Practice Maths

Trigonometric Graphs and Exact Values

Key Terms

Exact values
come from the unit circle and the special triangles (45°-45°-90° and 30°-60°-90°).
For y = sin(x) and y = cos(x): period = 2π, amplitude = 1, range [−1, 1].
y = sin(x): starts at (0, 0), reaches maximum 1 at x = π/2, returns to 0 at x = π, minimum −1 at x = 3π/2, back to 0 at x = 2π.
y = cos(x): starts at (0, 1), falls to 0 at x = π/2, minimum −1 at x = π, back to 0 at x = 3π/2, returns to 1 at x = 2π.
y = cos(x) is y = sin(x) shifted left by π/2: cos(x) = sin(x + π/2).
For y = tan(x): period = π, no amplitude (unbounded), vertical asymptotes at x = π/2 + nπ, range ℝ.
Symmetry
sin is odd: sin(−x) = −sin(x). cos is even: cos(−x) = cos(x).
Supplementary angles
sin(π − x) = sin(x); cos(π − x) = −cos(x).
Complementary angles
sin(π/2 − x) = cos(x); cos(π/2 − x) = sin(x).
Exact Value Table
θ (rad) θ (deg) sinθ cosθ tanθ
00°010
π/630°½√3/21/√3 = √3/3
π/445°√2/2√2/21
π/360°√3/2½√3
π/290°10undefined
2π/3120°√3/2−½−√3
3π/4135°√2/2−√2/2−1
5π/6150°½−√3/2−1/√3
π180°0−10
Memory trick: sin values for 0, 30, 45, 60, 90 are √0/2, √1/2, √2/2, √3/2, √4/2 = 0, 1/2, √2/2, √3/2, 1.

Graphs of y = sin(x) (blue) and y = cos(x) (orange) for x ∈ [0, 2π]

x y π/2 π 3π/2 2π 1 −1 y=sin(x) y=cos(x) max min 1
Function Period Amplitude Domain Range
y = sin(x)2π1ℝ[−1, 1]
y = cos(x)2π1ℝ[−1, 1]
y = tan(x)πnonex ≠ π/2 + nπℝ
Hot Tip To find exact values of related angles, use the ASTC rule (All Students Take Calculus): in the 1st quadrant All trig ratios are positive; in the 2nd quadrant only Sin is positive; in the 3rd quadrant only Tan is positive; in the 4th quadrant only Cos is positive. Always find the reference angle (the acute angle to the x-axis) first, then apply the correct sign.

Worked Example 1 — Exact values using the unit circle

Question: Find the exact value of sin(5π/6), cos(5π/6), and tan(5π/6).

Step 1: 5π/6 is in the 2nd quadrant (between π/2 and π). Reference angle = π − 5π/6 = π/6.

Step 2: From the table: sin(π/6) = 1/2, cos(π/6) = √3/2, tan(π/6) = 1/√3.

Step 3: In the 2nd quadrant, sin is positive, cos and tan are negative.

sin(5π/6) = 1/2   cos(5π/6) = −√3/2   tan(5π/6) = −1/√3 = −√3/3

Worked Example 2 — Reading key features from a graph

Question: For y = sin(x) over [0, 2π], state: (a) the x-values where y = 0, (b) the x-value where y reaches its maximum, (c) the x-value where y reaches its minimum.

(a) y = 0 at x = 0, π, 2π.

(b) Maximum y = 1 at x = π/2.

(c) Minimum y = −1 at x = 3π/2.

Where Exact Values Come From: Deriving, Not Memorising

The exact trig values at 30°, 45°, and 60° come from two special right triangles. Rather than memorising a table, derive them:

45° triangle: Start with a square of side length 1. Cut it diagonally. The diagonal has length √(1² + 1²) = √2. The resulting right triangle has legs 1, 1 and hypotenuse √2. So sin(45°) = 1/√2 = √2/2, cos(45°) = √2/2, tan(45°) = 1.

