Learn how to find the amplitude, period, phase shift, vertical shift, and the equation of trigonometric functions from their graphs. These key properties help in analyzing and understanding the behavior of sine and cosine functions.
Each question includes a graph of a trigonometric function along with step-by-step solutions for better comprehension.
To build a solid foundation, start with these interactive tutorials on the vertical shift of trigonometric functions.
For more in-depth review of the properties of trigonometric functions, visit this page: Trigonometric Function Properties.
a) Find equations for max and min in terms of \( a \) and \( d \) (assume \( a \gt 0 \) )
b) Find equations for \( a \) and \( d \) in terms of max and min.
a) Start with the range of function \( \sin(bx + c) \) which is:
\[ -1 \leq \sin(bx + c) \leq 1 \]Multiply all terms of the above inequality by \( a \) (where \( a > 0 \)) to obtain:
\[ - a \leq a \sin(bx + c) \leq a \]Add \( d \) to all terms of the above inequality to obtain:
\[ - a + d \leq a \sin(bx + c) + d \leq a + d \]The above inequality indicates that \( y = a \sin(bx + c) + d \) has a maximum value and a minimum value given by:
\[ \color{red}{ \text{max} = d + a \quad \text{and} \quad \text{min} = d - a } \]b) If we assume that the formulas \( \text{max} = d + a \) and \( \text{min} = d - a \) are equations in two variables \( d \) and \( a \), we may easily solve them for \( d \) and \( a \) to obtain:
\[ \color{red}{ d = \dfrac{\text{max} + \text{min}}{2} \quad \text{and} \quad a = \dfrac{\text{max} - \text{min}}{2} } \]
First the maximum max and the minimum min of the function shown in the graph below are equal to: \[ \text{max} = 0 \quad \text{and} \quad \text{min} = -4 \]
Using the formulas obtained in the last problem, we have: \[ d = \dfrac{\text{max} + \text{min}}{2} = -2 \quad \text{and} \quad a = \dfrac{\text{max} - \text{min}}{2} = 2 \]
We next find the period \( p \) from the graph of the function. \[ p = \dfrac{4\pi}{3} \]
Assuming \( b \) positive, the period is given by \( \dfrac{2\pi}{b} \). Hence the equation \[ \dfrac{2\pi}{b} = \dfrac{4\pi}{3} \] which gives \[ b = \dfrac{3}{2} \]
We can now write the function as \[ y = a \sin(bx + c) + d = 2 \sin\left( \dfrac{3x}{2} + c \right) - 2 \]
One way to determine \( c \) is to use a point on the given graph. For example for \( x = 0 \), \( y = 0 \) according to the graph. Hence \[ 2 \sin(0 + c) - 2 = 0 \] \[ \sin(c) = 1 \] gives: \( c = \dfrac{\pi}{2} + 2k\pi \), where \( k = 0, \pm 1, \pm 2, \ldots \)
Use \( c = \dfrac{\pi}{2} \) to write an expression for the function as: \[ \color{red}{y = 2 \sin\left( \dfrac{3x}{2} + \dfrac{\pi}{2} \right) - 2} \]
Find the constants \( a \), \( b \), \( c \), and \( d \) for the curve \[ y = a \sin(bx + c) + d \]
graphed below.
First the maximum max and the minimum min of the function shown in the graph below are equal to: \[ \text{max} = \dfrac{18}{5} \quad \text{and} \quad \text{min} = -\dfrac{6}{5} \]
Using the formulas obtained in the last problem, we have: \[ d = \dfrac{\text{max} + \text{min}}{2} = \dfrac{6}{5} \quad \text{and} \quad a = \dfrac{\text{max} - \text{min}}{2} = \dfrac{12}{5} \]
We next find the period \( p \) from the graph of the function. Since the points \( \left( \dfrac{3}{5}, \dfrac{18}{5} \right) \) and \( \left( \dfrac{7}{5}, -\dfrac{6}{5} \right) \) delimit half a cycle, the period \( p \) is given by twice the difference between the x coordinates of these points. Hence \[ p = 2 \left( \dfrac{7}{5} - \dfrac{3}{5} \right) = \dfrac{8}{5} \] Assuming \( b \) positive, the period is given by the formula \( \dfrac{2\pi}{b} \). Hence the equation \[ \dfrac{2\pi}{b} = \dfrac{8}{5} \] which gives \[ b = \dfrac{5\pi}{4} \]
We use the values of \( a \), \( b \), and \( d \) found above to write the function as \[ y = \dfrac{12}{5} \sin\left( \dfrac{5\pi x}{4} + c \right) + \dfrac{6}{5} \]
Use the point \( \left( \dfrac{3}{5}, \dfrac{18}{5} \right) \) by setting \( x = \dfrac{3}{5} \) and \( y = \dfrac{18}{5} \) in the equation and simplify. \[ \dfrac{18}{5} = \dfrac{12}{5} \sin\left( \dfrac{5\pi \cdot \dfrac{3}{5}}{4} + c \right) + \dfrac{6}{5} \] \[ \dfrac{12}{5} \sin\left( \dfrac{3\pi}{4} + c \right) = \dfrac{18}{5} - \dfrac{6}{5} \] \[ \dfrac{12}{5} \sin\left( \dfrac{3\pi}{4} + c \right) = \dfrac{12}{5} \] \[ \sin\left( \dfrac{3\pi}{4} + c \right) = 1 \]
We now solve the above trigonometric equation.
\[ \dfrac{3\pi}{4} + c = \dfrac{\pi}{2} + 2k\pi, \quad \text{where } k = 0, \pm1, \pm2, \ldots \] gives: \( c = \dfrac{\pi}{2} - \dfrac{3\pi}{4} = -\dfrac{\pi}{4} \), (we have used the solution for \( k = 0 \))We now write the final expression for the function as: \[ \color{red}{y = \dfrac{12}{5} \sin\left( \dfrac{5\pi x}{4} - \dfrac{\pi}{4} \right) + \dfrac{6}{5}} \]
The temperature \( T \), in degree Celsius, during the day is approximated by the function \[ T(t) = -8\cos\left(\dfrac{\pi t}{12}\right) + 30^{\circ} \] where \( t \) is the time in hours and \( t = 0 \) corresponds to 12:00 am.
a) Find the period of \( T \).
b) Find the maximum value of \( T \).
c) Find \( T \) at 12 am, 6 am, 12 pm, 6 pm and graph \( T \) over one period starting from 12 am.
a) Period \( P \) is given by: \[ P = \dfrac{2\pi}{\pi/12} = 24 \text{ hours} \]
b) Maximum value of \( T \) is given by:
\[ \max = 30 + 8 = 38^\circ \]c) Find values of \( T \) at 12 am, 6 am, 12 pm, 6 pm:
12 am corresponds to \( t = 0 \), hence
\[ T(12 \, \text{am}) = -8\cos(0) + 30^\circ = 22^\circ \]6 am corresponds to \( t = 6 \), hence
\[ T(6 \, \text{am}) = -8\cos\left(\dfrac{6\pi}{12}\right) + 30^\circ = 30^\circ \]12 pm corresponds to \( t = 12 \), hence
\[ T(12 \, \text{pm}) = -8\cos\left(\dfrac{12\pi}{12}\right) + 30^\circ = 38^\circ \]6 pm corresponds to \( t = 18 \), hence
\[ T(6 \, \text{pm}) = -8\cos\left(\dfrac{18\pi}{12}\right) + 30^\circ = 30^\circ \]Graph shown below.
