Thread: Pre-Calculus: Trigonometric forumlas and more

No need for a full out walktrhough - a brief description of what I need to do will suffice (AKA: enough to get me rolling on the right track, with enough advice to ensure I can do it all)


1. Show that the coordinates of point (x,y) rotated counter-clockwise by an angle of θ around the origin have the new coordinates (xcosθ - ysinθ, xsinθ + ycosθ). Draw a picture that clearly illustrates your derivation, and provide plenty of explanation. Hint: consider the distance from the origin to (x,y) as "d", and use the angle sum formulas.

2. Use this result to find the equation of the hyperbola xy = 2 rotated 45° counterclockwise. Compute the vertices, foci, and asymptotes of each of these hyperbolas.

3. Derive the equation of the general conic
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
when rotated counterclockwise through an angle θ. Hint: the coefficient of x^2 becomes (A cos^2θ + Bsinθcosθ + Csin^2θ)

4. Show that rotating the equation through an angle of θ where tan 2θ = B/(A-C) will eliminate that pesky xy term.

5. For each of the following, find:
-the angle of rotation that eliminates the xy term
-the equation resulting from that rotation
-the vertices, foci, and, if appropriate, asymptotes of each equation

a. x^2 + 4xy + y^2 - 3 = 0
b. 3x^2 - 10xy + 3y^2 - 32 = 0
c. x^2 + 4xy + 4y^2 + 5(root5y) + 5 = 0

I know it's lengthy and pretty annoying, but any help at all, no matter how little, is greatly appreciated. Thanks in advance!

Originally Posted by Sconts

1. Show that the coordinates of point (x,y) rotated counter-clockwise by an angle of θ around the origin have the new coordinates (xcosθ - ysinθ, xsinθ + ycosθ). Draw a picture that clearly illustrates your derivation, and provide plenty of explanation. Hint: consider the distance from the origin to (x,y) as "d", and use the angle sum formulas.
...

Hello,

I've added a screenshot of the complete work. Sorry!

EB

Hello, Sconts!

I can help you with #1 . . .

1. Show that the coordinates of point (x,y) rotated counter-clockwise by an angle of θ
around the origin have the new coordinates (x·cosθ - y·sinθ, x·sinθ + y·cosθ).
Draw a picture that clearly illustrates your derivation, and provide plenty of explanation.
Hint: consider the distance from the origin to (x,y) as "d", and use the angle-sum formulas.

Code:

        |                 P'
        |                 *(x',y')
        |               * :
        |             *   :
        |           *     :
        |      d  *       :     P
        |       *         :     *(x,y)
        |     *     d     *     :
        |   * θ     *     :     :
        | *   *  φ        :     :
      - * - - - - - - - - + - - + - X
        O

Point P(x,y) make angle φ with the x-axis.
Point P'(x',y') makes angle P'OP = θ.
Let d = OP = OP'.

The coordinates of P are: .x = d·cosφ, .y = d·sinφ .[1]

The coordinates of P' are: .x' = d·cos(θ + φ), .y' = d·sin(θ + φ)


We have: .x' .= .d·cos(θ + φ) .= .d·cosφ·cosθ - d·sinφ·sinθ
. . . . . . . . . . . . . . . . . . . . . . . . .↓ . . . - . - . . .↓
Substitute [1]: . . . . . . . . .x' .= . .x · cosθ . .- . .y · sinθ

. . Therefore: .x' .= .x·cosθ - y·sinθ


We have: .y' .= .d·sin(θ + φ) .= .d·cosφ·sinθ + d·sinφ·cosθ
. . . . . . . . . . . . . . . . . . . . . . . . .↓ . - . - . - . - .↓
Substitute [1]: . . . . . . . . .y' .= . .x · sinθ . .+ . .y · cosθ

. . Therefore: .y' .= .x·sinθ + y·cosθ