Algebra 2 (HS pathway)

Trigonometric Functions: Unit Circle, Graphs, and Identities

Algebra 2

✓Define and evaluate trigonometric functions for any angle using the unit circle.
✓Graph transformations of sine, cosine, and tangent functions, identifying amplitude, period, phase shift, and vertical shift.
✓Verify fundamental trigonometric identities using algebraic manipulation.
✓Apply trigonometric identities to simplify expressions and solve problems.

Key concepts

The Unit Circle

The unit circle is a circle with a radius of 1 unit centered at the origin (0,0) in the Cartesian coordinate system. For any angle θ measured counterclockwise from the positive x-axis, the coordinates (x,y) of the point where the terminal side of the angle intersects the unit circle define the cosine and sine of θ: x = cos θ and y = sin θ. The tangent of θ is defined as tan θ = y/x, provided x ≠ 0. The unit circle helps visualize trigonometric values for special angles (e.g., 0°, 30°, 45°, 60°, 90° and their radian equivalents) and understand the signs of trigonometric functions in different quadrants.

cos θ = x, sin θ = y, tan θ = y/x (where x ≠ 0)

Graphs of Sine and Cosine Functions

The graphs of sine and cosine functions are periodic, meaning their patterns repeat over a regular interval. The basic sine function, y = sin x, starts at the origin, increases to a maximum, crosses the x-axis, decreases to a minimum, and returns to the x-axis over one period. The basic cosine function, y = cos x, starts at its maximum value, decreases to a minimum, and returns to its maximum over one period. Both have a period of 2π (or 360°) and an amplitude of 1. Transformations of these functions are described by the general forms y = A sin(Bx - C) + D and y = A cos(Bx - C) + D, where: |A| is the amplitude, 2π/|B| is the period, C/B is the phase shift (horizontal shift), and D is the vertical shift (midline y=D).

y = A sin(Bx - C) + D, y = A cos(Bx - C) + D

Graphs of Tangent Functions

The graph of the tangent function, y = tan x, is also periodic, but its period is π (or 180°). Unlike sine and cosine, the tangent function has vertical asymptotes where cos x = 0 (i.e., at x = π/2 + nπ, where n is an integer). The function increases from negative infinity to positive infinity between consecutive asymptotes. Transformations are described by y = A tan(Bx - C) + D, where: |A| affects the vertical stretch, π/|B| is the period, C/B is the phase shift, and D is the vertical shift.

y = A tan(Bx - C) + D

Fundamental Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variable for which the functions are defined. These identities are crucial for simplifying expressions, solving trigonometric equations, and proving other identities. Key categories include Reciprocal Identities, Quotient Identities, and Pythagorean Identities.

Reciprocal Identities: csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ\nQuotient Identities: tan θ = sin θ/cos θ, cot θ = cos θ/sin θ\nPythagorean Identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = csc²θ

Key facts to remember

1The coordinates on the unit circle are (cos θ, sin θ).
2The period of y = sin x and y = cos x is 2π; the period of y = tan x is π.
3Amplitude is |A|, Period is 2π/|B| (for sin/cos) or π/|B| (for tan), Phase Shift is C/B, Vertical Shift is D.
4Pythagorean Identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = csc²θ.
5Reciprocal Identities: csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.
6Quotient Identities: tan θ = sin θ/cos θ, cot θ = cos θ/sin θ.
7Reference angles help determine trigonometric values in all quadrants.

Worked examples

Example 1

Evaluate sin(2π/3), cos(2π/3), and tan(2π/3) using the unit circle.

ILocate the angle 2π/3 radians on the unit circle. This angle is in Quadrant II.

IIIdentify the reference angle. The reference angle for 2π/3 is π - 2π/3 = π/3.

IIIRecall the coordinates for π/3 on the unit circle: (cos(π/3), sin(π/3)) = (1/2, √3/2).

IVAdjust the signs based on Quadrant II, where x is negative and y is positive. So, for 2π/3, the coordinates are (-1/2, √3/2).

VTherefore, sin(2π/3) = y = √3/2.

VIcos(2π/3) = x = -1/2.

VIItan(2π/3) = y/x = (√3/2) / (-1/2) = -√3.

