Pre-Calculus Spring Semester Exam Formulas

y = sin⁻¹(x) = arcsin(x)

Domain: [-1, 1] Range: [-π/2, π/2] - down left, up right

y = cos⁻¹(x) = arccos(x)

Domain: [-1, 1] Range: [0, π] - up left, down right

How to find the 2nd triangle if you know there's 2 triangles

Ex: C = 24°, b = 16 units, and c = 9 units - use law of sines to find B₁ - 180 - B₁ = B₂

Converting rectangular coordinates to polar coordinates

Ex: if (x,y) = (-3,-3,) - use pythagorean theorem to get r —> r² = x² + y² so r = 3√2 - use tan to get θ —> tan(θ) = y/x —> θ = tan⁻¹(-3/-3) = π/4 (+ 180° to get tan in right quadrant —> (-3, -3) is in Quadrant III) = 3π/4 so (-3,-3) in polar coordinates is (3√2, 5π/4) or (-3√2, π/4) or (3√2, -3π/4)

Can factor out numbers, even negative 1

Examples: 12 - 6√3 = 6(2 - √3) 2sin²x/cos x - 1 = 2(1 - cos²x)/cos x - 1 —> -2(cos²x - 1)/cos x - 1 cos²x - 1 = (cos x + 1)(cos x - 1)

How to know how many triangles you have - Only works for ASS - If you have to use Law of Cosines to solve, don't do this ——> Law of cosines will give you 1 triangle, if you get an error, there's 0 triangles

If the angle opposite the 2nd side that you're given has both an acute and obtuse version, that when summed with the angle given are both less than 180°, you have 2 triangles - have an acute triangle if: angle given + angle found is less than 180° - have an obtuse triangle if: 180° - (angle given + angle found) < 180° - Use law of sines to find angles - acute: less than 90° - obtuse: greater than 90° - helps to make a table to be be organized

Know how to factor by grouping

Know how to factor by grouping

Matrix addition is the same as vector addition

Matrix addition is the same as vector addition

Pay attention to if number is squared or not to know if can use the identity

Pay attention to if number is squared or not to know if can use the identity

r = 1 + 5cosθ

Period: 2π/(number in front of θ, if no number is in front of θ, number is 1) —> 2π/1 Interval: Period/4 —> 2π/4 = π/2 Midline: 1 (first number in equation) Amplitude: 5 (number in front of cos) - if interval is 2π, need to keep going until reach 2π

How to know if the square root in half angle identities is + or - Sin 1st quadrant: ___ 2nd quadrant: ___ 3rd quadrant: ___ 4th quadrant: ___ Cos 1st quadrant: ___ 2nd quadrant: ___ 3rd quadrant: ___ 4th quadrant: ___ Tan 1st quadrant: ___ 2nd quadrant: ___ 3rd quadrant: ___ 4th quadrant: ___

Sin 1st quadrant: Sin + 2nd quadrant: Sin + 3rd quadrant: Sin - 4th quadrant: Sin - Cos 1st quadrant: Cos + 2nd quadrant: Cos - 3rd quadrant: Cos - 4th quadrant: Cos + Tan 1st quadrant: Tan + 2nd quadrant: Tan - 3rd quadrant: Tan + 4th quadrant: Tan -

Sin(A + B) =

Sin(A) * Cos(B) + Cos(A) * Sin(B)

Sin(A - B) =

Sin(A) * Cos(B) - Cos(A) * Sin(B)

Sin half-angle identity

Sin(x) = ± √(1 - Cos(2x))/√2 - If want Sin(x/2), make Cos(2x) —-> Cos(x)

Tan half-angle identity

Tan(x) = ± √(1 - Cos(2x))/√(1 + Cos(2x)) - If want Tan(x/2), make each Cos(2x) —-> Cos(x)

Direction of a vector If ->v = <x, y>, then

Tan(θ) = y/x - so θ = Tan⁻¹(y/x) - Tan is in quadrant I and IV - if not, have to add 180° - θ is the direction angle for ->v

Work

Work = force * distance = Nm - force in Newtons (N) - distance in meters (m) - Nm means Joules (J)

Work in Vectors

Work = |->F||->d|Cosθ = ->F • ->d - F is the force - d is the distance

Vector Scalar Multiplication If ->v = <x, y> and a∈R, then - i.e. a is a real # - i. e. a is a scalar

a->v = a<x, y> = <ax, ay>

Determinant of [a b| |c d]

ad - bc - Notation: det (A) or |A| —> not absolute value

If Proj->u->v,

all ->u and ->v in the original formula are interchanged

Law of Cosines

c² = a² + b² - 2abCos(C) or b² = a² + c² - 2acCos(B) or a² = b² + c² -2bcCos(A) - basically first is in angle, the rest is in line

