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

To find the inverse function, we start with the equation:

$y = \frac{3 + e^{2x}}{4}.$

Interchanging $x$ and $y$, we have:

$x = \frac{3 + e^{2y}}{4}.$

Multiplying each side by 4 gives:

$4x = 3 + e^{2y}.$

Rearranging, we get:

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

Taking the natural logarithm of both sides:

$2y = \ln(4x - 3)$

Finally, dividing by 2, we find the inverse function:

$f^{-1}(x) = \frac{1}{2} \ln(4x - 3).$

To sketch the graphs:

1. For $y = \cos 2x$, plot the cosine function which oscillates between -1 and 1 with a period of $\pi$.
2. For $y = \frac{x + 1}{2}$, this is a linear function with a slope of $\frac{1}{2}$ and crosses the y-axis at 0.5.

The sketches should include key points and the intersections within the range $-\pi \leq x \leq \pi$.

By observing the graphs from part (b), count the intersections of the two functions $y = 2 \cos 2x$ and $y = x + 1$. Each intersection represents a solution to the equation. Given the oscillatory nature of the cosine function and the linear nature of $y = x + 1$, there will be multiple intersections; therefore, estimate the number of solutions visually from the graph.

Assuming the function $f(x) = 2 \cos 2x - (x + 1)$, find $f'(x)$ and apply Newton's method. Start the iteration with an initial guess $x_0 = 0.4$:

1. Calculate $f(0.4)$ and $f'(0.4)$.
2. Use Newton's formula:
   $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$

Repeat until desired accuracy is achieved, rounding the final approximation to three decimal places.

To prove the identity, utilize the double angle formulas. Recall that:
$\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$ and $\cos 2\theta = 2\cos^2 \theta - 1$. This leads to:

$\tan^2 \theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta},$

which shows the relationship given that $\cos 2\theta \neq -1$.

Using the identity:
$\tan \frac{\pi}{8} = \sqrt{\frac{1 - \cos \frac{\pi}{4}}{1 + \cos \frac{\pi}{4}}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}}.$
Simplifying leads to the exact value of $\tan \frac{\pi}{8} = \sqrt{2} - 1$.