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Reciprocal Identitiescsc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x 
Halfangle identitiessin (x/2) = radical((1  cos x) /2)
cos (x/2) = radical((1 + cos x) / 2)
tan (x/2) = (sin x / 1 + cos x) 
TransformationsTranslations
g(x) = f(x) + k is the graph of f(x) translated k units up when k > 0 and k units down when k < 0.
g(x) = f(x − h) is the graph of f(x) translated h units right when h > 0 and h units left when h < 0.
Reflections
g(x) = f(x) is the graph of f(x) reflected in the xaxis.
g(x) = f(x) is the graph of f(x) reflected in the yaxis.
Dilations
g(x) = a · f(x) is the graph of f(x) expanded vertically if a > 1 and compressed vertically if 0 < a < 1.
g(x) = f(ax) is the graph of f(x) compressed horizontally if a > 1 and expanded horizontally if 0 < a < 1. 
Domain and rangeDomain: The domain of a function is the set of all possible input values (often the "x" variable), which produce a valid output from a particular function. It is the set of all real numbers for which a function is mathematically defined.
Range: The range is the set of all possible output values (usually the variable y, or sometimes expressed as f(x)), which result from using a particular function. 
  Cotangent/Tangenttan x = sinx/cosx
cot x = cosx/sinx 
Doubleangle identitiescos2x = cos^{2}x  sin^{2}x
sin2x = (2sinx)(cosx)
tan2x = (2tanx / 1  tan^{2}x) 
Parent functionsconstant function
f(x) = a graph is a horizontal line
identity function
f(x) = x points on graph have coordinates (a, a)
quadratic function
f(x) = x^{2} graph is Ushaped
cubic function
f(x) = x^{3} graph is symmetric about the origin
square root function
f(x) = sqrt(x) graph is in first quadrant
reciprocal function
f(x) = 1/x graph has two branches
absolute value function
f(x) = │x│ graph is Vshaped 
  Pythagorean Identitiessin^{2}x + cos^{2}x = 1
tan^{2}x + 1 = sec^{2}x
1 + cot^{2}x = csc^{2}x 
You can convert the first identity into the second and third by dividing both sides by cos^{2}x or sin^{2}x.
Exponential and logarithmicLogarithmic
y = ln x
Exponential
y = b^{x} 

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Jessica j, 23:08 11 Mar 20
This really helped, thank you
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