math 7: 2.5-2.8 quiz review

graph of cosine

D: (-∞, ∞) R: [-1, 1] x-int: (π/2 + πk, 0) y-int: (0, 1) max: (2πk, 1) min: (π + 2πk, -1) symmetry: even (about y-axis) bounded above and below

graph of sine

D: (-∞, ∞) R: [-1, 1] x-int: (πk, 0) y-int: (0, 0) max: (π/2 + 2πk, 1) min: (3π/2 + 2πk, -1) symmetry: odd (about origin) bounded above and below

tan⁻¹x

D: (-∞, ∞), R: (-π/2, π/2) x and y values swap: (-1, -π/4), (1, π/4), (0, 0), looks like cubed root graph

sinx

D: (-∞, ∞), Range: [-1, 1] won't pass HLT to be 1:1, so create new domain new domain: [-π/2, π/2] pick points within new domain (will help when graphing inverse): (-π/2, -1), (0, 0), (π/2, 1)

cosx

D: (-∞, ∞), Range: [-1, 1] won't pass HLT to be 1:1, so create new domain new domain: [0, π] pick points within new domain (will help when graphing inverse): (0, 1), (π/2, 0), (π, -1)

cos⁻¹x

D: [-1, 1], Range: [0, π] x and y values swap: (1, 0), (0, π/2), (-1, π), looks like reflected cubic

graph of tangent

D: x≠π/2 + πk R: (-∞, ∞) x-int: (πk, 0) y-int: (0, 0) symmetry: odd (about origin) bounded: no asymptotes: x=π/2 + πk

tanx

D: x≠π/2 + πk, Range: (-∞, ∞) won't pass HLT to be 1:1, so create new domain new domain: (-π/2, π/2) pick points within new domain (will help when graphing inverse): (-π/4, -1), (π/4, 1), (0, 0)

inverse trig functions

a function has an inverse if it is 1-to-1. if function isn't 1-to-1, find inverse by restricting domain

sin⁻¹x

domain and range swap: D: [-1, 1], R: [-π/2, -π/2] x and y values in points swap, graph looks similar to cubic

even vs. odd trig functions

even: cos(-θ) = cos(θ) and sec(-θ) = sec(θ) odd: sin(-θ) = -sin(θ), csc(-θ) = -csc(θ), tan(-θ) = -tan(θ), cot(-θ) = -cot(θ); aka all of the rest besides cos/sec

problems with cos(2θ), (θ/2), (2θ-π/2)

first isolate trig function to one side then replace inside of parentheses w/ A solve for A as normal (ex. sinA = 1, A = π/2 + 2πk) then, replace A with OG terms (2θ = π/2 + 2πk) and solve for theta by isolating (in this ase divide whole expression by 2 INCLUDING GENERALIZATION) note: only manipulate generalization when isolating theta, don't add π/2 to generalization, just divide by 2 (ex. 2θ - π/2)

coterminal angles

if 2 angles in standard position have the same terminal side, then can add or subtract one revolution (360 or 2π); angles have infinitely many possible degrees/radians (ex. π/3 ±2π)

generalizing

if question doesn't ask for period of [0, 2π], must generalize easy way is to add 2πk to every answer (can sometimes condense into just πk depending on answers)

note for graphing

if sketch is hard to tell from equation, plug in points

reference angle

is the acute angle formed by the terminal side and x-axis; to find do x-axis minus the angle (ex. 180-150=30) note: positive angles move counter-clockwise, negative angles move clockwise

solving trig functions

isolate the variable to one side

converting degrees to radians

multiply by π/180 (and swapped if converting radians to degrees)

unit circle

see image

inverse properties

sin(sin⁻¹x) = x D: [-1, 1] sin⁻¹(sinx) = x D: [-π/2, π/2] (Q1 and 4) cos(cos⁻¹x) = x D: [-1, 1] cos⁻¹(cosx) = x D: [0, π] (Q1 and 2) tan(tan⁻¹x) = all reals tan⁻¹(tanx) = (-π/2, π/2) (Q 1 and 4) note: must ensure answers are in correct quadrants note: all inverses on inside correspond to new domains, all inverses on outside correspond to restricted domain of regular function note: always try to solve form inside out

half-angle identities

sin(u/2) = ±√(1-cosu)/(2) cos(u/2) = ±√(1+cosu)/(2) tan(u/2) = 1. ±√(1-cosu)/(1+cosu) 2. (1-cosu)/(sinu) 3. (sinu)/(1+cosu) note: for tan, use 2 or 3 to avoid±; to identify if it's + or -, find the quadrant the given angle is in and whether it's being evaluated for sin (y) or cos (x)

double angle identities

sin2u = 2sinucosu tan2u = (2tanu)/(1-tan²u) cos2u = 1. cos²u-sin²u 2. 2cos²u-1 3. 1-2sin²u note: can use sum formula to prove

SohCahToa

sine = opp/hyp, cosine = adj/hyp, tangent = opp/adj cosecant = hyp/opp, secant = hyp/adj, cotangent = adj/opp

trigonometric functions on graph/unit circle

sine = y/r, cosine = x/r, tangent = y/x aka sin/cos on unit circle cosecant = r/y, secant = r/x, cotangent = x/y note: radius (r) is always 1 on unit circle, to find x, y, or r use pythagorean theorem

power reducing formula

sin²u = (1-cos2u)/2 cos²u = (1+cos2u)/2 tan²u = (1-cosu)/(1+cosu) note: same equations as half-angles, just w/o √

pythagorean identities

sin²θ + cos²θ = 1 1 + tan²θ = sec²θ 1 + cot²θ = csc²θ note: all pythag. identities can be proven using og: sin²θ + cos²θ = 1

reciprocal identities

sinθ = 1/cscθ cosθ = 1/secθ tanθ = 1/cotθ cscθ = 1/sinθ secθ = 1/cosθ cotθ = 1/tanθ

cofunction identities

sinθ=cos(π/2-θ) cosθ=sin(π/2-θ) tanθ=cot(π/2-θ) cotθ=tan(π/2-θ) secθ=csc(π/2-θ) cscθ=sec(π/2-θ) note: draw a triangle, label the points 90, θ, and β, then β is 90-θ; this proves the cofunction identities above

when solving equations w/ even/odd functions...

start by using hypothesis/second half of prompt to find answer; anything w/ cos or sec will always be the original answer (b/c of even function)

sums and differences formula

sums: sin(α+β) = sinαcosβ+cosαsinβ cos(α+β) = cosαcosβ-sinαsinβ tan(α+β) = (tanα+tanβ)/(1-tanαtanβ) differences: sin(α-β) = sinαcosβ-cosαsinβ cos(α-β) = cosαcosβ+sinαsinβ tan(α-β) = (tanα-tanβ)/(1+tanαtanβ)

quotient identities

tanθ = sinθ/cosθ cotθ = cosθ/sinθ

note for proofs

when proving equation, pick the more complicated side

2.5

xxxxx

2.6

xxxxx

2.7

xxxxx

2.8

xxxxx

transformations of sin and cos

y=Asin(Bx-C)+D or y=Acos(Bx-C)+D amplitude/height = |A| (tells what the max y-value is) period: 2π/B (how often the graph repeats) phase shift: C/B (shift left or right) vertical shift: D midline: y=D (middle of graph/new x-axis)

transformations of tan

y=Atan(Bx-C)+D amplitude/height = |A| (tells what the max y-value is) period: π/B (how often the graph repeats) phase shift: C/B (shift left or right) vertical shift: D midline: y=D (middle of graph/new x-axis) asymptotes: left=C/B-1/2(π/B), right=C/B+1/2(π/B)