Find Trigonometric Functions Given Their Graphs With Phase Shift (2)

Learn how to find the amplitude, period, and phase shift of trigonometric functions by analyzing their graphs. This guide provides step-by-step questions with detailed solutions and clear explanations to help you master these concepts. For additional support, explore our interactive tutorials on understanding the phase shift, the period and the vertical shifts of trigonometric functions.

Questions

Question 1

Find the amplitude, period and phase shift for the curves shown below, then write the function in the form \( y = a \sin(bx + c) \).

a b c d e

Solution

Amplitude: \[ |a| = 2 \] We reproduce the graph of 1.a below and note the following:

\( 4 \) small divisions = \( \pi \), and therefore \( 1 \) small division = \( \dfrac{\pi}{4} \)

One period = \( 16 \) small divisions; Hence: \[ 1 \text{ period} = 16 \times \dfrac{\pi}{4} = 4\pi \]

Phase shift: It is the shift between the graphs of \( y = a \sin(bx) \) and \( y = a \sin(bx + c) \) and is defined by \( -\dfrac{c}{b} \).

In the graph of 1.a the phase shift is equal to \( -\dfrac{\pi}{4} \) as shown below (1 small division to the left):

We now use the results found above to write an equation of the form \( y = a \sin(bx + c) \) to the graph in 1.a:

\( |a| = 2 \), hence \( a = \pm 2 \). Let \( a = 2 \).

\( 1 \) period = \( 4\pi = \dfrac{2\pi}{b} \) (assuming \( b > 0 \)). Hence \[ b = \dfrac{2\pi}{4\pi} = \dfrac{1}{2} \]

Phase shift = \( -\dfrac{\pi}{4} = -\dfrac{c}{b} \).

Substitute \( b \) by its value to find: \[ c = b \cdot \dfrac{\pi}{4} = \dfrac{1}{2} \cdot \dfrac{\pi}{4} = \dfrac{\pi}{8} \] Equation of graph 1.a \[ y = 2 \sin \left( \dfrac{x}{2} + \dfrac{\pi}{8} \right) \]

b) Graph in 1.b:

Amplitude: \( |a| = 1.5 \)

One period: \( 4 \)

Phase shift: \( 1 \) unit to the right \( = 1 \)

We now use the results found above to write an equation of the form \( y = a \sin(bx + c) \) to the graph in 1.b

\( |a| = 1.5 \), hence \( a = \pm 1.5 \), let \( a = 1.5 \)

One period: \( 4 = \dfrac{2\pi}{b} \) (assuming \( b > 0 \)). Hence \( b = \dfrac{\pi}{2} \)

Phase shift: \( 1 = -\dfrac{c}{b} \)

Substitute \( b \) by its value to find: \[ c = -b = -\dfrac{\pi}{2} \] Equation of graph 1.b \[ y = 1.5 \sin\left( \dfrac{\pi x}{2} - \dfrac{\pi}{2} \right) \]

c) Graph in 1.c:

Amplitude: \( |a| = 10 \)

1 small division \( = \dfrac{\pi}{5} \), 1 period \( = 8 \) divisions

Hence, 1 period \( = \dfrac{8\pi}{5} \)

Phase shift \( = 2 \) divisions \( = \dfrac{2\pi}{5} \)

We now use the results found above to write an equation of the form \( y = a \sin(bx + c) \) to the graph in 1.c \( |a| = 10 \), hence \( a = \pm 10 \), let \( a = 10 \)

1 period \( = \dfrac{8\pi}{5} = \dfrac{2\pi}{b} \) (assuming \( b > 0 \)). Hence \( b = \dfrac{5}{4} \)

Phase shift \( = \dfrac{2\pi}{5} = -\dfrac{c}{b} \)

Substitute \( b \) by its value to find: \[ c = -\dfrac{2\pi b}{5} = -\dfrac{\pi}{2} \] Equation of graph 1.c \[ y = 10 \sin\left(\dfrac{5x}{4} - \dfrac{\pi}{2}\right) \]

d) Graph in 1.d:

Amplitude: \( |a| = 3 \)

1 small division = \( \dfrac{\pi}{12} \), 1 period = 16 divisions

Hence 1 period = \( 16 \times \dfrac{\pi}{12} = \dfrac{4\pi}{3} \)

Phase shift = 2 divisions to the left = \( -2 \times \dfrac{\pi}{12} = -\dfrac{\pi}{6} \)

We now use the results found above to write an equation of the form \( y = a \sin(bx + c) \) to the graph in 1.d

\( |a| = 3 \), hence \( a = \pm 3 \), let \( a = 3 \)

1 period = \( \dfrac{4\pi}{3} = \dfrac{2\pi}{b} \) (assuming \( b > 0 \)). Hence \( b = \dfrac{3}{2} \)

Phase shift = \( -\dfrac{\pi}{6} = -\dfrac{c}{b} \)

Substitute \( b \) by its value to find: \( c = \dfrac{\pi b}{6} = \dfrac{\pi}{4} \)

e) Graph in 1.e:

Amplitude: \( |a| = 2 \)

1 small division = \( \dfrac{\pi}{12} \), 1 period = 8 divisions

Hence 1 period = \( 8 \times \dfrac{\pi}{12} = \dfrac{2\pi}{3} \)

Phase shift = 1 division to the right = \( \dfrac{\pi}{12} \)

We now use the results found above to write an equation of the form \( y = a \sin(bx + c) \) to the graph in 1.e

\( |a| = 2 \), hence \( a = \pm 2 \), let \( a = 2 \)

1 period = \( \dfrac{2\pi}{3} = \dfrac{2\pi}{b} \) (assuming \( b > 0 \)). Hence \( b = 3 \)

Phase shift = \( \dfrac{\pi}{12} = -\dfrac{c}{b} \)

Substitute \( b \) by its value to find: \( c = -\dfrac{\pi b}{12} = -\dfrac{\pi}{4} \)

As an exercise graph each of the functions found above and compare its graph to the given graph.

