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5.3 Trigonometric Graphs

Read amplitude and period from sine and cosine.

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Ideas in this lesson

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Lesson brief and rules

5.3 Trigonometric Graphs

For $y = A \sin (Bx)$ , amplitude is $|A|$ and period is $\frac{2\pi }{\left|B\right|}$ when $B \ne 0$ .

Remember
A vertical shift changes the midline, not the amplitude.

Worked example

1

01 / Define a period and the least positive period, including the requirement that the domain is unchanged by translation. Use unit-circle coordinates to construct both graphs on

[0,2\pi ]

. Give their zeros, extrema and parity; list the least periods of all six trigonometric functions. Complete the quarter-cycle table and evaluate

\sin(\dfrac{13 \pi }{6}),\ \cos(\dfrac{13 \pi }{6})

. Explain separately why the constant function

f(x)=1

has no least positive period. Curve

1

\sin{\left(x \right)}

. Curve

2

\cos{\left(x \right)}

Show the requested calculations and reasoning. Written explanations and sketches require teacher review; choosing a summary does not verify that work. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

02 / A period

p>0

satisfies

D+p=D

and

f(x+p)=f(x)

for every

x\in D

. The least positive such number, when it exists, is the fundamental period.

03 / One full circuit of the unit circle adds

2\pi

to the input without changing either coordinate. The first return of both sine value and orientation, or cosine value and orientation, occurs after

2\pi

04 /

Input	Sine	Cosine
$0$	$0$	$1$
$\dfrac{\pi }{2}$	$1$	$0$
$\pi$	$0$	$-1$
$\dfrac{3 \pi }{2}$	$-1$	$0$
$2 \pi$	$0$	$1$

05 /

\sin x=0\iff x=n\pi

; sine maxima are at

x=\dfrac{\pi }{2}+2n\pi

and minima at

x=\dfrac{3\pi }{2}+2n\pi

\cos x=0\iff x=\dfrac{\pi }{2}+n\pi

; cosine maxima are at

2n\pi

and minima at

(2n+1)\pi

, for integers

n

06 / Reflection of the terminal point gives

\sin(-x)=-\sin x

and

\cos(-x)=\cos x

. Sine, cosine, secant and cosecant have least period

2\pi

; tangent and cotangent have least period

\pi

on their pole-excluded domains.

07 / Subtract

2 \pi

from the translated angle to obtain

\dfrac{\pi }{6}

. The values are

\dfrac{1}{2}

and

\dfrac{\sqrt{3}}{2}

08 / The constant function on all real inputs has every positive number as a period. Given any proposed least period, half of it is a smaller positive period, so no least one exists.

Worked example 2 · G01

Define a period and the least positive period, including the requirement that the domain is unchanged by translation. Use unit-circle coordinates to construct both graphs on $[0,2\pi ]$ . Give their zeros, extrema and parity; list the least periods of all six trigonometric functions. Complete the quarter-cycle table and evaluate $\sin(\dfrac{25 \pi }{6}),\ \cos(\dfrac{25 \pi }{6})$ . Explain separately why the constant function $f(x)=2$ has no least positive period. Curve $1$ : $\sin{\left(x \right)}$ . Curve $2$ : $\cos{\left(x \right)}$ .

A period $p>0$ satisfies $D+p=D$ and $f(x+p)=f(x)$ for every $x\in D$ . The least positive such number, when it exists, is the fundamental period.
One full circuit of the unit circle adds $2\pi$ to the input without changing either coordinate. The first return of both sine value and orientation, or cosine value and orientation, occurs after $2\pi$ .

