan \left( \frac { \beta } { 2 } \right) = 0$

If $L = \sin^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)$ and $M = \cos^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)$
(1) $L = -\frac{1}{2\sqrt{2}} + \frac{1}{2}\cos\frac{\pi}{8}$
(2) $L = \frac{1}{4\sqrt{2}} - \frac{1}{4}\cos\frac{\pi}{8}$
(3) $M = \frac{1}{4\sqrt{2}} + \frac{1}{4}\cos\frac{\pi}{8}$
(4) $M = \frac{1}{2\sqrt{2}} + \frac{1}{2}\cos\frac{\pi}{8}$