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

To find the inverse function, we first switch $x$ and $y$ in the function definition:

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

Next, we solve for $y$.

1. Multiply both sides by 4:
   $4x = 3 + e^{2y}$
2. Isolate $e^{2y}$:
   $e^{2y} = 4x - 3$
3. Take the natural logarithm of both sides:
   $2y = \ln(4x - 3)$
4. Finally, divide by 2:
   $y = \frac{1}{2}\ln(4x - 3)$

Thus, the inverse function is:

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

To sketch the graphs of $y = \cos 2x$ and $y = \frac{x + 1}{2}$, follow these steps:

1. Graph of $y = \cos 2x$:

   • The function oscillates between -1 and 1 with a period of $\pi$.
   • Key points include:
     • $x = -\pi, y = 1$
     • $x = -\frac{\pi}{2}, y = 0$
     • $x = 0, y = 1$
     • $x = \frac{\pi}{2}, y = 0$
     • $x = \pi, y = 1$

2. Graph of $y = \frac{x + 1}{2}$:

   • This is a straight line with a slope of $\frac{1}{2}$ and a y-intercept of $\frac{1}{2}$.
   • It intersects the y-axis at $(0, \frac{1}{2})$ and the x-axis at $(-1, 0)$.

Combine the two graphs on the same axes, ensuring proper labeling and scaling.

To determine the number of solutions for the equation $2 \cos 2x = x + 1$ in the interval $-\pi \leq x \leq \pi$, we analyze both graphs:

1. Behavior of $y = 2 \cos 2x$:

   • Ranges from -2 to 2, oscillating between these values.

2. Behavior of $y = x + 1$:

   • A straight line that crosses through points (0, 1), (-1, 0), and through the axis from -2 to 2 over the interval.

From the sketches, we identify intersection points representing solutions. Count the intersections visually observed to find how many solutions exist.

Applying Newton's method for the function $f(x) = 2 \cos 2x - (x + 1)$:

1. First Derivative:
   $f'(x) = -4 \sin 2x - 1$

2. Approximate using $x = 0.4$:
   $f(0.4) \approx 0, f'(0.4) \approx -1.32$

3. Newton's Iteration formula:
   $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
   Apply this iteratively to refine the approximation of the solution, ensuring adducts reach sufficient precision (3 decimal places).

To prove the identity, start from the double angle formulas:

1. Recall:
   $\tan \theta = \frac{\sin \theta}{\cos \theta}$
   $\sin 2\theta = 2\sin \theta\cos \theta \quad \text{and} \quad \cos 2\theta = \cos^2 \theta - \sin^2 \theta$

2. Now express $\tan^2 \theta$:
   $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$
   Using double angle identities, arrive at the formula. Confirm through algebraic manipulation:
   Combine and simplify to yield the desired result.

To compute $\tan \frac{\pi}{8}$, use the half angle formula:

1. Half-angle formula:
   $\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}$
   Let $\theta = \frac{\pi}{4}$, therefore:
   $\tan \frac{\pi}{8} = \tan \frac{\frac{\pi}{4}}{2}$
   where $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

Using this to evaluate yields a numerical expression that reduces to: $\tan \frac{\pi}{8} = \sqrt{2 - \sqrt{2}}$.