Input	Sine	Cosine
$0$	$0$	$1$
$\dfrac{\pi }{2}$	$1$	$0$
$\pi$	$0$	$-1$
$\dfrac{3 \pi }{2}$	$-1$	$0$
$2 \pi$	$0$	$1$

\sin x=0\iff x=n\pi

; sine maxima are at

x=\dfrac{\pi }{2}+2n\pi

and minima at

x=\dfrac{3\pi }{2}+2n\pi

\cos x=0\iff x=\dfrac{\pi }{2}+n\pi

; cosine maxima are at

2n\pi

and minima at

(2n+1)\pi

, for integers

n

.
Reflection of the terminal point gives

\sin(-x)=-\sin x

and

\cos(-x)=\cos x

. Sine, cosine, secant and cosecant have least period

2\pi

; tangent and cotangent have least period

\pi

on their pole-excluded domains.
Subtract

4 \pi

from the translated angle to obtain

\dfrac{\pi }{6}

. The values are

\dfrac{1}{2}

and

\dfrac{\sqrt{3}}{2}

.
The constant function on all real inputs has every positive number as a period. Given any proposed least period, half of it is a smaller positive period, so no least one exists.
Written-work review:

Give the domain-preserving definition and distinguish a period from the least period.
Construct both full base cycles and the exact quarter-cycle table, including zeros, extrema and parity.
Reduce the translated angle and justify the constant-function exception.

Read the graphs

Read periodic base graphs

Define a period and the least positive period, including the requirement that the domain is unchanged by translation. Use unit-circle coordinates to construct both graphs on

[0,2\pi ]

. Give their zeros, extrema and parity; list the least periods of all six trigonometric functions. Complete the quarter-cycle table and evaluate

\sin(\dfrac{13 \pi }{6}),\ \cos(\dfrac{13 \pi }{6})

. Explain separately why the constant function

f(x)=1

has no least positive period. Curve

1

\sin{\left(x \right)}

. Curve

2

\cos{\left(x \right)}

Input	Sine	Cosine
$0$	$0$	$1$
$\dfrac{\pi }{2}$	$1$	$0$
$\pi$	$0$	$-1$
$\dfrac{3 \pi }{2}$	$-1$	$0$
$2 \pi$	$0$	$1$

\sin x=0\iff x=n\pi

; sine maxima are at

x=\dfrac{\pi }{2}+2n\pi

and minima at

x=\dfrac{3\pi }{2}+2n\pi

\cos x=0\iff x=\dfrac{\pi }{2}+n\pi

; cosine maxima are at

2n\pi

and minima at

(2n+1)\pi

, for integers

n

.
Reflection of the terminal point gives

\sin(-x)=-\sin x

and

\cos(-x)=\cos x

. Sine, cosine, secant and cosecant have least period

2\pi

; tangent and cotangent have least period

\pi

on their pole-excluded domains.
Subtract

2 \pi

from the translated angle to obtain

\dfrac{\pi }{6}

. The values are

\dfrac{1}{2}

and

\dfrac{\sqrt{3}}{2}

Give the domain-preserving definition and distinguish a period from the least period.
Construct both full base cycles and the exact quarter-cycle table, including zeros, extrema and parity.
Reduce the translated angle and justify the constant-function exception.

Answer: Sine/cosine least period $2 \pi$ ; parities odd/even; translated values $\dfrac{1}{2}$ and $\dfrac{\sqrt{3}}{2}$ .

Transform heights and absolute outputs

Graph

y=\sin{\left(x \right)} + 3

over

[0,2\pi ]

. Describe the transformations, give the full range and least period, and locate all extrema in this interval. For an absolute output, explain why it is not a single sine or cosine curve. Curve

1

\sin{\left(x \right)} + 3

The transformation is vertical translation. Starting from $-1\le \sin{\left(x \right)}\le1$ , transform the entire inequality, reversing signs when multiplying by a negative number.
This gives the exact range $[2,4]$ . Its maximum is not an amplitude when the midline is nonzero.
Vertical translation and nonzero scaling preserve the base fundamental period $2\pi$ . The sinusoidal amplitude is $1.0$ .
Exact guide points: $\left( 0, \ 3\right)$ , $\left( \dfrac{\pi }{4}, \ \dfrac{\sqrt{2}}{2} + 3\right)$ , $\left( \dfrac{\pi }{2}, \ 4\right)$ , $\left( \dfrac{3 \pi }{4}, \ \dfrac{\sqrt{2}}{2} + 3\right)$ , $\left( \pi , \ 3\right)$ , $\left( \dfrac{5 \pi }{4}, \ 3 - \dfrac{\sqrt{2}}{2}\right)$ , $\left( \dfrac{3 \pi }{2}, \ 2\right)$ , $\left( \dfrac{7 \pi }{4}, \ 3 - \dfrac{\sqrt{2}}{2}\right)$ , $\left( 2 \pi , \ 3\right)$ . All maximum coordinates in the closed interval are $\left( \dfrac{\pi }{2}, \ 4\right)$ . All minimum coordinates are $\left( \dfrac{3 \pi }{2}, \ 2\right)$ . These lists include every repeated extremum and both closed endpoints when applicable.
Written-work review:

