Trigonometry Functions Comprehensive Notes

6.1 Radian Measure

  • Angles are often measured in degrees, but radians provide a unitless, pure number representation.
  • Arc length 'a' subtends angle θ\theta at the center of a circle with radius r.
  • θ\theta is proportional to 'r' and 'a': θ=ar\theta = \frac{a}{r}.
  • Since 'a' and 'r' have units of measurement (e.g., cm, m), θ\theta is a real number without units.
  • Example 1: Arc length of 6 cm in a circle with a 4 cm radius:
    θ=ar=6cm4cm=32radians\theta = \frac{a}{r} = \frac{6cm}{4cm} = \frac{3}{2} radians
  • One radian is the angle subtended by an arc with a length equal to the radius of the circle (a = r).
    θ=ar=rr=1radian\theta = \frac{a}{r} = \frac{r}{r} = 1 radian
  • Relationship between radians and degrees:
    • A full circle (360 degrees) has a circumference of 2πr2\pi r.
    • θ=ar=2πrr=2πradians\theta = \frac{a}{r} = \frac{2\pi r}{r} = 2\pi radians
    • Therefore, 2πradians=360∘2\pi radians = 360^\circ
    • πradians=180∘\pi radians = 180^\circ
    • 1radian=180∘π≈57.3∘1 radian = \frac{180^\circ}{\pi} \approx 57.3^\circ
  • Example 2: Convert to radians:
    • 20°: 20∘=(20∘)π180∘=π9radians≈0.35radians20^\circ = (20^\circ) \frac{\pi}{180^\circ} = \frac{\pi}{9} radians \approx 0.35 radians
    • 155°: 155∘×π180∘=155π180=31π36rad155^\circ \times \frac{\pi}{180^\circ} = \frac{155\pi}{180} = \frac{31\pi}{36} rad
    • 252°: 252∘×π180∘=252π180=126π90=63π45=7π5rad252^\circ \times \frac{\pi}{180^\circ} = \frac{252\pi}{180} = \frac{126\pi}{90} = \frac{63\pi}{45} = \frac{7\pi}{5} rad
  • Example 3: Convert 5π6\frac{5\pi}{6} radians to degrees.
    5π6radians=5π(180∘)6π=5(180∘)6=5(30∘)=150∘\frac{5\pi}{6} radians = \frac{5\pi(180^\circ)}{6\pi} = \frac{5(180^\circ)}{6} = 5(30^\circ) = 150^\circ
  • To convert from degrees to radians, multiply by π180∘\frac{\pi}{180^\circ}.
  • To convert from radians to degrees, multiply by 180∘π\frac{180^\circ}{\pi}.
  • Angles in Standard Position:
    • Start on the positive x-axis (Initial Arm).
    • Measure counterclockwise to the Terminal Arm.
    • Negative angles are measured clockwise from the Initial Arm.

