Algebra 2 Graphs of Trigonometric Functions and Identities Worksheet Solutions (PDF)

Algebra 2 math worksheet solutions: Graphs of Trigonometric Functions and Identities – Maths PDF

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This algebra 2 trig graphs answers for Algebra 2 covers graphs of trigonometric functions and identities (Common Core F-TF.B.5, F-TF.C.8, F-IF.C.7e), with a free PDF to print.

Every worksheet problem on graphs of trigonometric functions and the Pythagorean identity is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where using B instead of 2π/B for the period can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.

Algebra 2 · Common Core State Standards: F-TF.B.5, F-TF.C.8, F-IF.C.7e

Algebra 2 trig graphs answers: exercises 1 to 10

Exercise 1 – Read the parameters

Amplitude \(|3| = 3\). Period \(\frac{2\pi}{2} = \pi\). Midline \(y = 1\).

Amplitude 3; period π; midline y = 1

Exercise 2 – Reflection and stretch

Amplitude \(|-2| = 2\). \(B = \frac{1}{2}\), so the period is \(\frac{2\pi}{1/2} = 4\pi\).

The negative sign reflects the graph over the x-axis, so it starts at its minimum.

Amplitude 2; period 4π; reflected

Exercise 3 – Phase shift

The form \(x – C\) with \(C = \frac{\pi}{3}\) shifts the graph right. The midline is \(y = 2\).

Right π/3; midline y = 2

Exercise 4 – Equation from a graph

Amplitude 2, period π.

Amplitude \(\frac{1 – (-3)}{2} = 2\). Midline \(D = -1\). \(B = \frac{2\pi}{\pi} = 2\).

\(y = 2\sin(2x) – 1\)

Exercise 5 – Write from features

\(A = 4\), \(B = \frac{2\pi}{\pi} = 2\), \(D = 2\).

\(y = 4\sin(2x) + 2\)

Exercise 6 – Maximum and minimum

\(\cos x\) ranges from \(-1\) to \(1\), so \(5\cos x – 3\) ranges from \(-5 – 3 = -8\) to \(5 – 3 = 2\).

Maximum 2; minimum -8

Exercise 7 – Ferris wheel

\(h(0) = 25 – 20 = 5\). \(h(15) = 25 – 20\cos\pi = 45\).

\(h(10) = 25 – 20\cos\frac{2\pi}{3} = 25 + 10 = 35\).

Period: \(\frac{2\pi}{\pi/15} = 30\) seconds.

5 ft, 45 ft, 35 ft; period 30 s

Exercise 8 – Tides

Amplitude 4. \(B = \frac{\pi}{6}\), so the period is \(\frac{2\pi}{\pi/6} = 12\) hours.

\(y(3) = 4\sin\frac{\pi}{2} + 6 = 10\).

Amplitude 4 ft; period 12 hours; 10 ft at 3 AM

Exercise 9 – Temperature

Midline 65, amplitude 15: maximum \(80\), minimum \(50\).

\(T(15) = 15\sin\frac{\pi}{2} + 65 = 80\): the high occurs at 3 PM.

High 80°F, low 50°F; T(15) = 80°F

Exercise 10 – Find B from the period

\(\frac{2\pi}{B} = \frac{\pi}{2}\), so \(B = 4\).

\(B = 4\)

Exercises 11 and up

Exercise 11 – Identity at an angle

\(\left(\frac{1}{2}\right)^{2} + \left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{1}{4} + \frac{3}{4} = 1\).

Verified

Exercise 12 – Find cosine in quadrant II

The right triangle in the unit circle: cos² θ + sin² θ = 1.

\(\cos^{2}\theta = 1 – \frac{9}{25} = \frac{16}{25}\). In quadrant II cosine is negative: \(\cos\theta = -\frac{4}{5}\).

\(\tan\theta = \frac{3/5}{-4/5} = -\frac{3}{4}\).

cos = \(-\frac{4}{5}\), tan = \(-\frac{3}{4}\)

Exercise 13 – Find sine in quadrant III

\(\sin^{2}\theta = 1 – \frac{25}{169} = \frac{144}{169}\). Sine is negative in quadrant III: \(\sin\theta = -\frac{12}{13}\).

\(\tan\theta = \frac{-12/13}{-5/13} = \frac{12}{5}\).

sin = \(-\frac{12}{13}\), tan = \(\frac{12}{5}\)

Exercise 14 – Quadrant IV

\(\cos^{2}\theta = 1 – \frac{1}{4} = \frac{3}{4}\). Cosine is positive in quadrant IV: \(\frac{\sqrt{3}}{2}\).

\(\tan\theta = \frac{-1/2}{\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}\).

cos = \(\frac{\sqrt{3}}{2}\), tan = \(-\frac{\sqrt{3}}{3}\)

Exercise 15 – Simplify with the identity

\(\sin^{2}x = 1 – \cos^{2}x = (1 – \cos x)(1 + \cos x)\). Cancel \(1 – \cos x\).

\(1 + \cos x\)

Exercise 16 – Prove an identity

\((1 – \sin x)(1 + \sin x) = 1 – \sin^{2}x\). By the identity, \(1 – \sin^{2}x = \cos^{2}x\).

Both sides equal cos² x

Exercise 17 – Cosine is shifted sine

y = sin x (blue) and y = cos x (orange) both have period 2π and amplitude 1.

Shifting the sine graph left \(\frac{\pi}{2}\) moves its maximum from \(x = \frac{\pi}{2}\) to \(x = 0\), which is where cosine has its maximum 1.

The two graphs coincide after a left shift of π/2

Exercise 18 – Daylight

Amplitude 2.5, midline 12: maximum \(14.5\) and minimum \(9.5\) hours.

Period \(\frac{2\pi}{2\pi/365} = 365\) days.

14.5 h, 9.5 h; period 365 days

Exercise 19 – Find the error

The period is \(\frac{2\pi}{B}\), not \(B \cdot \pi\). With \(B = 3\): \(\frac{2\pi}{3}\). The graph completes three cycles in \(2\pi\).

\(\frac{2\pi}{3}\)

Exercise 20 – Spring

Amplitude \(\frac{18 – 6}{2} = 6\), midline \(\frac{18 + 6}{2} = 12\), \(B = \frac{2\pi}{3}\).

Starting at the minimum means a reflected cosine: \(y(0) = -6 + 12 = 6\).

\(y(t) = -6\cos\left(\frac{2\pi}{3}t\right) + 12\)

Key facts behind this algebra 2 trig graphs answers

Key facts

Sine and cosine have period 2π and range -1 to 1.
y = A sin(B(x – C)) + D: amplitude |A|, period 2π/B, shift C, midline D.
Model cycles with a sinusoid: midline = average of max and min.
sin² θ + cos² θ = 1.
The quadrant gives the sign of the missing trig value.

How to use this algebra 2 trig graphs answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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