Identify every vertical transformation and compute the exact full range.
Draw every extremum on the stated interval, including endpoint extrema.
Explain the folded-output least period and why an absolute-value wave is not one sinusoid.

Answer: Range $[2,4]$ ; least period $2 \pi$ .

Recover a formula from an actual graph

The plotted curve is exactly one complete sinusoidal cycle. Read its extrema, endpoints and initial direction. Recover the amplitude, period, horizontal shift and midline, then give an equation in sine or cosine form. Use the exact marked coordinates, not rounded grid ticks.

Point	Horizontal coordinate	Vertical coordinate
$P1$	$\dfrac{\pi }{6}$	$1$
$P2$	$\dfrac{5 \pi }{12}$	$4$
$P3$	$\dfrac{2 \pi }{3}$	$1$
$P4$	$\dfrac{11 \pi }{12}$	$-2$
$P5$	$\dfrac{7 \pi }{6}$	$1$

Curve

1

: unidentified sinusoid.

Normalize the frequency to be positive while retaining the $\sin$ basis and signed outside coefficient. Since sine is odd and cosine is even, the equivalent factored model is $y=1+3\sin{\left(2\left(x-\dfrac{\pi }{6}\right)\right)}$ .
$A=3,\quad k=2,\quad h=\dfrac{\pi }{6},\quad d=1$ . Thus $T=\dfrac{2\pi }{k}=\pi$ .
In this retained $\sin$ basis with the stated signed coefficient $A$ , the phase is $b=k h=\dfrac{\pi }{3}$ , whereas the horizontal shift is $\dfrac{\pi }{6}$ . A phase in radians is not a horizontal shift in input units.
The five quarter-cycle coordinates are $\left( \dfrac{\pi }{6}, \ 1\right)$ , $\left( \dfrac{5 \pi }{12}, \ 4\right)$ , $\left( \dfrac{2 \pi }{3}, \ 1\right)$ , $\left( \dfrac{11 \pi }{12}, \ -2\right)$ , $\left( \dfrac{7 \pi }{6}, \ 1\right)$ . The range is $[-2,4]$ .
This nonzero-amplitude graph repeats after $\pi$ . Any smaller period would give a smaller period of the base sine or cosine after undoing the nonzero vertical scale and input scale.
Curve $1$ : unidentified sinusoid.

Written-work review:

Read actual extrema, exact marked endpoints and initial direction from the unidentified graph.
Recover a formula and verify it against all five marked points.
Explain why a single point would not determine the sinusoid.

Answer: Amplitude $3$ ; least period $\pi$ ; shift $\dfrac{\pi }{6}$ ; midline $y=1$ ; factored-argument phase $\dfrac{\pi }{3}$ .

Choose honest windows and domains

Choose and justify a viewing window and sampling step that resolve the curve for

y=2 \sin{\left(48 x \right)}

. State its domain and least positive period, draw every visible branch and mark excluded poles separately. Explain what a default wide or narrow view can hide. Curve