30–60 triangle: Start with an equilateral triangle of side length 2. All angles are 60°. Cut it in half vertically: you get a right triangle with hypotenuse 2, base 1 (half the equilateral side), and height √(2² − 1²) = √3. The 30° angle is at the top (opposite the short side of 1), and the 60° angle is at the base (opposite the tall side of √3). So sin(30°) = 1/2, cos(30°) = √3/2, sin(60°) = √3/2, cos(60°) = 1/2.

The Unit Circle: Extending Trig Beyond 90°

The unit circle (radius = 1, centred at origin) provides the definition of trig functions for ALL angles, not just acute angles. A point on the unit circle at angle θ (measured anticlockwise from the positive x-axis) has coordinates (cosθ, sinθ). This is the definition — not just a convenient coincidence.

Why does this work for the basic case? For a right triangle with hypotenuse r at angle θ, the horizontal side is r cosθ and the vertical side is r sinθ. With r = 1, these become exactly the coordinates. The power of the unit circle is that it extends seamlessly to angles beyond 90°: at 120°, the x-coordinate is negative (cos 120° = −1/2) and the y-coordinate is positive (sin 120° = √3/2).

Reading the Trig Graphs: Features and Behaviour

The sine and cosine graphs both have period 2π (one complete oscillation) and amplitude 1 (maximum distance from the midline). They are essentially the same curve shifted horizontally: cos(x) = sin(x + π/2).

For y = sin(x): starts at (0, 0), rises to maximum 1 at (π/2, 1), returns to 0 at (π, 0), falls to minimum −1 at (3π/2, −1), returns to 0 at (2π, 0). For y = cos(x): starts at the maximum (0, 1), falls to 0 at (π/2, 0), reaches minimum −1 at (π, −1), returns to 0 at (3π/2, 0), back to maximum at (2π, 1). Knowing these key points lets you sketch either graph accurately without plotting many values.

The Tangent Graph: Different in Kind

The tangent function tan(x) = sin(x)/cos(x) behaves very differently from sine and cosine. Because it involves division by cos(x), it is undefined wherever cos(x) = 0, which happens at x = π/2 + nπ for integer n. At these values, tan(x) has vertical asymptotes.

Between each pair of consecutive asymptotes, tan(x) runs from −∞ to +∞, crossing zero at x = 0, π, 2π, … The period is π (not 2π), because the pattern repeats every half-rotation. Tan has no amplitude (it is unbounded) and its range is all real numbers. The graph has a characteristic S-shape between asymptotes, steepest at the zeros.

ASTC: Which Quadrant, Which Sign

The mnemonic “All Students Take Calculus” (or “All Stations To Central”) gives the sign of each trig function in each quadrant: All positive (1st), Sine positive (2nd), Tangent positive (3rd), Cosine positive (4th). This follows directly from the unit circle: in the second quadrant, x is negative (so cos < 0) but y is positive (so sin > 0), and tan = sin/cos < 0.

To find the exact value of sin(150°): 150° is in the 2nd quadrant, so sin is positive. The reference angle is 180° − 150° = 30°. So sin(150°) = +sin(30°) = 1/2. This approach works for any angle that has a 30°, 45°, or 60° reference angle.

Exam Tip: sin(30°) = 1/2, NOT cos(30°). The 30° angle is opposite the shortest side of the 30–60–90 triangle, and sin = opposite/hypotenuse = 1/2 (short side over hypotenuse 2). The 60° angle is opposite the longer side √3, so sin(60°) = √3/2. If you confuse these, sketch the triangle: the angle you are working with determines which side is “opposite”.
Exam Tip: Whenever you are unsure of the sign of a trig value for an angle in a particular quadrant, sketch the unit circle quickly and mark which quadrant the angle lies in. The coordinates (cosθ, sinθ) of the point on the circle tell you the signs immediately: positive x-coordinate means cos > 0, positive y-coordinate means sin > 0. This takes 10 seconds and eliminates sign errors.

Mastery Practice

See Answers ➔
  1. Fluency

    Find the exact value of each expression.

    1. (a) sin(π/6)
    2. (b) cos(π/4)
    3. (c) tan(π/3)
    4. (d) sin(π/2)
    5. (e) cos(0)
    6. (f) tan(0)
  2. Fluency

    Find the exact value using reference angles and the ASTC rule.

