Learn how to sketch the sine and cosine functions of the general forms \[ y = a \sin\bigl(k (x - d)\bigr) + c \quad \text{and} \quad y = a \cos\bigl(k (x - d)\bigr) + c \] with clear explanations, step-by-step examples, and detailed solutions. This introduction covers key concepts such as amplitude, period, phase shift, and vertical translation to help you master graphing trigonometric functions.
Amplitude = \( |a| \)
Period = \( \dfrac{2\pi}{|k|} \)
Horizontal Shift (translation) = \( d \), to the left if \(-d\) is positive and to the right if \(-d\) is negative.
Vertical Shift (translation) = \( c \), up if \(c\) is positive and down if \(c\) is negative.
In order to sketch transformed sine and cosine functions, we need to know how to sketch basic sine and cosine functions. The unit circle (radius = 1) gives the values of \(\sin(x)\) and \(\cos(x)\) at 5 key points which can be used to graph more complex sine and cosine functions. The coordinates of any point on the unit circle give the cosine and sine of the angle in standard position corresponding to that point.
The angle in standard position \(x = 0|) corresponds the point \[ (1, 0) = (\cos(0), \sin(0)) \]
The angle in standard position \(x = \pi/2\) (or 90°) corresponds the point \[ (0, 1) = (\cos(\pi/2), \sin(\pi/2)) \]
The angle in standard position \(x = \pi\) (or 180°) corresponds the point \[ (-1, 0) = (\cos(\pi), \sin(\pi)) \]
The angle in standard position \(x = 3\pi/2\) (or 270°) corresponds the point \[ (0, -1) = (\cos(3\pi/2), \sin(3\pi/2)) \]
The angle in standard position \(x = 2\pi\) (or 360°) corresponds the point \[ (1, 0) = (\cos(2\pi), \sin(2\pi)) \] as shown below.
Sketch the graph of \[ y = 2 \cos(x) + 1 \] over one period.
Amplitude = \( |2| = 2 \)
Period = \( 2\pi \)
Vertical Shift (translation) = \( 1 \), up 1 unit.
Horizontal Shift (translation) = \( 0 \)
Three steps to graph the function \( y = 2 \cos(x) + 1 \)
1) We start by sketching \[ y = \cos(x) \] using the values of \( x \) and \( y \) from the unit circle (blue graph below).
\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 0 & \dfrac{\pi}{2} & \pi & \dfrac{3\pi}{2} & 2\pi \\ \hline y & 1 & 0 & -1 & 0 & 1 \\ \hline \end{array} \]2) We then sketch \( y = 2 cos(x) \) by streching \( y = cos(x) \) by 2 (green graph below)
3) and finally \( y = 2 cos(x) + 1 \) by shifting up 1 unit (red graph below).
Sketch the graph of \[ y = - 2 sin(x) - 1 \] over one period.
Amplitude = \( |-2| = 2 \)
Period = \( 2\pi \)
Vertical Shift (translation) = \( -1 \), down 1 unit.
Horizontal Shift (translation) = \( 0 \)
Three steps to graph the function \( y = - 2 \sin(x) - 1 \)
1) We start by sketching \( y = \sin(x) \) using the values of \( x \) and \( y \) from the unit circle (blue graph below).
\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 0 & \dfrac{\pi}{2} & \pi & \dfrac{3\pi}{2} & 2\pi \\ \hline y & 0 & 1 & 0 & -1 & 0 \\ \hline \end{array} \]2) We then sketch \( y = - 2 \sin(x) \) by streching \( y = sin(x) \) by 2 and reflecting it on the x-axis(green graph below)
3) and finally sketch \( y = - 2 sin(x) - 1 \) by shifting down 1 unit (red graph below).
Sketch the graph of \( y = 3 \cos(2x + \dfrac{\pi}{3}) - 1 \) over one period.
Graphing Parameters
Amplitude = \( |3| = 3 \)
Period = \( \dfrac{2\pi}{2} = \pi \)
Vertical Shift (translation) = \( -1 \), down 1 unit.