1

2 \sin{\left(48 x \right)}

The proposed window is $[- \dfrac{\pi }{24},\dfrac{\pi }{24}]\times[-2.5,2.5]$ . The plotted horizontal width is $\dfrac{\pi }{12}$ .
Two complete cycles fit the horizontal span. A step no larger than one eightieth of the period resolves extrema rather than joining one sample per cycle into an aliased line.
Use $\Delta x\le \dfrac{\pi }{1920}$ for the trigonometric curve and split at every listed domain boundary. No finite plot alone establishes a global period.
The domain is $\mathbb{R}$ . The base trig period divided by the input frequency gives $\dfrac{\pi }{24}$ , with domain translation invariance checked as well.
The coarse alternative uses a sampling step equal to a whole period. Its samples repeat the same phase and hide intervening extrema, zeros and, for pole-bearing functions, entire branches. Thus it is not a nonmisleading resolved graph, even though its span and domain are correct. A different well-resolved window is acceptable; the supplied rectangle is not unique.
For this actual proposed rectangle, the coarse grid starting at its left edge gives at $- \dfrac{\pi }{24}$ , output $0$ ; at $0$ , output $0$ ; at $\dfrac{\pi }{24}$ , output $0$ . The repeated values omit interior extrema or intervening branches. The good proposal uses $80$ steps per trig period instead.
Written-work review:

Justify both the window and sampling scale from the actual oscillation period.
Preserve every domain gap and pole; distinguish output squaring from an input square.
For the approximation subtype, evaluate every polynomial term and compare the computed result with cosine.

Answer: Least period $\dfrac{\pi }{24}$ ; proposed horizontal span $\dfrac{\pi }{12}$ ; sampling step $\dfrac{\pi }{1920}$ ; domain $\mathbb{R}$ .

Add oscillations point by point

Plot the two summands and their sum on the same axes for

f(x)=\dfrac{x}{2}

g(x)=\sin{\left(x \right)}

. Complete a four-column table at

0

\dfrac{\pi }{5}

\dfrac{\pi }{2}

\pi

, explain ordinate addition, and decide whether the sum is periodic, nonperiodic or constant. Relate harmonic sums to sound; distinguish changing amplitude from changing oscillation spacing. Curve

1

\dfrac{x}{2}

. Curve

2

\sin{\left(x \right)}

. Curve

3

\dfrac{x}{2} + \sin{\left(x \right)}

At any fixed horizontal coordinate, add the two vertical coordinates; do not add the horizontal coordinates.

Input	First summand	Second summand	Sum
$0$	$0$	$0$	$0$
$\dfrac{\pi }{5}$	$\dfrac{\pi }{10}$	$\dfrac{\sqrt{10 - 2\,\sqrt{5}}}{4}$	$\dfrac{\pi }{10} + \dfrac{\sqrt{10 - 2\,\sqrt{5}}}{4}$
$\dfrac{\pi }{2}$	$\dfrac{\pi }{4}$	$1$	$\dfrac{\pi }{4} + 1$
$\pi$	$\dfrac{\pi }{2}$	$0$	$\dfrac{\pi }{2}$

The linear part is unbounded while the sine part is bounded, so a translated repeating graph is impossible.
Finite harmonic sums model combinations of oscillations. Amplitude modulation changes the envelope height; frequency modulation changes local cycle spacing. These are different mechanisms, not interchangeable labels for any non-sinusoidal sum.
Written-work review:

Compute every requested row of the two-summand-and-sum table.
Draw all three curves on shared axes and explain ordinate addition.
Justify repetition, constant cancellation or nonperiodic drift and distinguish amplitude from frequency modulation.

Answer: Sum at $\dfrac{\pi }{5}$ : $\dfrac{\pi }{10} + \dfrac{\sqrt{10 - 2\,\sqrt{5}}}{4}$ ; classification: nonperiodic.

Prove and interpret envelopes

Graph all three functions

a(x)=\dfrac{x^{2}}{2}

-a(x)=- \dfrac{x^{2}}{2}

, and

F(x)=\dfrac{x^{2} \cos{\left(2 \pi x \right)}}{2}

on the same axes. State the real domain; prove the correctly ordered envelope inequality, identify when the curves touch, and explain any swapping of upper and lower boundaries. Evaluate the product and absolute envelope at