Answer

sin(2π/3) = √3/2, cos(2π/3) = -1/2, tan(2π/3) = -√3

Remember that sine is positive in Quadrants I and II, cosine is positive in Quadrants I and IV, and tangent is positive in Quadrants I and III.

Example 2

Graph one full period of the function y = 3 sin(2x - π) + 1. Identify the amplitude, period, phase shift, and vertical shift.

IThe function is in the form y = A sin(Bx - C) + D. Here, A = 3, B = 2, C = π, D = 1.

IIAmplitude = |A| = |3| = 3.

IIIPeriod = 2π/|B| = 2π/2 = π.

IVPhase Shift = C/B = π/2. This means the graph shifts π/2 units to the right.

VVertical Shift = D = 1. This means the midline is y = 1.

VITo graph, find the starting point of one period: Bx - C = 0 => 2x - π = 0 => x = π/2.

VIIFind the ending point of one period: Bx - C = 2π => 2x - π = 2π => 2x = 3π => x = 3π/2.

VIIIThe five key points for a sine curve (start, quarter, middle, three-quarter, end) will be at x = π/2, 3π/4, π, 5π/4, 3π/2.

9Calculate the y-values for these points:

10At x = π/2 (start): y = 3 sin(0) + 1 = 1 (midline)

11At x = 3π/4 (quarter): y = 3 sin(π/2) + 1 = 3(1) + 1 = 4 (maximum)

12At x = π (middle): y = 3 sin(π) + 1 = 3(0) + 1 = 1 (midline)

13At x = 5π/4 (three-quarter): y = 3 sin(3π/2) + 1 = 3(-1) + 1 = -2 (minimum)

14At x = 3π/2 (end): y = 3 sin(2π) + 1 = 3(0) + 1 = 1 (midline)

15Plot these points and draw a smooth curve through them, extending if necessary to show the periodic nature.

Answer

Amplitude = 3, Period = π, Phase Shift = π/2 to the right, Vertical Shift = 1 (midline y=1). The graph starts at (π/2, 1), rises to (3π/4, 4), crosses the midline at (π, 1), falls to (5π/4, -2), and returns to the midline at (3π/2, 1).

Always factor out B from the argument (Bx - C) to correctly identify the phase shift as C/B.

Example 3

Verify the identity: (sec θ - cos θ) / sin θ = tan θ.

IStart with the left-hand side (LHS) and transform it into the right-hand side (RHS).

IILHS = (sec θ - cos θ) / sin θ

IIIRewrite sec θ in terms of cos θ: = (1/cos θ - cos θ) / sin θ

IVFind a common denominator in the numerator: = (1/cos θ - cos²θ/cos θ) / sin θ

VCombine the terms in the numerator: = ((1 - cos²θ) / cos θ) / sin θ

VIApply the Pythagorean Identity sin²θ + cos²θ = 1, so 1 - cos²θ = sin²θ: = (sin²θ / cos θ) / sin θ

VIIRewrite division by sin θ as multiplication by 1/sin θ: = (sin²θ / cos θ) * (1 / sin θ)

VIIICancel out a sin θ term: = sin θ / cos θ

9Recognize the Quotient Identity: = tan θ

10Since LHS = tan θ = RHS, the identity is verified.

Answer

The identity is verified by transforming the left-hand side into the right-hand side using reciprocal, Pythagorean, and quotient identities.

When verifying identities, it's generally easier to start with the more complex side and simplify it. Avoid moving terms across the equals sign.

Common mistakes

✗Confusing the x and y coordinates for cosine and sine on the unit circle.
✗Incorrectly calculating the phase shift by not factoring out B from the argument (e.g., using C instead of C/B).
✗Making algebraic errors when simplifying expressions or verifying identities, such as incorrectly squaring terms or distributing.
✗Forgetting the vertical asymptotes for tangent and cotangent graphs.
✗Assuming an identity is true without showing all steps of algebraic transformation.

Exam tips

★Memorize the unit circle values for common angles (0, π/6, π/4, π/3, π/2, etc.) and their corresponding (x,y) coordinates.
★Practice graphing transformations by identifying the amplitude, period, phase shift, and vertical shift first, then plotting key points.
★When verifying identities, try converting all terms to sine and cosine first if you get stuck, and always work one side of the equation at a time.
★Pay close attention to signs of trigonometric functions in different quadrants when using the unit circle.

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