Chapter 14: Trigonometry

Trigonometric identities, basic rules, drawing graphs, etc.

Radians and Conversion between Radians/Degrees

Angles are positive when measured anticlockwise and negative when measured clockwise. One radian (1c) is the angle subtended at the center of a unit circle by an arc of length 1 unit.

Conversions:

• Degrees → Radians: Multiply by π/180​.

• Radians → Degrees: Multiply by 180/π.​

Note: 1 radian ≈ 57.3.

Defining Sin/Cos using Unit Circle

The coordinates of a point P(θ) on the unit circle are given by:

• x = cos(θ)

• y = sin(θ)

Thus, P(θ) = (cos(θ), sin(θ)).

Key Properties:

• cos(2π + θ) = cos(θ), sin(2π + θ) = sin(θ) (periodicity).

• At θ = π, P(π) = (-1,0) → cos(π) = -1, sin(π) = 0.

• If θ is an odd multiple of π/2, sine is ±1 and cosine is 0.

• If θ is an even multiple of π, cosine is ±1 and sine is 0.

Defining Tan using Unit Circle

The tangent of an angle θ is given by:

tan(θ) = sin(θ) / cos(θ)

• Tan(θ) is undefined when cos(θ) = 0, i.e., at θ = ±π/2, ±3π/2, etc.

• The domain of tan(θ) is R \ {θ : cos(θ) = 0}.

Key Symmetry Properties of Trigonometric Functions

Key Symmetry Properties

1. Second Quadrant (π - θ)

   • sin(π - θ) = sin(θ)

   • cos(π - θ) = -cos(θ)

   • tan(π - θ) = -tan(θ)

2. Third Quadrant (π + θ)

   • sin(π + θ) = -sin(θ)

   • cos(π + θ) = -cos(θ)

   • tan(π + θ) = tan(θ)

3. Fourth Quadrant (2π - θ)

   • sin(2π - θ) = -sin(θ)

   • cos(2π - θ) = cos(θ)

   • tan(2π - θ) = -tan(θ)

4. Negative Angles

   • sin(-θ) = -sin(θ)

   • cos(-θ) = cos(θ)

   • tan(-θ) = -tan(θ)

Signs of Circular Functions in Quadrants

• 1st Quadrant (0 to π/2): All functions are positive.

• 2nd Quadrant (π/2 to π): Only sin is positive.

• 3rd Quadrant (π to 3π/2): Only tan is positive.

• 4th Quadrant (3π/2 to 2π): Only cos is positive.

Mnemonic: "All Students Take Calculus"
(A = All, S = Sin, T = Tan, C = Cos)

Exact values for Trig Functions

θ	sin(θ)	cos(θ)	tan(θ)

0

0

1

0

π/6 (30°)

1/2

√3/2

1/√3

π/4 (45°)

√2/2

√2/2

1

π/3 (60°)

√3/2

1/2

√3

π/2 (90°)

1

0

undef

Wavelength, Period and Amplitude

A function which repeats itself regularly is called a periodic function, and the interval between the repetitions is the period of the function (also called the wavelength).

The distance between the ‘mean position’ and the maximum position is the amplitude.

y = sin (x)

• Period: 2π (the graph repeats every 2π units).

• Amplitude: 1 (distance from the mean position to the maximum).

• Maximum value: 1, Minimum value: -1.

• Starts at: (0,0) (sin 0 = 0)

y = cos (x)

• Period: 2π (the graph repeats every 2π units).

• Amplitude: 1 (distance from the mean position to the maximum).

• Maximum value: 1, Minimum value: -1.

• Starts at: (0,1) (cos 0 = 1)

y = a sin (nt)

• Obtained from y = sin (x)

• Amplitude: a

• Period: 2π / n

• Transformation from y = sin t:

  • Dilation of factor a from the t-axis (affects amplitude).

  • Dilation of factor 1/n from the y-axis (affects period).

• Maximal Domain: R

• Range: [-a, a]

• Starts at (0,0)

y = a cos (nt)

• Obtained from y = cos (x)

• Amplitude: a

• Period: 2π / n

• Transformation from y = sin t:

  • Dilation of factor a from the t-axis (affects amplitude).

  • Dilation of factor 1/n from the y-axis (affects period).

• Maximal Domain: R

• Range: [-a, a]

• Starts at (0,1)

Special: Reflection of a sin or cos (nt)

Reflection in the y-axis:

• y = cos x remains the same.

• y = sin x becomes y = -sin x.

• The point with coordinates (t, y) is mapped to the point with coordinates (t/n, ay).

Solving sin t = b and cos t = b

• Find the principal solution using inverse functions:
  t = sin⁻¹(b) or t = cos⁻¹(b).

• Use symmetry properties:

  • sin(π - t) = sin t (2nd quadrant)

  • cos(2π - t) = cos t (4th quadrant)

• General solutions:

  • sin t = b → t = θ + 2kπ or π - (θ + 2kπ)

  • cos t = b → t = ±θ + 2kπ

Example:
Solve sin θ = 1/2 for 0 ≤ θ ≤ 2π.

• θ = π/6 (1st quadrant)

• θ = π - π/6 = 5π/6 (2nd quadrant)

• Solutions: θ = π/6, 5π/6

Solving a sin(nt) = b and a cos(nt) = b

1. First, substitute x = nt.

2. Work out the interval in which solutions for x are required.

3. Rewrite as sin(nt) = b/a or cos(nt) = b/a.

4. Solve for x first, then divide by n to get t.

5. Extend solutions using periodicity.

Example:
Solve 2sin(2θ) = -√3 for -π ≤ θ ≤ π.