\dfrac{1}{11}

to four decimals. Curve

1

\dfrac{x^{2}}{2}

. Curve

2

- \dfrac{x^{2}}{2}

. Curve

3

\dfrac{x^{2} \cos{\left(2 \pi x \right)}}{2}

$|\cos{\left(2 \pi x \right)}|\le1$ implies $|F(x)|=|a(x)|\,|\cos{\left(2 \pi x \right)}|\le|a(x)|$ . Therefore $-|a(x)|\le F(x)\le|a(x)|$ on $\mathbb{R}$ .
The two drawn curves $a(x)$ and $-a(x)$ exchange upper/lower roles wherever $a(x)$ changes sign. Multiplying an inequality by a negative envelope without reversing signs is invalid.
Touching occurs wherever $a(x)=0$ or the oscillating factor equals $1$ or $-1$ . The actual upper/lower assignment depends on the sign of $a(x)$ .
At $x=\dfrac{1}{11}$ , the product is $0.0035$ and the absolute envelope is $0.0041$ . Envelope extrema are not automatically extrema of the product.
The displayed window illustrates the proved bound; it does not prove a global extrema claim.
Written-work review:

Draw the envelope, its negative and the product together on the correct domain.
Prove the ordered absolute-envelope inequality, reversing boundaries when the envelope changes sign.
Identify touching conditions and evaluate both product and bound without confusing their extrema.

Answer: Product $0.0035$ ; absolute envelope $0.0041$ ; bound $-|a(x)|\le F(x)\le|a(x)|$ .

Find complete and global extrema

Find the minimum and maximum of

f(x)=2 \sin{\left(2 x \right)} + 2 \cos{\left(x \right)}

\mathbb{R}

. Give locations or exact candidate equations as well as values, rounded to four decimals where needed. Check every endpoint when the interval is bounded. Use algebra or the supplied stationary-point conditions in the solution to justify completeness, rather than trusting sampled extrema. Curve

1

2 \sin{\left(2 x \right)} + 2 \cos{\left(x \right)}

The function has period $2\pi$ . The stationary condition is $-\sin x+2\cos(2x)=0$ . Setting $u=\sin x$ gives $4u^{2}+u-2=0$ , whose two roots are $u=\dfrac{-1\pm\,\sqrt{33}}8$ .
Both roots lie in $[-1,1]$ . Each gives two sine angles in one cycle. The four candidate coordinates are $\left( - \operatorname{asin}{\left(\dfrac{1}{8} - \dfrac{\sqrt{33}}{8} \right)}, \ 3.52034518609\right)$ , $\left( \operatorname{asin}{\left(\dfrac{1}{8} - \dfrac{\sqrt{33}}{8} \right)} + \pi , \ -3.52034518609\right)$ , $\left( \operatorname{asin}{\left(\dfrac{1}{8} + \dfrac{\sqrt{33}}{8} \right)} + \pi , \ 0.738017459656\right)$ , $\left( - \operatorname{asin}{\left(\dfrac{1}{8} + \dfrac{\sqrt{33}}{8} \right)} + 2 \pi , \ -0.738017459656\right)$ . The cycle endpoints both have output $2$ .
Every global extremum is attained in a closed full cycle. Comparing all four stationary outputs and both endpoints gives minimum $-3.52034519$ and maximum $3.52034519$ . The points attaining each value are selected explicitly: minimum at $\operatorname{asin}{\left(\dfrac{1}{8} - \dfrac{\sqrt{33}}{8} \right)} + \pi$ and maximum at $- \operatorname{asin}{\left(\dfrac{1}{8} - \dfrac{\sqrt{33}}{8} \right)}$ , together with every $2n\pi$ translate. Separate summand extrema are not simultaneous.
Written-work review:

List all valid algebraic or stationary candidates and all bounded-interval endpoints.
Evaluate and compare every candidate, reporting both extrema and their locations or defining equations.
For the rational-decay case, prove parity, uniqueness of the central-lobe minimum and a strict tail bound.

Answer: Minimum $-3.5203$ ; maximum $3.5203$ .

Find every graphical intersection

Use the graph of

y=\sin{\left(x \right)}

and

y=\dfrac{1}{5}

to solve

\sin{\left(x \right)}=\dfrac{1}{5}

[0,2 \pi ]

. List all solutions in increasing order to two decimals. Also give every input in this interval where the left side is at least the right side. Curve