    1. (a) sin(2π/3)
    2. (b) cos(3π/4)
    3. (c) tan(5π/6)
    4. (d) sin(7π/6)
    5. (e) cos(5π/3)
    6. (f) tan(7π/4)
  3. Fluency

    State the period, amplitude, domain and range of each function.

    1. (a) y = sin(x)
    2. (b) y = cos(x)
    3. (c) y = tan(x)
    4. (d) y = 3 sin(x)
    5. (e) y = cos(2x)
  4. Fluency

    For y = sin(x) on [0, 2π], find the x-coordinates where each value occurs.

    1. (a) sin(x) = 0
    2. (b) sin(x) = 1
    3. (c) sin(x) = −1
    4. (d) sin(x) = 1/2   (two values)
    5. (e) sin(x) = −√3/2   (two values)
  5. Understanding

    Reading and interpreting trig graphs.

    Use your knowledge of y = sin(x) and y = cos(x) to answer the following.
    1. (a) For y = cos(x) on [0, 2π], state the x-values where y = 0, y = 1 and y = −1.
    2. (b) Describe how y = cos(x) is related to y = sin(x) as a transformation.
    3. (c) For y = tan(x), state the x-values of the vertical asymptotes on [−π, 2π].
    4. (d) On the same set of axes, which function (sin or cos) starts at y = 0? Which starts at y = 1?
  6. Understanding

    Using symmetry properties.

    Symmetry rules: sin(π − x) = sin(x)   cos(π − x) = −cos(x)   sin(−x) = −sin(x)   cos(−x) = cos(x)
    1. (a) Use sin(π − x) = sin(x) to find sin(5π/6) given sin(π/6) = 1/2.
    2. (b) Use cos(π − x) = −cos(x) to find cos(2π/3) given cos(π/3) = 1/2.
    3. (c) Find sin(−π/4) using the odd symmetry of sine.
    4. (d) Find cos(−5π/6) using the even symmetry of cosine.
  7. Understanding

    Exact values in expressions.

    1. (a) Evaluate sin²(π/4) + cos²(π/4). What does this confirm?
    2. (b) Evaluate sin(π/6) × cos(π/3) + cos(π/6) × sin(π/3). What trig identity does this represent?
    3. (c) Simplify: (sin(π/3))² + (cos(π/3))².
    4. (d) Find the exact value of (tan(π/4))² + 1. What identity does this confirm?
  8. Understanding

    Comparing functions over different domains.

    1. (a) How many full cycles does y = sin(x) complete over [0, 6π]?
    2. (b) How many times does y = cos(x) equal zero on [0, 4π]? List all x-values.
    3. (c) How many vertical asymptotes does y = tan(x) have on [0, 3π]?
    4. (d) On the graph of y = sin(x), identify all x ∈ [0, 2π] where the gradient is zero (i.e. the maximum and minimum points).
  9. Problem Solving

    Exact values in context.

    Challenge. A Ferris wheel of radius 10 m has its centre at height 12 m. A rider starts at the 3 o’clock position on the wheel. Their height h (m) above the ground as a function of angle θ from the starting position is h = 12 + 10 sin(θ).
    1. (a) Find the rider’s height when θ = π/2. Interpret your answer.
    2. (b) Find the rider’s height when θ = π. Interpret your answer.
    3. (c) At what angle θ is the rider at the lowest point of the wheel? Find the minimum height.
    4. (d) Give the exact height at θ = π/6 and θ = π/3.
  10. Problem Solving

    Proving an exact value result.

    Multi-step challenge.
    1. (a) Using the 30°-60°-90° triangle with hypotenuse 2 and opposite side 1, verify that sin(30°) = 1/2 and cos(30°) = √3/2.
    2. (b) Using the 45°-45°-90° isoceles right triangle with equal legs of length 1, verify that sin(45°) = cos(45°) = √2/2.
    3. (c) Using the Pythagorean identity sin²(x) + cos²(x) = 1, find sin(x) if cos(x) = 1/3 and x is in the first quadrant. Give an exact answer.
    4. (d) Find tan(x) using your answers to part (c).