Expert: Paul Klarreich - 5/20/2008

Question
Use identities for cos (C+D) and cos (C-D) to prove that cos A + cos B = 2cos((A+B)/2)cos((A-B)/2). Hence, find in terms of pi, the general solution of the equation:
cos(5t) + cos(t) = cos (3t) (where t= theta)

I proved the identities without any problems but im not sure how to find the general solutions.  I think it has something to with +/- theta + 2npi.

Thank you

Answer
Questioner:   Eliza
Category:  Advanced Math
Private:  No
 
Subject:  Trig functions
Question:  Use identities for cos (C+D) and cos (C-D) to prove that cos A + cos B = 2cos((A+B)/2)cos((A-B)/2). Hence, find in terms of pi, the general solution of the equation:
cos(5t) + cos(t) = cos (3t) (where t= theta)

I proved the identities without any problems but im not sure how to find the general solutions.  I think it has something to with +/- theta + 2npi.

Thank you
........................................
Hi, Eliza,

Yes, this is one of the semi-standard 'Sum-to-product' identities.  [There are also product-to-sum identities.]

So apply it to your equation: (I will skip a lot of parentheses.)

cos 5t + cos t = cos 3t

Use A = 5t,  B = t,  (A+B)/2 = 3t, (A-B)/2 = 2t.

cos 5t + cos t = 2 cos 3t cos 2t

Write your equation:

2 cos 3t cos 2t = cos 3t

2 cos 3t cos 2t - cos 3t = 0

Factor: (treat it as if it were a quadratic)

cos 3t(2 cos 2t - 1) = 0

Set each factor equal to zero:
...................
cos 3t = 0.

This gives

3t = pi/2 + 2npi,  3t = 3pi/2 + 2npi

t = pi/6 + 2npi/3,  t = pi/2 + 2npi/3
.....................

2 cos 2t - 1 = 0

cos 2t = 1/2

2t = pi/3 + 2npi,  2t = -pi/3 + 2npi (or 5pi/3)

t = pi/6 + npi,    t = - pi/6 + 2npi (or 5pi/6)

...........................
Yes, a lot of solutions.  That is how it is with tig equations. [They are a real pain.]