1

\sin{\left(x \right)}

. Curve

2

\dfrac{1}{5}

Intersections correspond to equal outputs only where both functions are defined. Dashed asymptotes are not intersections.
Sine increases from $0$ to $1$ on the first quadrant and then decreases to $0$ ; it is nonpositive on the second half-cycle. Thus exactly these two crossings occur.
The complete root list is $\operatorname{asin}{\left(\dfrac{1}{5} \right)}$ , $\pi - \operatorname{asin}{\left(\dfrac{1}{5} \right)}$ . Rounded only at the end: $0.20$ , $2.94$ .
The sine curve is at or above the target exactly on $[\operatorname{asin}{\left(\dfrac{1}{5} \right)},\pi - \operatorname{asin}{\left(\dfrac{1}{5} \right)}]$ . Include both crossing endpoints because the inequality is inclusive.
Written-work review:

List all admissible crossings on the complete closed interval, with correct rounding.
Use monotonicity, factorization or complete branch enumeration to prove no root was missed.
Exclude poles, retain endpoint roots and give inclusive threshold intervals when requested.

Answer: Solutions $\{0.20,\ 2.94\}$ .

Keep quotient holes and limits distinct

For

f(x)=\dfrac{1 - \cos{\left(2 x \right)}}{2 x}

, determine the full domain and parity, every horizontal-axis intercept, a global and near-zero graph, both tail limits, and the limit at zero. Explain why the limiting value is not

f(0)

. Curve

1

\dfrac{1 - \cos{\left(2 x \right)}}{2 x}

Curve

1

\dfrac{1 - \cos{\left(2 x \right)}}{2 x}

Denominator $\ne0$ , so the domain is $\mathbb{R}\setminus\{0\}$ . This domain is symmetric about zero.
The numerator is even and the denominator is odd. Consequently $f(-x)=-f(x)$ , proving odd parity.
Zeros require the numerator to vanish and the input to remain in the domain: $x=\dfrac{2n\pi }{2},\quad n\in\mathbb{Z}\setminus\{0\}$ . In particular zero itself is not an intercept.
$|f(x)|\le\dfrac{2}{2|x|}$ proves that both tails approach zero.
The bound $0\le1-\cos u\le\dfrac{u^{2}}{2}$ gives $|f(x)|\le\dfrac{4|x|}{4}\to0$ .
The hole is at $(0,0)$ . It stays open unless a new extended function is explicitly defined.
Written-work review:

State the symmetric punctured domain, parity and every nonzero zero family.
Draw both global and near-zero views with an open hole at the correct limiting height.
Prove both-tail bounds and the zero-input limit without assigning a value at the excluded input.

Answer: Parity odd; zero-input limit $0$ ; both-tail limit $0$ ; value at zero undefined.

Interpret periodic signals with units

A harbor water-height model above a fixed chart datum is

y(t)=\dfrac{5 \cos{\left(\dfrac{\pi t}{7} \right)}}{2} + \dfrac{3}{2}

, with time in seconds and output in meters. Find the period, frequency, midline, maximum, minimum and peak-to-trough change. Draw two cycles with units on the axes. Find the crest-to-trough height, not merely the amplitude. Curve

1

y(t)=\dfrac{5 \cos{\left(\dfrac{\pi t}{7} \right)}}{2} + \dfrac{3}{2}

Compare the coefficient of time with $\omega=\dfrac{2\pi }{T}$ . Thus $T=14$ seconds and $f=\dfrac{1}{T}=0.0714285714285714$ cycles per second.
The midline is $\dfrac{3}{2}$ meters and amplitude is $\dfrac{5}{2}$ meters. Therefore minimum $-1$ , maximum $4$ , and peak-to-trough difference $5$ meters.
The model has a nonzero baseline, so amplitude is not the maximum output.
Written-work review:

Derive period and reciprocal frequency with the correct input units.
Give baseline, both extrema and twice-amplitude spread and draw two cycles.
Complete the relevant millisecond conversion, numerical reference comparison or magnitude-brightness interpretation.

Answer: Period $14$ seconds; frequency $\dfrac{1}{14}$ cycles per second; minimum $-1$ and maximum $4$ meters.

