MATHEMATICSP2 QUESTION 6 Given the functions: $$ f(x)=\cos \left(x-60^{\circ}\right) $$ and $$ g(x)=\sin 3 x $$ for $$ x \in\left[-90^{\circ} ; 180^{\circ}\right] $$ 6.1 Write down: 6.1 .1 the amplitude of $$ f $$ 6.2 Determine the values of $$ x $$ for which $$ f(x)=g(x) $$ for $$ x \in\left[-90^{\circ} ; 180^{\circ}\right] $$ 6.3 On the same set of axes, sketch the graph of $$ f $$ and $$ g $$ for $$ x \in\left[-90^{\circ} ; 180^{\circ}\right] $$ in the SPECIAL ANSWER BOOK. Show ALL intercepts with the axes as well as turning and end points. 6.4 For which value(s) of $$ x $$ is $$ \frac{g(x)}{f(x)} $$ undefined for $$ x \in\left[-90^{\circ} ; 180^{\circ}\right] ? $$

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The amplitude of a cosine function $$ f(x) = A \cos(B(x - C)) + D $$ is given by the absolute value of $$ A $$. In this case, $$ f(x) = \cos(x - 60^{\circ}) $$ has an amplitude of 1, since $$ A = 1 $$.
To find the values of $$ x $$ for which $$ f(x) = g(x) $$, we set the equations equal to each other:

$$
\cos(x - 60^{\circ}) = \sin(3x)
$$

Using the identity $$ \sin(3x) = \cos(90^{\circ} - 3x) $$, we can rewrite the equation as:

$$
\cos(x - 60^{\circ}) = \cos(90^{\circ} - 3x)
$$

This leads to two cases:

1. $$ x - 60^{\circ} = 90^{\circ} - 3x + 360^{\circ}k $$
2. $$ x - 60^{\circ} = - (90^{\circ} - 3x) + 360^{\circ}k $$

where $$ k $$ is any integer.

Let's solve these equations for $$ x $$ in the interval $$ [-90^{\circ}, 180^{\circ}] $$.

**Case 1:**
$$
x - 60^{\circ} = 90^{\circ} - 3x
$$
$$
4x = 150^{\circ}
$$
$$
x = 37.5^{\circ}
$$

**Case 2:**
$$
x - 60^{\circ} = -90^{\circ} + 3x
$$
$$
-2x = -30^{\circ}
$$
$$
x = 15^{\circ}
$$

Now, we will check if these values are within the interval $$ [-90^{\circ}, 180^{\circ}] $$. Both $$ 37.5^{\circ} $$ and $$ 15^{\circ} $$ are valid solutions.

Thus, the values of $$ x $$ for which $$ f(x) = g(x) $$ are $$ x = 15^{\circ} $$ and $$ x = 37.5^{\circ} $$.
To sketch the graphs of $$ f(x) = \cos(x - 60^{\circ}) $$ and $$ g(x) = \sin(3x) $$ over the interval $$ [-90^{\circ}, 180^{\circ}] $$:

1. **For $$ f(x) $$**:
   - The function has a period of $$ 360^{\circ} $$.
   - The maximum value is 1 and the minimum value is -1.
   - The x-intercepts occur when $$ \cos(x - 60^{\circ}) = 0 $$, which happens at $$ x = 60^{\circ} + 90^{\circ}k $$.
   - The turning points are at $$ x = 60^{\circ} + 180^{\circ}k $$.

2. **For $$ g(x) $$**:
   - The function has a period of $$ 120^{\circ} $$.
   - The maximum value is 1 and the minimum value is -1.
   - The x-intercepts occur when $$ \sin(3x) = 0 $$, which happens at $$ x = 0^{\circ} + 40^{\circ}k $$.

3. **Intercepts and Turning Points**:
   - For $$ f(x) $$: Intercepts at $$ x = 60^{\circ}, 240^{\circ} $$ (only $$ 60^{\circ} $$ is in the interval).
   - For $$ g(x) $$: Intercepts at $$ x = 0^{\circ}, 40^{\circ}, 80^{\circ}, 120^{\circ} $$.

The sketch should include these points and the general shape of the cosine and sine functions. Ensure to label the axes and mark the intercepts and turning points clearly.
The expression $$ \frac{g(x)}{f(x)} $$ is undefined when $$ f(x) = 0 $$.

From the earlier analysis, $$ f(x) = \cos(x - 60^{\circ}) = 0 $$ occurs at:

$$
x - 60^{\circ} = 90^{\circ} + 360^{\circ}k \quad \text{or} \quad x - 60^{\circ} = -90^{\circ} + 360^{\circ}k
$$

Solving these gives:

1. $$ x = 150^{\circ} + 360^{\circ}k $$
2. $$ x = -30^{\circ} + 360^{\circ}k $$

In the interval $$ [-90^{\circ}, 180^{\circ}] $$, the only relevant solution is $$ x = -30^{\circ} $$.

Thus, $$ \frac{g(x)}{f(x)} $$ is undefined at $$ x = -30^{\circ} $$.
6.1.1: Amplitude of $$ f $$ is 1. 6.2: Values of $$ x $$ for which $$ f(x) = g(x) $$ are $$ 15^{\circ} $$ and $$ 37.5^{\circ} $$. 6.4: $$ \frac{g(x)}{f(x)} $$ is undefined at $$ x = -30^{\circ} $$.

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