Use an addition or subtraction formula to find the solutions of the equation that are in the interval [0, 𝜋).

Question

sin 5t cos 2t = sin 2t cos 5t

t=?

Answer 1

sin(5t)cos(2t) - cos(5t)sin(2t) = sin(5t-2t) = sin(3t) = 0

3t = kπ, where k = 0, 1, -1, 2, -2, ...

t = kπ/3

If k = 0, then t = 0

If k = 1, then t = π/3

If k = 2, then t = 2π/3

All other choices for k give solutions that lie outside of the interval [0,π)

Answer 2

sin 5t=(sin 2t+3t)= sin2t sin3t+cos 2tcos3t

sin 5t cos2t= (sin2t cos3t+cos 2t sin3t)cos 2t

Next

cos 5t = cos(2t+3t)= cos2t cos3t - sin 2t sin3t

sin 2t cos 5t=sin 2t(cos2t cos3t - sin 2t sin3t)

sin 2t(cos2t cos3t - sin 2t sin3t)= (sin2t cos3t+cos 2t sin3t)cos 2t

Cancel : sin^2 2t sin 3t = Cos^2 2t sin 3t

0= (sin^2 2t-cos^2 2t )sin 3t and cos^2 2t=1- sin^2 2t

0= (2sin2 2t -1)(sin 3t)

0 =sin 3t 3t =0 or 3t =pi or 3t = 2pi or 3t =3pi (need to go to 3pi) Get t=0, pi/3, or pi

or

2sin^2 2t -1 =0 Factor a^2-1 =a+1 a-1

sin^2 2t =1/2 ----> 2t = pi/6, 5pi/6, ---> t=pi/12, or 5pi/12

Answer 3

sin 5t cos 2t - sin 2t cos 5t = 0

sin (5t - 2t) = 0

sin 3t = 0

3t = 0

t = 0