Prove composition periodicity and parity

Graph and classify each function by domain-aware parity and genuine least positive period, or prove nonperiodicity:

f_1(x)=\left|{\sin{\left(x \right)}}\right|

f_2(x)=\sin{\left(\left|{x}\right| \right)}

. A finite repeating-looking plot is evidence for a conjecture, not a proof. Curve

1

\left|{\sin{\left(x \right)}}\right|

. Curve

2

\sin{\left(\left|{x}\right| \right)}

$|\sin(1(x+\dfrac{\pi }{1}))|=|\sin(x)|$ proves the first period. Its zeros are spaced by $\pi$ , so no smaller positive period can preserve that zero set.
Both expressions are even, but their values differ on negative inputs. If the second had period $P>0$ , its positive-input sine tail forces $P=2n\pi$ . For small $\varepsilon>0$ , values at $-\varepsilon$ and $P-\varepsilon$ would be opposite nonzero numbers. This contradicts periodicity.
Written-work review:

State the complete actual domain before claiming parity.
Draw every composition and give a valid period-minimality or nonperiodicity proof for each.
For the nested comparison, prove the strict global inequality using both sign cases rather than graph samples.

Answer: First: even, least period $\pi$ ; second/result: even, nonperiodic.

Find all equivalent phase families

The plotted sinusoid has midline

-1

, amplitude

3

, and one cycle from

- \dfrac{1}{3}

- \dfrac{1}{3} + 2 \pi

, beginning at a downward midline crossing. For the fixed positive frequency

1

, find every pair

(A,b)

in both forms

y=-1+A\sin(x-b)

and

y=-1+A\cos(x-b)

. The amplitude

|A|

is nonnegative. Include both signs of the coefficient

A

and all integer phase families, then prove completeness. Curve

1

: unidentified downward-starting sinusoid.

The graph gives $y=- 3 \sin{\left(x + \dfrac{1}{3} \right)} - 1$ . The midline and frequency are fixed; the maximum and minimum force $|A|=3$ .
For sine, $A=-3$ requires $b=- \dfrac{1}{3}+2n\pi$ . Changing the sign requires adding $\pi$ to the phase, so $A=3$ requires $b=- \dfrac{1}{3} + \pi +2n\pi$ .
Using $-\sin u=\cos(u+\dfrac{\pi }{2})$ , the positive-coefficient cosine phase is $b=- \dfrac{\pi }{2} - \dfrac{1}{3}+2n\pi$ . Negating its coefficient adds $\pi$ to that phase, giving $b=- \dfrac{1}{3} + \dfrac{\pi }{2}+2n\pi$ .
Completeness: the nonnegative amplitude $|A|$ is fixed by the extrema, leaving only the two coefficient signs. Expanding the sine or cosine difference and comparing the coefficients of $\sin(x)$ and $\cos(x)$ fixes the phase modulo $2\pi$ . Thus there are no other coefficient or phase families for the specified frequency and midline. Horizontal shifts are these phases divided by $1$ .
Written-work review:

Recover the amplitude magnitude, positive frequency and midline from the actual graph.
Give both coefficient signs for both sine and cosine with all integer phase families.
Prove completeness by comparing sine and cosine coefficients and distinguish phases from horizontal shifts.

Answer: Sine $(A,b)=(-3,- \dfrac{1}{3}+2n\pi )$ , $(A,b)=(3,- \dfrac{1}{3} + \pi +2n\pi )$ ; cosine $(A,b)=(3,- \dfrac{\pi }{2} - \dfrac{1}{3}+2n\pi )$ , $(A,b)=(-3,- \dfrac{1}{3} + \dfrac{\pi }{2}+2n\pi )$ , $n\in\mathbb{Z}$ .

15 questions with a worked answer for each.Start practice ↗

Textbook reference: Stewart, Redlin & Watson, Precalculus: Mathematics for Calculus, 7th ed. (Cengage, 2016). The explanations and practice here are original, organized by textbook section.

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5.3 Trigonometric Graphs

Ideas in this lesson

5.3 Trigonometric Graphs

Read the graphs

Read periodic base graphs

Transform heights and absolute outputs

Recover a formula from an actual graph

Choose honest windows and domains

Add oscillations point by point

Prove and interpret envelopes

Find complete and global extrema

Find every graphical intersection

Keep quotient holes and limits distinct

Interpret periodic signals with units

Prove composition periodicity and parity

Find all equivalent phase families