Even Identity and Even Trig Functions f is even if

f (-x) = f(x) - cos(x) - sec(x) - If a function is even, its reciprocal is even

Odd Identity and Odd Trig Functions f is odd if

f(-x) = -f(x) - sin(x) - tan(x) - csc (x) - cot (x) - arcsin(x) - arctan (x) - If a function is odd, its reciprocal is odd - just think it's everything except for everything related to cos(x)

Can go down a degree by

multiplying x²/y * 1/x/1/x = x/y/x

Equation for Polar Graphing

r = a + bcosθ or r = a + bsinθ

sin(θ) = _____ csc(θ) = _____ cos(θ) = _____ sec (θ) = _____ tan(θ) = _____ cot(θ) = _____

sin(θ) = y/r csc(θ) = r/y cos(θ) = x/r sec (θ) = r/x tan(θ) = y/x cot(θ) = x/y

Co-function Identities

sin(π/2 - θ) = cos(θ) cos(π/2 - θ) = sin(θ) tan(π/2 - θ) = cot(θ) csc(π/2 - θ) = sec(θ) sec(π/2 - θ) = csc(θ) cot(π/2 - θ) = tan(θ)

Pythagorean Identities

sin²(x) + cos²(x) = 1 tan²(x) + 1 = sec²(x) (can get this by dividing original equation by cos²(θ)) cot²(x) + 1 = csc²(x) ((can get this by dividing original equation by sin²(x)) - need to pay attention that it's sin², not sin - DON'T FORGET THAT CAN REARRANGE TAN² AND COT² LIKE SIN²

Law of Sines

sin∠A/a = sin∠B/b = sin∠C/c

Dot Product ->u • ->v =

u₁ * v₁ + u₂ * v₂

Converting polar coordinates to rectangular coordinates

x = rcosθ y = rsinθ Ex: (r,θ) = (4,π/3) x = 4cos(π/3) = 2 y = 4sin(π/3) = 2√3 so (x,y) = (2,2√3)

y = a × sin(b θ ± h) ± k

y = a × sin(b θ ± h) ± k - a = vertical stretch/compression - (if a is negative) = reflection over the x-axis (switch y-values) - b = horizontal stretch/compression - (if b is negative) = reflection over the y-axis (switch x-values) - h = horizontal shift - (+h, move left) - (-h, move right) - k = vertical shift - Period = 2π/|b| - Amplitude = |a| - sin starts at middle, goes up - (-sin starts at middle, goes down) - cos starts at top, goes down - (-cos starts at bottom, goes up)

Magnitude of a vector

|->v| = |<x,y>| = √x² + y²

Period of Tan and Cot

π/b

Convert radians to angle

# of radians * (180°/π) Example: convert 5π/4 radians to degrees (5π/4) * (180°/π) = 225°

Proj->v->u =

(->u • ->v/|v|²)->v = (->u • ->v/->v • ->v)->v

Common technique to get a unit vector

(->u)/(|->u|) - wil always be a unit vector in the same direction as ->u

Vector Subtraction If ->v = <x₁, y₁> and ->u = <x₂, y₂>, then

(->v) - (->u) = <(x₁ - x₂),(y₁ - y₂)>

cos²x =

(1 + cos(2x))/2

tan²x =

(1 - cos(2x))/(1 + cos(2x))

sin²x =

(1 - cos(2x))/2

tan(2x) =

(2tan(x))/(1 - tan²(x))

Convert angle to radians

(Angle) * (π/180°) Example: convert 225° to radians 225° * (π/180°) = 225π/180 - now simplify = 5π/4

(Tan²x - 1) =

(Tan x + 1)(Tan x - 1)

Tan(A + B) =

(Tan(A) + Tan(B))/(1 - Tan(A) * Tan(B))

Tan(A - B) =

(Tan(A) - Tan(B))/(1 + Tan(A) * Tan(B))

(Root over everything) √2/2 =

(√2)/(√2)

Identity Matrices (I)

- 1s on diagonal (top left to bottom right) - 0s everywhere else

When can use sum/difference of angles identities

- Have an angle like 75°, break it up into 30° and 45° and use sum of angles identity - After using sum of angles identity, look at unit circle to get actual number and solve - Can use sum of angles identity on something like sin(3x) - Ex: sin(3x) = sin(2x + x) —> ((Sin (A + B) Identity)) —> sin(2x)cos(x) + cos(2x)sin(x)

