1. **State the problem:** We have two trigonometric functions: $f(x) = a \cos x$ and $g(x) = \sin bx$ defined on $x \in [-180^\circ, 180^\circ]$. We need to find: - 8.1.1 The values of $a$ and $b$. - 8.1.2 The range of $f$. - 8.1.3 The period of $g$. 2. **Determine $a$ and $b$:** - From the graph, $f(x) = a \cos x$ has maximum value approximately 2 and minimum approximately -2. - Since $\cos x$ ranges from -1 to 1, $a$ must be 2 to scale the cosine wave to these values. - For $g(x) = \sin bx$, the graph shows peaks at $y = 1.5$ and troughs at $-1.5$, so amplitude is 1.5. - The sine function normally has amplitude 1, so amplitude here is 1.5, but since $g(x) = \sin bx$ only has $b$ affecting frequency, amplitude is 1.5, so amplitude is not affected by $b$. - To find $b$, note the period $T$ of $g$ is the length between repeating points. - From the graph, $g(x)$ peaks at about $-135^\circ$ and again at $45^\circ$, so period $T = 45 - (-135) = 180^\circ$. - The period of $\sin bx$ is $\frac{360^\circ}{b}$, so: $$ T = \frac{360^\circ}{b} = 180^\circ \implies b = \frac{360^\circ}{180^\circ} = 2 $$ 3. **Range of $f$:** - Since $f(x) = 2 \cos x$ and $\cos x \in [-1,1]$, the range is: $$ [-2, 2] $$ 4. **Period of $g$:** - As found, period $T = \frac{360^\circ}{b} = \frac{360^\circ}{2} = 180^\circ$ **Final answers:** - $a = 2$ - $b = 2$ - Range of $f$ is $[-2, 2]$ - Period of $g$ is $180^\circ$