6.2 Angles and the Cartesian Plane

  • Primary Trigonometric Ratios:
    • sin⁡θ=yr\sin \theta = \frac{y}{r}
    • cos⁡θ=xr\cos \theta = \frac{x}{r}
    • tan⁡θ=yx\tan \theta = \frac{y}{x}
  • Reciprocal Trigonometric Ratios:
    • csc⁡θ=ry\csc \theta = \frac{r}{y}
    • sec⁡θ=rx\sec \theta = \frac{r}{x}
    • cot⁡θ=xy\cot \theta = \frac{x}{y}
  • Where x2+y2=r2x^2 + y^2 = r^2 and r > 0.
  • SOH CAH TOA:
    • sin⁡θ=opphyp\sin \theta = \frac{opp}{hyp}
    • cos⁡θ=adjhyp\cos \theta = \frac{adj}{hyp}
    • tan⁡θ=oppadj\tan \theta = \frac{opp}{adj}
  • Radian measure on the Cartesian Plane uses Special Triangles:
    • 45°-45°-90° Triangle
    • 30°-60°-90° Triangle
  • Example 1: Determine the exact primary trig ratios of π3\frac{\pi}{3}.
    sin⁡π3=32;cos⁡π3=12;tan⁡π3=3\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}; \cos \frac{\pi}{3} = \frac{1}{2}; \tan \frac{\pi}{3} = \sqrt{3}
  • Example 2: Determine the exact value of sin⁡(π2)=1\sin (\frac{\pi}{2}) = 1
  • Example 3: Determine the exact value of cot⁡(3π2)=0\cot(\frac{3\pi}{2}) = 0
  • Example 4: Determine the exact value of cos⁡5π4=−12=−22\cos \frac{5\pi}{4} = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}
  • Note: Negative x-axis and y-axis indicate negative lengths.
  • Example 5: If tan⁡θ=−3\tan \theta = -\sqrt{3}, determine the possible values of θ\theta.
    tan⁡θ=−3=yx=−31or3−1\tan \theta = -\sqrt{3} = \frac{y}{x} = \frac{-\sqrt{3}}{1} or \frac{\sqrt{3}}{-1}
    The two angles corresponding to these side lengths are 5π3\frac{5\pi}{3} and 2π3\frac{2\pi}{3} or 300° and 120°
    Therefore, θ=5π3or2π3\theta = \frac{5\pi}{3} or \frac{2\pi}{3}
  • CAST Acronym:
    • C (Cosine positive) in Quadrant IV
    • A (All positive) in Quadrant I
    • S (Sine positive) in Quadrant II
    • T (Tangent positive) in Quadrant III
    • If a primary trig ratio is positive, its reciprocal is also positive (and vice versa for negative).

6.3 Graphs of Primary Trig Ratios

  • Period: The change in the independent variable (θ\theta) for one complete cycle of the function.
  • Amplitude: The vertical distance from the axis of a sinusoidal function to the maximum or minimum value.
    • Amp=max−min2Amp = \frac{max - min}{2}
  • Axis: The horizontal line halfway between the maximum and minimum values.
    • Axis=max+min2Axis = \frac{max + min}{2}

6.4 Graphing Transformations of Trig Functions

  • General transformed function: y=af(k(x−d))+cy = af(k(x - d)) + c
  • Method 1: Point Transformation:
    • Transform parent points (x,y)→(xk+d,ay+c)(x, y) \rightarrow (\frac{x}{k} + d, ay + c)
  • Example 1: Sketch f(x)=3sin⁡(2(x+π4))−1f(x) = 3\sin(2(x + \frac{\pi}{4})) - 1
    • a=3,k=2,d=−π4,c=−1a = 3, k = 2, d = -\frac{\pi}{4}, c = -1
    • Transformed points: (x2−π4,3y−1)( \frac{x}{2} - \frac{\pi}{4}, 3y - 1)
    • (0,0)→(−π4,−1)(0, 0) \rightarrow (-\frac{\pi}{4}, -1)
    • (π2,1)→(0,2)(\frac{\pi}{2}, 1) \rightarrow (0, 2)
    • (π,0)→(π4,−1)(\pi, 0) \rightarrow (\frac{\pi}{4}, -1)
    • (3π2,−1)→(π2,−4)(\frac{3\pi}{2}, -1) \rightarrow (\frac{\pi}{2}, -4)
    • (2π,0)→(3π4,−1)(2\pi, 0) \rightarrow (\frac{3\pi}{4}, -1)
  • Method 2: Sketching Using Key Features
    • Period: Period=2π∣k∣Period = \frac{2\pi}{|k|}
    • Axis: y=cy = c
    • Amplitude: ∣a∣|a|
    • Maximum value: axis+ampaxis + amp
    • Minimum value: axis−ampaxis - amp
  • Example: For f(x)=3sin⁡(2(x+π4))−1f(x) = 3\sin(2(x + \frac{\pi}{4})) - 1
    • Period: 2π2=π\frac{2\pi}{2} = \pi
    • Axis: y=−1y = -1
    • Amplitude: ∣3∣=3|3| = 3
    • Maximum: −1+3=2-1 + 3 = 2
    • Minimum: −1−3=−4-1 - 3 = -4
  • Word Problem Example: h(t)=10sin⁡(2πt+1.5π)+15h(t) = 10\sin(2 \pi t + 1.5\pi) + 15, where 0≤t≤30 \le t \le 3
    • h(t)=10sin⁡(2π(t+0.75))+15h(t) = 10\sin(2\pi(t + 0.75)) + 15
    • Amplitude = 10
    • Axis: y = 15
    • Max = 25
    • Min = 5
    • Period = 2πk=2π2π=1\frac{2\pi}{k} = \frac{2\pi}{2\pi} = 1

