Test 2 Flashcards
Original graph for y = sin(x)
x | 0 | π/2 | π | 3π/2 | 2π |
y | 0 | 1 | 0 | -1 | 0 |
Original graph for y = cos(x)
x | 0 | π/2 | π | 3π/2 | 2π |
y | 1 | 0 | -1 | 0 | 1 |
Equations for y = cos(x) & y = sin(x)
A(trig F)(ωx - φ) + V
How to find period
2π / ω = P for sin & cos; π / ω = P for tan
How do you know whether to use cos/sin from a graph on the test?
cos starts above/below 0;
sin goes through origin
Equation for amplitude from graph
(max - min) / 2
Equation for period from graph
2π/(cycle end - cycle start)
Original graph for y = tan(x)
x | -π/2 | -π/4 | 0 | π/4 | π/2 |
y | NA | -1 | 0 | 1 | NA |
Graph for tan (where are asymptotes & period too)
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asymptotes @ ±xπ/2; period = πGraph for cot (where are the asymptotes, period, & x-intercept too)
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asymptotes @ ±xπ & 0; period = π; x-int = π/2How to draw sec graph & its asymptotes
- Draw cos graph
- At each crest & troph, reciprocate it into a bunch of parabolas sticking up & down
- Points on cos graph w/ a y-val of 0 will be vertical asymptotes (so ±xπ/2)
How to draw csc graph & its asymptotes
- Draw sin graph
- At each crest and troph, reciprocate it into a bunch of parabolas sticking up & down
- Points on sin graph w/ a y-val of 0 will be vertical asymptotes (so ±xπ & 0)
Order of transformations
period, phase shift, amplitude, vertical shift
What happens if ω is negative?
Undef for sin & tan; if it’s in cos, use even function property and…take the negative away lmao
How do you know whether Amp is positive or negative?
If the cycle begins or goes straight to a negative troph, it’s a negative Amp
If it begins/goes straight to positive troph, it’s positive
Original domain & range of sin⁻¹(x)
D = [-π/2, π/2] R = [-1, 1]
f⁻¹(f(x)) –> sin⁻¹(sin(x)) –> what within what boundaries
x radians
-π/2 ≤ x ≤ π/2
f(f⁻¹(x)) –> sin(sin⁻¹(x)) –> what within what boundaries
x ratio
-1 ≤ x ≤ 1
What if angle in an inverse trig F breaks the pi limits?
take the ref angle instead but MAKE SURE THE SIGN OF THE REF ANGLE IS EQUAL TO THE ONE THATS PLUGGED IN OK??
What if ratio in inverse trig F breaks [-1, 1] limits?
it’s undefined
Original domain & range of cos⁻¹(x)
D = [-1, 1] R = [0, π]
f⁻¹(f(x)) –> cos⁻¹(cos(x)) –> what within what boundaries
x radians
0 ≤ x ≤ π
f(f⁻¹(x)) –> cos(cos⁻¹(x)) –> what within what boundaries
x ratio
-1 ≤ x ≤ 1
Original domain & range of tan⁻¹(x), and vertical asymptotes
D = (-∞, ∞) R = [-π/2, π/2] VA = ±π/2
f⁻¹(f(x)) –> tan⁻¹(tan(x)) –> what within what boundaries
x radians
-π/2 ≤ x ≤ π/2
f(f⁻¹(x)) –> tan(tan⁻¹(x)) –> what within what boundaries
x ratio
-∞ ≤ x ≤ ∞ (so all the time)
should you designate if an inverse of a function = f⁻¹(x) on a test?
YES FREAKING DO IT
Original domain & range of sec⁻¹(x), and vertical asymptotes
D = (-∞, -1)⋃(1, ∞) R = [0, π] VA = xπ/2
Original domain & range of csc⁻¹(x), and vertical asymptotes
D = (-∞, -1)⋃(1, ∞) R = [-π/2, π/2] VA = 0
Original domain & range of cot⁻¹(x), and vertical asymptotes
D = ∞
R = [0, π]
No VA
Range group for [0, π]
cos
sec
cot
Range group for [-π/2, π/2]
sin
csc
tan
How to find inverse of csc & sec
Just flip the ratio inside the parentheses & use the corresponding range group sin or cos function
Steps to take if you get a negative ratio for cot
- Take the ratio, flip it, and put it in tan⁻¹
- Plug into calc and get an angle
- If in degrees, subtract from 180; if in radians, subtract from xπ
How to write something like sin( tan⁻¹(u)) in an expression in terms of u
- Take the ratio u/1 as tangent and put it as the lengths of triangle sides, then work from there
