1. **Problem statement:** Sketch the functions $f(x) = 2\sin x$ and $g(x) = \sin x + 1$ for $x \in (0^\circ, 360^\circ)$. Find the period of $f$ and the amplitude of $g$. 2. Given $f(x) = a \cos x$ and $g(x) = \sin x + q$, find $a$, $q$, the range of $g$, and solve $g(x) - f(x) = 2$. --- ### Step 1: Sketching the functions - $f(x) = 2\sin x$ is a sine wave with amplitude 2. - $g(x) = \sin x + 1$ is a sine wave shifted up by 1. ### Step 2: Period of $f(x)$ - The period of $\sin x$ is $360^\circ$. - Multiplying by 2 (amplitude) does not change the period. - Therefore, period of $f(x)$ is $360^\circ$. ### Step 3: Amplitude of $g(x)$ - Amplitude is the coefficient of $\sin x$, which is 1. - Vertical shift $+1$ does not affect amplitude. - So amplitude of $g$ is 1. ### Step 4: Finding $a$ and $q$ from the graph - $f(x) = a \cos x$ has amplitude $a$. - From the graph, amplitude $a = 1$. - $g(x) = \sin x + q$ is shifted up by $q$. - From the graph, vertical shift $q = 1$. ### Step 5: Range of $g(x)$ - Since $\sin x$ ranges from $-1$ to $1$, adding $q=1$ shifts range to $0 \leq y \leq 2$. ### Step 6: Solve $g(x) - f(x) = 2$ - Substitute: $\sin x + 1 - a \cos x = 2$ with $a=1$. - Simplify: $\sin x + 1 - \cos x = 2$. - Rearrange: $\sin x - \cos x = 1$. ### Step 7: Solve $\sin x - \cos x = 1$ - Use identity: $\sin x - \cos x = \sqrt{2} \sin(x - 45^\circ)$. - So $\sqrt{2} \sin(x - 45^\circ) = 1$. - Divide both sides: $\sin(x - 45^\circ) = \frac{1}{\sqrt{2}} = \sin 45^\circ$. - Solutions in $0^\circ < x < 360^\circ$: - $x - 45^\circ = 45^\circ \Rightarrow x = 90^\circ$ - $x - 45^\circ = 180^\circ - 45^\circ = 135^\circ \Rightarrow x = 180^\circ$ --- **Final answers:** 1. - a) Sketch as described. - b) Period of $f$ is $360^\circ$. - c) Amplitude of $g$ is 1. 2. - a) $a = 1$ - b) $q = 1$ - c) Range of $g$ is $0 \leq y \leq 2$ - d) $x = 90^\circ, 180^\circ$ satisfy $g(x) - f(x) = 2$.