How to set up matrix when asks: x + y = 3 y + z = 8 x + z = 1

- If there no variable, it's 0 x y z [1 1 0] |0 1 1| [1 0 1] [3] |8| [1]

Using calculator to solve matrices example problem

- basically: [x] |y| = A⁻¹B [z] - B is your matrix for you system - C is your matrix for your answers

Graph y = cot(x)

- cot asymptotes at 0, π, 2π without shifts - curve/line at half your period - for Cot, know asymptote at 0 without any shifts, so add/subtract π/b to original asymptote to get next asymptote - if shifts, do the shift first then half the period - arrow goes up left to down right - Cot is just Tan flipped and everything shifted right π/2 - -cot(x) arrow goes down left to up right - need to memorize this

Solving a linear system using matrices color coded

- found the inverse of [5 1| |8 2] by using the formula for the inverse of a 2 x 2 matrix

Graph y = tan(x)

- is sin(x)/cos(x) - tan asymptotes at π/2, 3π/2 without shifts - tan line starts at 0 without shifts - 1st asymptote is always at half the period - add a period to the 1st asymptote to get to the 2nd asymptote - if shifts, do the shift first then half the period - arrow goes down left to up right - need to memorize this - -tan(x) arrow goes down right to up left

Graph y = sec(x)

- is y = 1/cos(x) - graph y = cos(x), then do reciprocals - there's a vertical asymptote every time graph hits 0

Graph y = csc(x)

- is y = 1/sin(x) - graph y = sin(x), then do reciprocals - there's a vertical asymptote every time graph hits 0

Rows and columns

- rows are vertical - columns are horizontal - say "1, 2, 9, 15" Ex: 2 x 2 matrix - first 2: # of rows - second 2: # of columns

Rearrangement matrix

- swap place of a and d - switch signs of b and c - up (+) - down (-) - down (-) - up (+)

Common sense things for when solving Trig Equations

- tan(x) = sin(x)/cos(x) - cot(x) = cos(x)/sin(x) - 1 = cos(x)/cos(x) —> anything over itself - sec(x) = 1/cos(x) - csc(x) = 1/sin(x) - sin²(x) + cos²(x) = 1 - tan²(x) + 1 = sec²(x) - cot²(x) + 1 = csc²(x) - multiplying by conjugate

Answer roots like this

-(√2)/4 - (√6)/4, NOT -(√8)/4

Standard basis vectors

->i = <1,0> ->j = <0,1>

If ->u • ->v = 0, then

->u • ->v is orthogonal (90°)

Vector Addition If ->v = <x₁, y₁> and ->u = <x₂, y₂>, then

->v + ->u = <(x₁ + x₂),(y₁ + y₂)

Area

1/2abSinC

sin(2x) =

2Sin(x)Cos(x) - can also be Sin(x)2Cos(x) - think of the 2 as being factored out but not multiplying —> (2sin(x)2cos(x)) is wrong

3 x 3 Determinant Procedure 1. _______ 2. _______ 3. _______

3 x 3 Determinant Procedure 1. pick any row or column 2. for each #, eliminate current row and column 3. add/subtract based on sign chart

->v = <x,y> =

<|->v|Cosθ, |->v|Sinθ>

A x A⁻¹ = I and A⁻¹ x A = I - A⁻¹ is A inverse

A x A⁻¹ = I and A⁻¹ x A = I - A⁻¹ is A inverse

(Solving for x) If Ax = B, then - A is a 2 x 2 - x is a 2 x 1 - B is a 2 x 1

A⁻¹Ax ——> x = A⁻¹B - Can't do BA⁻¹ because the inside numbers are different (m ≠ m) - So must multiply by A⁻¹ on the left

How to get matrices on the calculator

Calculator: - 2nd - x⁻¹ - select matrix - x⁻¹ - (for fraction of matrix) - Math - Enter - Enter

cos(2x) =

Cos(2x) = Cos²(x) - Sin²(x) OR Cos(2x) = 1 - 2Sin²(x) OR Cos(2x) = 2Cos²(x) - 1

Cos(A - B) =

Cos(A) * Cos(B) + Sin(A) * Sin(B)

Cos(A + B) =

Cos(A) * Cos(B) - Sin(A) * Sin(B)

Cos half-angle identity

Cos(x) = ±√(1 + Cos(2x))/√2 - If want Cos(x/2), make Cos(2x) —-> Cos(x)

y = tan⁻¹(x) = arctan(x)

Domain: (-∞, +∞) Range: [-π/2, π/2] - asymptote at range