The double-angle identity for sine is written as

\(\displaystyle \sin 2\theta= 2\sin\theta \cos\theta\)

and we know that 

\(\displaystyle \cos\theta = \frac{4}{5}\)

Using \(\displaystyle x^{2}+y^{2}=r^{2}\), we see that \(\displaystyle (4)^{2} + (y)^{2} = (5)^{2}\), which gives us 

\(\displaystyle y=\pm3\)

Since we know \(\displaystyle \theta\) is between \(\displaystyle \frac{3\pi}{2}\) and \(\displaystyle 2\pi\), sin \(\displaystyle \theta\) is negative, so \(\displaystyle y= -3\). Thus,

\(\displaystyle \sin \theta = -\frac{3}{5}\).

Finally, substituting into our double-angle identity, we get

\(\displaystyle \sin 2\theta = 2 \sin\theta \cos\theta = 2 \cdot\bigg(-\frac{3}{5}\bigg)\cdot\frac{4}{5} = -\frac{24}{25}\)

The half-angle identity for sine is:

\(\displaystyle \sin \bigg(\frac{x}{2}\bigg) = \pm\sqrt\frac{1-\cos x}{2}\)

If our half-angle is \(\displaystyle 15^{\circ}\), then our full angle is \(\displaystyle 30^{\circ}\). Thus,

\(\displaystyle \sin 15^{\circ}= \pm\sqrt\frac{1-\cos30^{\circ}}{2}\)

The exact value of \(\displaystyle \cos 30^{\circ}\) is expressed as \(\displaystyle \frac{\sqrt{3}}{2}\), so we have

\(\displaystyle \sin 15^{\circ} = \pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\)

Simplify under the outer radical and we get

\(\displaystyle \sin 15^{\circ} = \pm\sqrt\frac{2-\sqrt{3}}{4}\)

Now simplify the denominator and get

\(\displaystyle \sin 15^{\circ} = \pm\frac{\sqrt{2-\sqrt{3}}}{2}\)

Since \(\displaystyle 15^{\circ}\) is in the first quadrant, we know sin is positive. So,

\(\displaystyle \sin 15^{\circ} = \frac{\sqrt{2-\sqrt{3}}}{2}\)

Write the half angle identity for cosine.

\(\displaystyle cos^2(\theta) =\frac{1+cos(2\theta)}{2}\)

Replace theta with two theta.

\(\displaystyle \theta=2\theta\)

Therefore:

\(\displaystyle 2cos^2(2\theta)= 2\left[\frac{1+cos(2 \times 2\theta)}{2}\right] = 1+cos(4\theta)\)

The key here is to use the half-angle identity for \(\displaystyle cos^2(x)\) to convert it and make it much easier to work with.

\(\displaystyle acos^2(x) = \frac{a}{2}[1 + cos(x)]\)

In this case, \(\displaystyle a = 8\), so therefore...

\(\displaystyle 8cos^2(x) = \frac{8}{2}[1 + cos(x)] = 4 + 4cos(x)\)

Consequently, \(\displaystyle 8cos^2(x)\) has an amplitude of \(\displaystyle 4\).

Because \(\displaystyle 22.5^{\circ}=\frac{45^{\circ}}{2}\), we can use the half-angle formula for cosines to determine \(\displaystyle cos(22.5^{\circ})\).

In general,

\(\displaystyle cos(\frac{\Theta}{2})= \sqrt{\frac{1}{2}(1+cos\Theta)}\)

for \(\displaystyle 0^{\circ}\leq\Theta< 360^{\circ}\).

For this problem,

\(\displaystyle cos(22.5^{\circ})= \sqrt{\frac{1}{2}(1+cos(45^{\circ}))}\)

                      \(\displaystyle = \sqrt{\frac{1}{2}(1+\frac{\sqrt{2}}{2})}\)

                      \(\displaystyle =\sqrt{\frac{1}{2}(\frac{2+\sqrt{2}}{2})}\)

                      \(\displaystyle =\sqrt{\frac{1}{4}(2+\sqrt{2})}\)

                      \(\displaystyle =\frac{1}{2}\sqrt{(2+\sqrt{2})}\)

Hence, 

\(\displaystyle cos(22.5^{\circ})=\frac{1}{2}\sqrt{2+\sqrt{2}}\) 

Let \(\displaystyle x=\frac{5\pi}{6}\); then

\(\displaystyle \cos(\frac{5\pi}{12})=\cos(\frac{x}{2})\).

We'll use the half-angle formula to evaluate this expression.

\(\displaystyle \cos(\frac{x}{2})=\pm \sqrt{\frac{1+\cos x}{2}}\)

Now we'll substitute \(\displaystyle \frac{5\pi}{6}\) for \(\displaystyle x\).

\(\displaystyle \begin{align*} \cos(\frac{5\pi}{12})=\cos(\frac{\frac{5\pi}{6}}{2})&=\pm\sqrt{\frac{1+\cos(\frac{5\pi}{6})}{2}}\\ &=\pm\sqrt{\frac{1+\frac{-\sqrt{3}}{2}}{2}}\\ &=\pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\ &=\pm\sqrt{\frac{2-\sqrt{3}}{2}}\\ &=\pm\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right) \end{align*}\)

\(\displaystyle \frac{5\pi}{6}\) is in the first quadrant, so \(\displaystyle \cos(\frac{5\pi}{6})\) is positive. So

\(\displaystyle \cos(\frac{5\pi}{12})=\frac{\sqrt{2-\sqrt{3}}}{2}\).

Using the half angle formula for tangent, 

\(\displaystyle \tan(\frac{\theta}{2})=\frac{\sin(\theta)}{(1+\cos(\theta))}\),

we plug in 30 for \(\displaystyle \theta\).

We also know from the unit circle that \(\displaystyle \sin(30)\) is \(\displaystyle 1/2\) and \(\displaystyle \cos(30)\) is \(\displaystyle \sqrt3/2\).

Plug all values into the equation, and you will get the correct answer. 

\(\displaystyle \\ \tan(\frac{30}{2})=\frac{\sin(30)}{(1+\cos(30))} \\ \\ \\ \tan(15)=\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}\)