6.5 The Graphs of the Reciprocal Trig Functions

  • Zeros of original functions become vertical asymptotes in reciprocal functions.
  • Reciprocal functions have the same positive and negative intervals as the original.
  • Intervals of increase on the original become intervals of decrease on the reciprocal, and vice versa.
  • Example 1: Sketch f(x)=csc⁡θ+2f(x) = \csc \theta + 2
    • Sketch y=csc⁡θy = \csc \theta and translate key points up 2 units.

6.6 Modelling with Trig Functions

  • Example: A siren's frequency fluctuates with time.
    • Minimum frequency = 500 Hz
    • Maximum frequency = 1000 Hz
    • Maximum frequency occurs at t = 0 and t = 15
    • 6 maximums in 15 seconds.
    • Axis: F=750F = 750
    • Amplitude = 250
    • Period = 3 seconds
    • Period=2πk→k=2πperiod=2π3Period = \frac{2\pi}{k} \rightarrow k = \frac{2\pi}{period} = \frac{2\pi}{3}
    • F=250cos⁡(2π3t)+750F = 250\cos(\frac{2\pi}{3}t) + 750

6.7 Rates of Change in Trig Functions

  • Average and instantaneous rates of change are determined using previous strategies.
  • The tangent (IROC) is 0 at the max and min points.
  • The tangent is steepest at the axis of the function.
  • Example: Find the average rate of change of f(x)=2cos⁡(x−π3)+1f(x) = 2\cos(x - \frac{\pi}{3}) + 1 for the interval π6≤x≤π2\frac{\pi}{6} \le x \le \frac{\pi}{2}
    • AROC=f(x<em>2)−f(x</em>1)x<em>2−x</em>1AROC = \frac{f(x<em>2) - f(x</em>1)}{x<em>2 - x</em>1}
    • x1=π6x_1 = \frac{\pi}{6}
    • x2=π2x_2 = \frac{\pi}{2}
    • f(x1)=f(π6)=2cos⁡(π6−π3)+1=2cos⁡(−π6)+1=2(32)+1=3+1f(x_1) = f(\frac{\pi}{6}) = 2\cos(\frac{\pi}{6} - \frac{\pi}{3}) + 1 = 2\cos(-\frac{\pi}{6}) + 1 = 2(\frac{\sqrt{3}}{2}) + 1 = \sqrt{3} + 1
    • f(x2)=f(π2)=2cos⁡(π2−π3)+1=2cos⁡(π6)+1=2(32)+1=3+1f(x_2) = f(\frac{\pi}{2}) = 2\cos(\frac{\pi}{2} - \frac{\pi}{3}) + 1 = 2\cos(\frac{\pi}{6}) + 1 = 2(\frac{\sqrt{3}}{2}) + 1 = \sqrt{3} + 1
    • AROC=f(x<em>2)−f(x</em>1)x<em>2−x</em>1=(3+1)−(3+1)π2−π6=0π3=0AROC = \frac{f(x<em>2) - f(x</em>1)}{x<em>2 - x</em>1} = \frac{(\sqrt{3} + 1) - (\sqrt{3} + 1)}{\frac{\pi}{2} - \frac{\pi}{6}} = \frac{0}{\frac{\pi}{3}} = 0