Horizontal Shift: Because of the term \( \dfrac{\pi}{3} \), the graph is shifted horizontally. We first rewrite the given function as: \( y = 3 \cos\left[ 2\left( x + \dfrac{\pi}{6} \right) \right] - 1 \) and we can now write the shift as being equal to \( \dfrac{\pi}{6} \) to the left.
Three steps to graph the function \( y = 3 \cos(2x + \dfrac{\pi}{3}) - 1 \)
1) We start by sketching \( 3 \cos(2x) \) with minimum and maximum values -3 and +3 over one period \( = \pi \) (blue graph below).
2) We then sketch \( y = 3 \cos(2x) - 1 \) translating the previous graph 1 unit down (green graph below).
3)
We now shift the previous graph \( \dfrac{\pi}{6} \) to the left (red graph below) so that the sketched period starts at \( -\dfrac{\pi}{6} \) and ends at \( -\dfrac{\pi}{6} + \pi = \dfrac{5\pi}{6} \) which is one period \( = \pi \).
Sketch the graph of \[ y = -0.2 \sin\left(0.5x - \dfrac{\pi}{6} \right) + 0.1 \] over one period.
Graphing Parameters
Amplitude = \( | -0.2 | = 0.2 \)
Period = \( \dfrac{2\pi}{0.5} = 4\pi \)
Vertical Shift (translation) = 0.1, up 0.1 unit.
Horizontal Shift: Because of the term \(- \dfrac{\pi}{6}\), the graph is shifted horizontally. We first rewrite the given function as:
\[ y = -0.2 \sin\left( 0.5(x - \dfrac{\pi}{3}) \right) + 0.1 \]
and we can now write the shift as being equal to \(\dfrac{\pi}{3}\) to the right.
Three steps to graph the function \( y = -0.2 \sin\left(0.5x - \dfrac{\pi}{6} \right) + 0.1 \)
1) We start by sketching \[ y = -0.2 \sin(0.5x) \] with minimum and maximum values \(-0.2\) and \(0.2\) over one period \[ = 4\pi \] (blue graph below).
2) We then sketch \[ y = -0.2 \sin(0.5x) + 0.1 \] by translating the previous graph 0.1 unit up (green graph below).
3) We then shift the previous graph \(\dfrac{\pi}{3}\) to the right (red graph below) so that the sketched period starts at \(\dfrac{\pi}{3}\) and ends at \(\dfrac{\pi}{3} + 4\pi\), which is one period \[ = 4\pi \].
Graphing Parameters
Amplitude = \(\left| 2 \right| = 2\)
Vertical Shift (translation) = \(-2\), down 2 units.
Period = \(\dfrac{360^\circ}{2} = 180^\circ\)
Horizontal Shift: Because of the term \(-60^\circ\), the graph is shifted horizontally. We first rewrite the given function as \[ y = 2 \cos\left[ 2 \left( x - 30^\circ \right) \right] - 2 \] and we can now write the shift as being equal to \(30^\circ\) to the right.
Three steps to graph the function \( y = 2 \cos(2x - 60^\circ) - 2 \)
1) We start by sketching \[ y = 2 \cos(2x) \] with minimum and maximum values \(-2\) and \(+2\) over one period \(= 180^\circ\) (blue graph below).
2) We then sketch \[ y = 2 \cos(2x) - 2 \] translating the previous graph 2 units down (green graph below).
3) Finally, we shift the previous graph \(30^\circ\) to the right (red graph below) so that the sketched period starts at \(30^\circ\) and ends at \[ 30^\circ + 180^\circ = 210^\circ \] which is one period \(= 180^\circ\).
Three steps to graph the function \( y = -2 \sin\left(\dfrac{x}{3} + \dfrac{\pi}{3}\right) - 1 \)
1) We start by sketching \(-2 \sin\left(\dfrac{x}{3}\right)\) with minimum and maximum values \(-2\) and \(+2\) over one period \(= 6 \pi\) (blue graph below).
2) We then sketch \[ y = -2 \sin\left(\dfrac{x}{3}\right) - 1 \] translating the previous graph 1 unit down (green graph below).
3) We then shift the previous graph \(\pi\) to the left (red graph below) so that the sketched period starts at \(-\pi\) and ends at \(5\pi\) which is one period \(= 6\pi\).