• Rewrite: sin(2θ) = -√3/2.

• Substituting x = 2θ: Find solutions for x.

• Principal solutions: x = 4π/3, 5π/3.

• Divide by 2:

  • θ = 2π/3

  • θ = 5π/6

Graphs of y = a sin n(t ± ε) and y = a cos n(t ± ε)

1. Translations: The term ±ε shifts the graph horizontally along the t-axis. (They are translations of the graphs of

   y = a sin(nt) and y = a cos(nt) respectively.)

   • (t - ε) shifts right by ε.

   • (t + ε) shifts left by ε.

2. Transformations:

   • Amplitude = |a| (vertical stretch/shrink).

   • Period = (2π/n) (horizontal stretch/shrink).

   • Phase Shift = ±ε (horizontal translation).

3. Graph Shape: The shape remains the same but is transformed accordingly.

Examples

✅ Example 1: y = 3 sin(2t - π/4)

• Amplitude = 3, Period = π

• Phase shift: Right by π/4

Sketch graphs of y = a sin n(t ± ε) ± b and y = a cos n(t ± ε) ± b

Translations Along the t-axis:

From y = a sin n(t ± ε) and y = a cos n(t ± ε)!

• t + ε: Shift left by ε

• t - ε: Shift right by ε

Translations Along the y-axis:

• ±b: Shift up or down by b

Finding x-axis intercepts:

• Set y = 0 and solve for x.

• Example: y = √2 sin(x) + 1 → intercepts at 5π/4 and 7π/4

• Example: y = 2 cos(2x) - 1 → intercepts at π/6, 5π/6, 7π/6, etc.

Example: y = 3 sin(2t - π/4) + 2

• Amplitude = 3

• Period = π

• Phase Shift: Right by π/4

• Vertical Shift: Up by 2 units

Complementary Relationships

From the diagram, we have the following complementary relationships:

• sin(π/2 − θ) = cos(θ)

• cos(π/2 − θ) = sin(θ)

Similarly:

• sin(π/2 + θ) = cos(θ)

• cos(π/2 + θ) = −sin(θ)

Example:

Given sin(θ) = 0.3 and cos(α) = 0.8, find:

• cos(π/2 − α) = 0.8

• sin(π/2 − α) = cos(α) = 0.8

Full table (inc. Tan):

Angle	cos(θ)	sin(θ)	tan(θ)
π/2​−θ	sin(θ)	cos(θ)	cos⁡(θ)/sin⁡(θ)​
π/2+θ	-sin(θ)	cos(θ)	−cos⁡(θ)/sin⁡(θ)​
3π/2−θ	-sin(θ)	cos(θ)	−cos⁡(θ)/sin⁡(θ)​
3π/2+θ	sin(θ)	-cos(θ)	−cos⁡(θ)/sin⁡(θ)​

Pythagorean Identity

Consider a point P(θ) on the unit circle. By Pythagoras' theorem:

OP² = OM² + MP² → 1 = (cos(θ))² + (sin(θ))²

Thus, we have the Pythagorean identity:

cos²(θ) + sin²(θ) = 1

Pythagorean Identity Questions: Solving

1. Identify Quadrant

2. Draw triangle to identify unknown angles

3. Find values.

Example:

Given that cos(x) = -1/3 and π ≤ x ≤ 3π/2, we need to find sin(x) and tan(x).

Step 1: Use the Pythagorean Identity

The Pythagorean identity is:

sin²(x) + cos²(x) = 1

Substitute cos(x) = -1/3 into the identity:

sin²(x) + (-1/3)² = 1

sin²(x) + 1/9 = 1

Now, solve for sin²(x):

sin²(x) = 1 - 1/9

sin²(x) = 9/9 - 1/9 = 8/9

Take the square root of both sides:

sin(x) = ±√(8/9) = ±2√2/3

Since π ≤ x ≤ 3π/2 (third quadrant), sin(x) is negative, so:

sin(x) = -2√2/3

Step 2: Find tan(x)

We know that:

tan(x) = sin(x) / cos(x)

Substitute the values of sin(x) = -2√2/3 and cos(x) = -1/3:

tan(x) = (-2√2/3) / (-1/3) = 2√2

Final Answer:

• sin(x) = -2√2/3

• tan(x) = 2√2

The tangent function

Transformation: y = a tan(nt)

• Dilation by factor a from the t-axis (vertical scaling).

• Dilation by factor 1/n from the x-axis (horizontal scaling).

• Period: π/n.

• Asymptotes: t = (2k + 1)π / (2n), where k is an integer.

• Intercepts: The t-axis intercepts are t = kπ / n, where k is an integer.

• Range is R.

• Axis intercepts: find using solving tangent equation within a certain range.

Solving Tangent Equations

• Start with the equation:
  tan(2x) = √3

• Solve for 2x:
  The general solution for tan(θ) = √3 is:
  θ = π/3 + kπ, where k is an integer.

  So, for tan(2x) = √3:
  2x = π/3 + kπ

• Solve for x:
  Divide both sides by 2:
  x = π/6 + kπ/2, where k is an integer.

Optional: General solutions for Trigonometric Functions (Recap from 9 SEAL Extension)

For a ∈ [−1, 1], the general solution of the equation cos x = a is:

x = 2nπ ± cos−1(a), where n ∈ Z

For a ∈ R, the general solution of the equation tan x = a is:

x = nπ + tan−1(a), where n ∈ Z

For a ∈ [−1, 1], the general solution of the equation sin x = a is:

x = 2nπ + sin−1(a) or x = (2n + 1)π − sin−1(a), where n ∈ Z