Let $f(x) = \frac{3 + e^{2x}}{4}$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2009 - Paper 1

To find the inverse function, we set: $y = f(x) = \frac{3 + e^{2x}}{4}.$

Interchanging $x$ and $y$ gives: $x = \frac{3 + e^{2y}}{4}.$

Multiplying both sides by 4 yields: $4x = 3 + e^{2y}.$

Rearranging gives: $e^{2y} = 4x - 3.$

Taking the natural logarithm:

To sketch the graphs, we analyze $y = \cos 2x$, which oscillates between -1 and 1 with a period of $\pi$. The line $y = \frac{x + 1}{2}$ is a linear function with a slope of rac{1}{2}.

Plot points for both functions over the interval $-\pi \leq x \leq \pi$ and then sketch using these characteristics.

By observing the intersection points of the two graphs, we analyze how many times the oscillating function $y = 2\cos 2x$ intersects with the linear function $y = x + 1$ over the interval from $-\pi$ to $\pi$. Count the intersections to find the number of solutions.

Using Newton's method, we define: $f(x) = 2 \cos 2x - (x + 1)$

The derivative is: $f'(x) = -4 \sin 2x - 1.$

Using an initial approximation of $x_0 = 0.4$: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Iterate this step until $x_n$ converges to the desired precision of three decimal places.

We start from the double angle formula: $\cos 2\theta = 1 - 2\sin^2 \theta$

Now, we can express $an^2 \theta$ as: $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\sin^2 \theta}{1 - \sin^2 \theta}$

Substituting into the equation leads to: $\tan^2 \theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta}.$

Using the identity from part (c), we have: $\tan \frac{\pi}{8} = \sqrt{\tan^2 \frac{\pi}{8}} = \sqrt{\frac{1 - \cos \frac{\pi}{4}}{1 + \cos \frac{\pi}{4}}}.$

Knowing that $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, we substitute and simplify to find: $\tan \frac{\pi}{8} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}}.$

Calculating this yield the exact value.