Cos
Adjacent / Hypotenuse
Sin
Opposite / Hypotenuse
Tan
Opposite / Hypotenuse
Sec
1 / cos
hypotenuse / adjacent
Csc
1 / sin
hypotenuse / opposite
Cot
1 / tan
adjacent / opposite
cos / sin
Unit Circle
circle with the radius of 1 centered around the origin
positive direction is counter clock wise
negative direction is clock wise
Unit Circle coordinates
cos = x
sin = y
Unit Circle first quadrant
When flipping values from sin to sec or cos to csc you keep the negative and just flip the number
note
Pythagorean identities
sin^2 + cos^2 = 1
1 + tan^2 = sec^2
cot^2 + 1 = csc^2
How do you get the Pythagorean identities?
You take the first on of sin^2 + cos^2 = 1 and you divided it by sin and then by cos to get the other two
Exponential Function
form y = b^x
horizonal asymptote at y=0
Logarithmic Function
logbX
vertical asymptote at x=0
Natural Log
log with base e
ln
Common Log
log with base 10
log10X
logb(XY)
logb(X) + logb(Y)
logb(X / Y)
logb(X) - logb(Y)
logb(X^r)
rlogb(X)
logbB = 1
ln(e) = 1
log10 = 1
note
Velocity
change in x / change in t
change in position over change in time
Instantaneous velocity
the velocity at one particular point
Average Rate of Change
f(x2) - f(x1) / x2 - x1
Secant line
a line which passes through two points on the curve
can find its slope by using the average rate of change equation
To determined the rate of change at one point instead of the rate of change between two points we need the instantaneous rate of change
note
The limit is the instantaneous rate of change
note
limit
suppose f(x) is defined when x is near the number a (this means that f is defined on some open interval contain a, except possible at a itself)
lim f(x) = L
x→ a
the limit of f(x) as x approaches a equals L
lim f(x) = L
x → a-
the left hand limit of f(x) as x approaches a is equal to L if we can made the values of f(x) arbitrarily close to L by restricting x to sufficiently close to a with x < a
lim f(x) = L
x → a+
the right handed limit of f(x) as x approaches a is equal to L if we can make the value of f(x) arbitrarily close to L by restricting x to be sufficiently close to a with x > a
if lim f(x) = L
x → a
then lim f(x) = lim f(x) = L
x → a- x → a+
if lim f(x) does not = lim f(x) = L
x → a- x → a+
lim f(x) = DNE
x → a
The outputs for the limit is the y output for the function
note
lim f(x) = infinity
x → a
let f(x) be defined on both sides of a expect possibly at a itself
means that the values of f(x) can be made arbitrarily large ( as large as we please by taking x sufficiently close to a, but not equal to a
lim f(x) = - infinity
x → a
means that the values of f(x) can be made arbitrarily large and negative by taking x sufficiently close to a but not equal to a
There is a vertical asymptote if one of these statements is true
lim f(x) = infinity
x → a
lim f(x) = infinity
x → a-
lim f(x) = infinity
x → a+
lim f(x) = - infinity
x → a
lim f(x) = - infinity
x → a-
lim f(x) = - infinity
x → a+
if y = f(x) / g(x) and g(a) = 0 while f(a) does not = 0 then f(x) has a vertical asymptote at x = a
you plug it what the lim is approaching towards to the f(x) function if the numerator is equal to positive and the denominator equal to positive than the lim is approaching positive infinity. If the numerator is approaching positive and the denominator is zero you are approaching positive infinity. If the numerator is negative and the denominator is negative, you are approaching negative infinity. If the numerator is negative and the denominator is zero you are approaching neative infinity
This is only if there is one function in f(x)
lim [f(x) + g(x)]
x → a
lim f(x) + lim g(x)
x → a x → a
lim [f(x) - g(x)]
x → a
lim f(x) - lim g(x)
x → a x → a
lim c[f(x)]
x → a
lim f(x) * c
x → a
lim [f(x) * g(x)]
x → a
lim f(x) * lim g(x)
x → a x → a
lim [f(x) / g(x)]
x → a
lim f(x) / lim g(x) when g(x) is not equal to 0
x → a x → a
lim f(x)^n
x → a
[lim f(x) ]^n
x → a
lim square root (f(x))
x → a
square root [lim f(x)]
x → a
Direct Substitution Property
Let f(x) be a polynomial or a rational function and a is in the domain of f then
lim f(x) = f(a)
x → a