TRIG: 6.1,2,3,4,5,6,7

Area of a sector

A = θ/2π x πr² A = 1/2 θ r²

Multiplying trigonometric functions A) inverse on the inside: tan (tan⁻¹)(-5)) = -5 B) inverse on the outside: sin⁻¹ (sin (4π/3)) = -π/3

A) they cancel B) evaluate sin and then find the distance from the angle 0 THIS ONLY WORKS FOR VALUES INSIDE THE ACCEPTABLE RANGE

Sin ⁻¹ (π)

D.N.E. sin caps out at one; it can't be any bigger than one. sin (x) = y/r Y can't be bigger than the radius, just the same!

tan⁻¹ (x) or ARCtan (x)

Dom: (-∞,∞) Rng: (-π/2, π/2); The answer to any arctan problem is in the right half of the Unit Circle.

graph of tangent

Period: π Domain: ALL x ≠ π/2 + nπ Range: [-∞, ∞] x int: nπ, n must be an integer y int: 0,0 Vertical Asymptotes: x = π/2 + nπ Amplitude: not defined Symmetry: origin (odd)

All students take calculus

Q1: all functions Q2: sine Q3: tangent Q4: cosine

trick for remembering patterns for the sine and cosine functions

SINE 0° = √0/2 30° = √1/2 45° = √2/2 60° = √3/2 90° = √4/2 COSINE - sine reversed! TANGENT - sine / cosine!

unit circle

all of these conversions can be done using the conversion factor

Pythagorean Theorem: a² + o² = h²

a² + o² = h² sin² (θ) + cos² (θ) = o²/h² + a²/h² = 1 sin² (θ) + cos² (θ) = 1

Definitions of trigonometric functions of any angle

cos (θ) = x/r sin (θ) = y/r tan (θ) = y/x sec (θ) = r/x csc (θ) = r/y cot (θ) = x/y

even trigonometric functions; identity for even functions

cos, sec cos (-θ) = cos (θ)

center line Range (upper/lower bounds) period

d where y = A sin/cos (x) + d Rng = (d - |A|, d + |A|) Vertical transformations don't affect it; only horizontal stretches or compressions for y = A sin/cos (bx) + d, period = 2π/b; b > 0

arc length formula

s = rθ R = 4 in θ = 240° = 4π/3 4 in x 4π/3 = 16π/3 in.

sec (θ) = csc (θ) = cot (θ) =

sec (θ) = 1/cos (θ) = h/a csc (θ) = 1/sin (θ) = h/o cot (θ) = 1/tan (θ) = a/o = cos (θ)/sin (θ)

solving for trig functions

sin² (θ) + cos² (θ) = 1 1 + cot² (θ) = csc² (θ) tan² (θ) + 1 = sec² (θ)

pythagorean trigonometric identities

sin²θ + cos²θ = 1 1 + tan²θ = sec²θ 1 + cot²θ = csc²θ

supplementary

sum of angles is 180°

quotient trigonometric identities

tan = sin/cos cot = cos/sin

Guide for graphing asymptotes: when are functions zero?

tan = sin/cos sec = 1/cos asymptotes when cosine is 0 (multiples of π) cot = cos/sin csc = 1/sin asymptotes when sine is 0 (multiples of π/2)

reference angle

the acute angle θ' formed by the terminal side of θ and the horizontal axis

amplitude

the distance from the center line to the peak = |A| where y = A x cos (x) or sin (x)

difference between Sin ⁻¹ (x) and (sin (x))⁻¹

the first is arcsin (x), and the second is csc (x)

co-terminality

two angles have different measures but end up in the same place any angle + n(360) produces a coterminal angle, n MUST be an integer

angular speed ω

ω = θ/t = central angle / time

degrees to radians conversion

degrees ( π/180) = radians

how to use a reference angle

using the angle θ' (the reference angle) , you can evaluate all 6 trig functions. These will give you the same values as θ, except possibly in sign. To determine the sign of the function, you look at the quadrant in which θ lies.

Steps for graphing trigonometric functions

1. Write in standard form y = ±A cos or sin [b(x - c)] + d 2. Identify the center line (which is always d) 3. Identify max/min (using d and A [amplitude]) 4. Identify the starting point (normally x = 0, ascertained from (c)) 5. Identify the ending point using the period (2π/b) 6. Graph the 3 remaining points: the middle and the middle-middles (intercepts of the center line) 7. Graph Acronym :) Sarah's Chocolate Milk Seems Pretty Marvelous

degrees to degrees, minutes, seconds form conversion

350.355° 1. separate out the whole number, leaving the decimal. 2. multiply 0.355 x 60' (minutes) = 21.3' 3. separate out the whole number, leaving the decimal. 4. multiply 0.3 x 60'' (seconds) = 18'' 350° 21'18''

Cos⁻¹ or ARCcos (x)

Dom: [-1,1} Rng: [0,π] The answer to any ARCcos problem is in the top half of the Unit Circle.

