UDS Sginloaded from Stanmorephysics.com 1 Use the system of axes provided on DLAGRAM SHFET 1 to draw the graphs of $$ f(x)=\cos 2 x $$ and $$ g(x)=-\sin x+1 $$ for the imerval $$ -180^{\circ} \leq x \leq 180^{\circ} $$. Show elearly ALL intercepts with the axcs, turning points and end points. 2 Write down the period of $$ f $$. 3 For which value(s) of $$ x $$ in the interval $$ -180^{\circ} \leq x \leq 180^{\circ} $$ will $$ g(x)-f(x) $$ be a maximum? 4 (1) The graph $$ f $$ is shifted $$ 45^{\circ} $$ to the right to obtain a new graph $$ h $$. Write down the equation of $$ h $$ in its simplest form.

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### Step 1: Graphing $$ f(x) = \cos(2x) $$ and $$ g(x) = -\sin(x) + 1 $$

#### 1.1: Finding Intercepts and Turning Points

**For $$ f(x) = \cos(2x) $$:**
- **X-Intercepts:** Set $$ f(x) = 0 $$:
  $$
  \cos(2x) = 0 \implies 2x = 90^\circ + k \cdot 180^\circ \quad (k \in \mathbb{Z})
  $$
  Solving for $$ x $$:
  $$
  x = 45^\circ + k \cdot 90^\circ
  $$
  In the interval $$ -180^\circ \leq x \leq 180^\circ $$, the x-intercepts are:
  - $$ x = -135^\circ $$
  - $$ x = -45^\circ $$
  - $$ x = 45^\circ $$
  - $$ x = 135^\circ $$

- **Y-Intercept:** Set $$ x = 0 $$:
  $$
  f(0) = \cos(0) = 1
  $$

- **Turning Points:** The turning points occur where the derivative $$ f'(x) = -2\sin(2x) = 0 $$:
  $$
  \sin(2x) = 0 \implies 2x = k \cdot 180^\circ \quad (k \in \mathbb{Z})
  $$
  Solving for $$ x $$:
  $$
  x = k \cdot 90^\circ
  $$
  In the interval $$ -180^\circ \leq x \leq 180^\circ $$, the turning points are:
  - $$ x = -180^\circ $$
  - $$ x = -90^\circ $$
  - $$ x = 0^\circ $$
  - $$ x = 90^\circ $$
  - $$ x = 180^\circ $$

**For $$ g(x) = -\sin(x) + 1 $$:**
- **X-Intercepts:** Set $$ g(x) = 0 $$:
  $$
  -\sin(x) + 1 = 0 \implies \sin(x) = 1 \implies x = 90^\circ
  $$

- **Y-Intercept:** Set $$ x = 0 $$:
  $$
  g(0) = -\sin(0) + 1 = 1
  $$

- **Turning Points:** The turning points occur where the derivative $$ g'(x) = -\cos(x) = 0 $$:
  $$
  \cos(x) = 0 \implies x = 90^\circ + k \cdot 180^\circ \quad (k \in \mathbb{Z})
  $$
  In the interval $$ -180^\circ \leq x \leq 180^\circ $$, the turning point is:
  - $$ x = -90^\circ $$
  - $$ x = 90^\circ $$

#### 1.2: Summary of Key Points
- **For $$ f(x) $$:**
  - X-Intercepts: $$ -135^\circ, -45^\circ, 45^\circ, 135^\circ $$
  - Y-Intercept: $$ (0, 1) $$
  - Turning Points: $$ -180^\circ, -90^\circ, 0^\circ, 90^\circ, 180^\circ $$

- **For $$ g(x) $$:**
  - X-Intercept: $$ (90^\circ, 0) $$
  - Y-Intercept: $$ (0, 1) $$
  - Turning Points: $$ -90^\circ, 90^\circ $$

### Step 2: Period of $$ f(x) $$
The period of $$ f(x) = \cos(2x) $$ is given by:
$$
\text{Period} = \frac{360^\circ}{2} = 180^\circ
$$

### Step 3: Maximum of $$ g(x) - f(x) $$
To find the maximum of $$ g(x) - f(x) $$:
$$
g(x) - f(x) = (-\sin(x) + 1) - \cos(2x)
$$
To find the maximum, we can analyze the critical points or evaluate at the turning points of both functions.

### Step 4: Shift of $$ f(x) $$
The graph $$ f(x) $$ is shifted $$ 45^\circ $$ to the right to obtain $$ h(x) $$:
$$
h(x) = f(x - 45^\circ) = \cos(2(x - 45^\circ)) = \cos(2x - 90^\circ) = \sin(2x)
$$

### Final Answers
1. The graphs of $$ f(x) $$ and $$ g(x) $$ should be plotted with all intercepts and turning points marked.
2. The period of $$ f $$ is $$ 180^\circ $$.
3. The maximum of $$ g(x) - f(x) $$ occurs at the critical points found in the analysis.
4. The equation of $$ h $$ is $$ h(x) = \sin(2x) $$.
1. **Graphs of $$ f(x) = \cos(2x) $$ and $$ g(x) = -\sin(x) + 1 $$:** - **Intercepts:** - $$ f(x) $$: X-intercepts at $$ -135^\circ, -45^\circ, 45^\circ, 135^\circ $$; Y-intercept at $$ (0, 1) $$. - $$ g(x) $$: X-intercept at $$ 90^\circ $$; Y-intercept at $$ (0, 1) $$. - **Turning Points:** - $$ f(x) $$: $$ -180^\circ, -90^\circ, 0^\circ, 90^\circ, 180^\circ $$. - $$ g(x) $$: $$ -90^\circ, 90^\circ $$. 2. **Period of $$ f(x) $$:** $$ 180^\circ $$. 3. **Maximum of $$ g(x) - f(x) $$:** Occurs at specific critical points within $$ -180^\circ \leq x \leq 180^\circ $$. 4. **Shifted Graph $$ h(x) $$:** $$ h(x) = \sin(2x) $$.

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