Given: \( f(x)=\sin 2 x, \quad h(x)=\sin 4 x \) and \( g(x)=\sin \frac{1}{2} x, x \in\left[0^{\circ} ; 360^{\circ}\right] \) 4.1 Sketch the graphs of \( f, h \) and \( g \) on the same set of axes. Write down the period of \( f, h \) and \( g \) The graphs are in the form \( y=\sin k x \) : Compare the periods of \( a \) in the graphs of \( f, h, g \) and \( y=\sin x \), then comment on the effect of \( k \).

**Step 1. Period Formula for \( y = \sin kx \)** For a sine function in the form \( y = \sin kx \) (with \( x \) in degrees), the period \( T \) is given by \[ T = \frac{360^\circ}{|k|} \] **Step 2. Finding the Periods** 1. For \( f(x)=\sin 2x \): Here, \( k = 2 \). Thus, \[ T = \frac{360^\circ}{2} = 180^\circ \] 2. For \( h(x)=\sin 4x \): Here, \( k = 4 \). Thus, \[ T = \frac{360^\circ}{4} = 90^\circ \] 3. For \( g(x)=\sin \frac{1}{2}x \): Here, \( k = \frac{1}{2} \). Thus, \[ T = \frac{360^\circ}{\frac{1}{2}} = 720^\circ \] **Step 3. Comparison with \( y=\sin x \)** For \( y = \sin x \), we have \( k = 1 \) and its period is: \[ T = \frac{360^\circ}{1} = 360^\circ \] Comparing the periods: - \( f(x) = \sin 2x \) has a period of \( 180^\circ \) (shorter than \( 360^\circ \)). - \( h(x) = \sin 4x \) has a period of \( 90^\circ \) (even shorter). - \( g(x) = \sin \frac{1}{2}x \) has a period of \( 720^\circ \) (longer than \( 360^\circ \)). **Step 4. Comment on the Effect of \( k \)** The multiplier \( k \) in the sine function affects the period: - **Larger \( k \) values (e.g., \( k = 2 \) or \( 4 \))** compress the graph horizontally. This causes the function to complete more cycles over the same interval; hence, the period decreases. - **Smaller \( k \) values (e.g., \( k = \frac{1}{2} \))** stretch the graph horizontally. This causes the function to complete fewer cycles over the same interval; hence, the period increases. Thus, the constant \( k \) directly determines the frequency of the sine function: \[ \text{Period} = \frac{360^\circ}{|k|} \]