graph of sine

Domain: (-∞, ∞) Range: [1, 1] x int: nπ, n must be an integer y int: 0,0 1 period: 2π

graph of cosine

Domain: (-∞, ∞) Range: [1, 1] x int: nπ/2, n must be an odd integer y int: 0,1 1 period: 2π

Unit circle

For any point on the unit circle, cos (θ) = x sin (θ) = y tan (θ) = y/x

graph of cosecant

Period: 2π Domain: ALL x ≠ nπ Range: ( -∞, -1 ] ∪ [ 1, ∞ ) x int: none y int: none Vertical Asymptotes: x = nπ, n must be an integer Amplitude: 1 Symmetry: origin (odd)

graph of secant

Period: 2π Domain: ALL x ≠ π/2 + nπ Range: ( -∞, -1 ] ∪ [ 1, ∞ ) x int: none y int: (0,1) Vertical Asymptotes: x = π/2 + nπ, n must be an integer Amplitude: 1 Symmetry: y-axis (even)

graph of cotangent

Period: π Domain: ALL x ≠ nπ Range: [-∞, ∞] x int: nπ/2, n must be an odd integer y int: 0,0 Vertical Asymptotes: x = nπ, n must be an integer Amplitude: not defined Symmetry: origin (odd)

bearing

an angle, measured clockwise from the north direction

Sin takes a(n) ________ and returns a(n) _________. Sin ⁻¹ takes a(n) ________ and returns a(n) _________.

angle, number number, angle

60° right triangle - 6 Trigonometric Functions

cos (45°) = 1/2 sin (45°) = √3/2 tan (45°) = √3 sec (45°) = 2 csc (45°) = 2√3/3 cot (45°) = √3/3

45° right triangle - 6 Trigonometric Functions

cos (45°) = √2/2 sin (45°) = √2/2 tan (45°) = 1 sec (45°) = √2 csc (45°) = √2 cot (45°) = 1

30° right triangle - 6 Trigonometric Functions

cos (45°) = √3/2 sin (45°) = 1/2 tan (45°) = √3/3 sec (45°) = 2√3/3 csc (45°) = 2 cot (45°) = √3

phase shift

equal to c/b y = a sin (or cos) (bx - c) The left and right endpoints of a one cycle interval can be determined by solving the equations bx - c = 0 and bx - c = 2π

How to solve for ARCsin of a number:

find the value of sin on the Unit Circle, and figure out which one fits into the range [-π/2, π/2]. next, find a way to arrive at the angle from the starting point that keeps you within [-π/2, π/2]. The distance you move is your answer. (Answer in radians!)

period

horizontal distance it takes a graph to complete one full rotation given by 2π/b when the function looks like y = a sin (or cos) bx

When you're multiplying two trigonometric functions, and you have one on the inside with an inverse,

make a triangle with the numbers given and solve for the outside function with the triangle. That's your answer.

Sin ⁻¹ or ARCsin (x)

means the same thing as Sin ⁻¹ (x). Domain: [-1,1] Rng: [-π/2, π/2] y = arcsin x if and only if sin y = x The answer to any ARCsin problem is in the right side of the Unit Circle.

coordinate grid; 9 parts quadrant 1 angle quadrant 2 angle quadrant 3 angle quadrant 4 angle

quadrant 1 angle: 0 < θ < 90°, 0 < θ < π/2° quadrant 2 angle: π/2 < θ < π quadrant 3 angle: π < θ < 3π/2 quadrant 4 angle: 3π/2 < θ < 2π

cofunctions of complementary angles are equal

sin (90°-θ) = cos θ cos (90°-θ) = sin θ tan (90°-θ) = cot θ cot (90°-θ) = tan θ sec (90°-θ) = csc θ csc (90°-θ) = sec θ

why are functions such as sine and cosine periodic?

sin (t + 2πn) = sin t cos (t + 2πn) = cos t

sine to cosine transformation cosine to sine transformation

sin (x) = cos (x - π/2) cos (x) = sin (x + π/2)

sin (θ) = cos (θ) = tan (θ) =

sin (θ) = o/h cos (θ) = a/h tan (θ) = o/a = sin (θ) / cos (θ)

reciprocal trigonometric identities

sin = 1/csc csc = 1/sin cos = 1/sec sec = 1/cos tan = 1/cot cot = 1/tan

odd trigonometric functions; identity for odd functions

sin, csc, tan, cot sin (-θ) = - sin (θ)

complementary

sum of angles is 90°

linear speed (v)

v = s/t velocity = distance / time s = rθ w = θ/t v = rω ( v = rω ) !!!

how to find 2 consecutive vertical asymptotes for cotangent

y = a cot (bx - c) bx - c = 0 bx - c = π

How to find 2 consecutive vertical asymptotes for tangent

y = a tan (bx - c) bx - c = π/2 bx - c = - π/2

1 radian

≈ 57° one radian is the measure of a central angle θ that intercepts an arc s equal in length to the radius r of the circle. Algebraically, this means that θ = s/r, (θ = 1 when s = r)