Home
Flashcards
Preview
Trig equations
Home
Get App
Take Quiz
Create
cos(a+b)
cos(a)cos(b) - sin(a)sin(b)
cos(a-b)
cos(a)cos(b) + sin(a)sin(b)
sin(a+b)
sin(a)cos(b) + cos(a)sin(b)
sin(a-b)
sin(a)cos(b) - cos(a)sin(b)
tan(a-b)
tan(a)-tan(b) / 1+tan(a)tan(b)
tan(a+b)
tan(a)+tan(b) / 1-tan(a)tan(b)
sin2(a)
2sin(a)cos(a)
2cos(a)sin(a)
cos2(a)
•cos^2(a) - sin^2(a)
•1 - 2sin^2(a)
•2cos^2(a) - 1
tan2(a)
2tan(a) / 1 - tan^2(a)
sin(a/2)
√[1-cos(a)] / 2
cos(a/2)
√[1+cos(a)] / 2
tan(a/2)
sin(a)/1+cos(a)
1-cos(a)/sin(a)
±√[(1-cos(a)/1+cos(a)]
Law of Sines
sin(A)/a = sin(B)/b = sin(C)/c
Law of Cosines
a^2 = b^2 + c^2 - 2(a)(b)cos(A)
Heron's Fomula (find the area of a triangle given 3 sides)
√[s(s-a)(s-b)(s-c)]
s= 1/2(a+b+c)
What is the magnitude of a vector?
The length of the line that is made.
V= <4,2>= 4i + 2j
Magnitude of a vector
|V|= √[a^2 + b^2] <--- Pythagorean theorem
With imaginary numbers, the x and y axis become the _____ and ______ axis, respectively.
Imaginary and Real
When adding two vectors the result
r
is
the line that can be drawn between the endpoints of each vector.
Dot Product of two vectors
v • w= ac + bd
Angle between two vectors
cosθ= (u•v)÷(|mag. u|)(|mag. v|)
Unit Vector=
magnitude of 1
Formula to find unit vector
u= <vector> / |mag. v|
imaginary number
i
=
√[-1]
i^2=
i^4=
-1
1
trig form of complex number equation z=x + yi
z= r(cosθ)+r(sinθ)= r(cosθ+sinθ)
For complex numbers:
sinθ=
cosθ=
tanθ=
y/r ; y=r(sinθ)
x/r ; x=r(cosθ)
y/x
How do you find r for a complex number equation?
Magnitude of z:
|z|= √[x^2 + y^2]
How is the argument (angle) measured?
ALWAYS
fromt the positive x-axis.
To multiply complex number equations:
multiply the modulus (r) and add the argument (the angles on the cosine and sine in the parenthesis).
To divide complex number equations:
divide the modulus (r) and subtract the argument (the angles on the cosine and sine in the parenthesis).
Demovire's theorem (multiplying a complex number equation by itself)
z^n= r^n(cos(nθ) + i sin(nθ))
Author
tenorsextets
ID
45322
Card Set
Trig equations
Description
look at title
Updated
2010-11-15T16:44:18Z
Show Answers
Home
Flashcards
Preview