Question 2

2.a 2.b 2.c

Solution

amplitude: \(= |a| = 4\)

We reproduce the graph of 2.a below and note the following:

One period \( = 3 π/ 2 \dfrac{2 \pi}{2} \)

Phase shift: It is the shift from the graphs of \( y = a \cos(bx) \) to the graph of \( y = a \cos(bx + c) \) and is defined by \[ -\dfrac{c}{b} \]

The graph of \( y = a \cos(bx) \) gives a maximum at \( x = 0 \). In the graph of 2.a, the maximum is shifted 3 small divisions to the right.

1 small division \( = \dfrac{\pi}{8} \)

Hence, phase shift = \( 3 \times \pi / 3 = \dfrac{3\pi}{8} \)

We now use the results found above to write an equation of the form \( y = a \cos(bx + c) \) that models the graph in the image above.

\(|a| = 4\), hence \( a = \pm 4 \); let \( a = 4 \)

1 period = \( \dfrac{3\pi}{2} = \dfrac{2\pi}{b} \) (assuming \( b > 0 \)). Hence \( b = \dfrac{4}{3} \)

Phase shift = \( \dfrac{3\pi}{8} = -\dfrac{c}{b} \)

Substitute \( b \) by its value to find: \[ c = -b \times \dfrac{3\pi}{8} = -\dfrac{\pi}{2} \]

Final equation of the given curve in 2.a: \[ y = 4 \cos\left(\dfrac{4x}{3} - \dfrac{\pi}{2}\right) \] b) Graph in 2.b:

amplitude: \( = |a| = 3 \)

One period = 1 (length on the x axis of one cycle)

Phase shift = 1 / 2 (half a unit to the right)

We now use the results found above to write an equation of the form y = a cos(bx + c) to the graph in 2.b

\(|a| = 3\), hence \(a = \pm 3\), let \(a = 3\)

1 period = 1 = \(\dfrac{2\pi}{b}\) (assuming \(b > 0\)). Hence \(b = 2\pi\)

Phase shift = \(\dfrac{1}{2} = -\dfrac{c}{b}\)

Substitute \( b \) by its value to find: \[ c = -\dfrac{b}{2} = -\pi\]

Final equation of the given curve in 2.b: \[ y = 3 \cos(2\pi x - \pi) \] c) Graph in 2.c:

amplitude: \( = |a| = 40 \)

1 small division = \( \dfrac{\pi}{2} \div 4 = \dfrac{\pi}{8} \)

1 period = 8 divisions

Hence 1 period = \( 8 \times \dfrac{\pi}{8} = \pi \)

Phase shift = 3 divisions (to the right) = \( 3 \times \dfrac{\pi}{8} = \dfrac{3\pi}{8} \)

We now use the results found above to write an equation of the form \( y = a \cos(bx + c) \) to the graph in 2.c

\( |a| = 40 \), hence \( a = \pm 40 \), let \( a = 40 \)

1 period = \( \pi = \dfrac{2\pi}{b} \) (assuming \( b > 0 \)). Hence \( b = 2 \)

Phase shift = \( \dfrac{3\pi}{8} = -\dfrac{c}{b} \)

Substitute \( b \) by its value to find: \[ c = -\dfrac{3\pi b}{8} = -\dfrac{3\pi}{4} \)]

Final equation of the given curve in 2.c: \[ y = 40 \cos\left(2x - \dfrac{3\pi}{4}\right) \]

As an exercise graph each of the functions found above and compare its graph to the given graph.

Question 3

The graph of the function \( y = sin(bx + c) \) has an x-intercept at \( x = \dfrac{6\pi}{5} \) and a maximum at \( x = \dfrac{11\pi}{5} \) as shown below.

b) Find b and c, and the equation of the graph.

Solution

a) The absolute value of the difference of the \( x \) coordinates of the points \(\left( \dfrac{6\pi}{5}, 0 \right)\) and \(\left( \dfrac{11\pi}{5}, 1 \right)\) gives the quarter of the period. Hence the period \( P \) is equal to: \[ P = 4\left( \dfrac{11\pi}{5} - \dfrac{6\pi}{5} \right) = 4\pi \] b) The period found above is also given by \[ P = \dfrac{2\pi}{|b|} \] Take \( b \) positive and solve the equation \( \dfrac{2\pi}{b} = 4\pi \) for \( b \) \[ b = \dfrac{1}{2} \] The phase shift of the graph given above is equal to \( \dfrac{6\pi}{5} \) and is also given by the formula \( -\dfrac{c}{b} \). Hence the equation: \[ -\dfrac{c}{b} = \dfrac{6\pi}{5} \] Solve for \( c \). \[ c = -\dfrac{6\pi b}{5} = -\dfrac{3\pi}{5} \] The equation of the graph is given by: \[ y = \sin\left( \dfrac{1}{2}x - \dfrac{3\pi}{5} \right) \]

References and Links

• Properties of The Six Trigonometric Functions
• High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers
• Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
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