Trig Identity Question Sort of

[s_j_sawyer]

Homework Statement

Okay so the objective here is to express

y(t) = cos(t - b) - cos(t)

in the form

y(t) = Asin(t - c)

where A and c are in terms of b.

Homework Equations

For easy reference, here is a table of identities:
http://www.sosmath.com/trig/Trig5/trig5/trig5.html

The Attempt at a Solution

Well, using the sum and difference formulas, I got that

y(t) = cost(cosb - 1) + sint*sinb

equating this to the desired expression gives

cost(cosb - 1) + sint*sinb = Asin(t - c)
cost(cosb - 1) + sint*sinb = A(sint)(cosc) - A(cost)(sinc)

So thus I determined that

cosb - 1 = -Asinc (1)
sinb = Acosc (2)

Squaring both sides and adding gave me, eventually,

A^2 = -2cosb + 1

So would A be +/- sqrt(-2cosb + 1) ?

Then I did almost the exact same thing for c simply by moving the -1 on the left side of (1) to the right:

cosb = -Asinc + 1 (1*)
sinb = Acosc (2)

Squaring and adding I got

A^2 - 2Asinc = 0

A - 2sinc = 0

sinc = A/2

so then would c = arcsin(A/2)?I don't even know if I am doing this right so any assistance would be great!

Thank you.

 

[rootX]

cost(cosb - 1) + sint*sinb = A(sint)(cosc) - A(cost)(sinc)

cosb - 1 = -Asinc (1)
sinb = Acosc (2)

Squaring and adding I got

A^2 - 2Asinc = 0

It is very hard to read it but squaring and ading (1) and (2) should give A^2 on the right side but looks like you simplified.

Yes, you are using the right approach. In case you missed second step is to divide (1) by (2) to get second equation.
A.tan(c) = ...