Trig prblm?

Question

Find sin(theta+beta)

if cos theta = -1/2 and pi tan beta = 1/3 abd 0

Answer 1

cos θ = - 1 / 2
θ = 240° (3rd quadrant)
tan β = 1 / 3
β = 18.4° (1st quadrant)
sin (240 + 18.4)°
sin 258.4° = - 0.98

Answer 2

Hi.

Since cos theta is negative and theta is between pi and 2pi, theta must be in the 3rd quadrant.

Beta, on the other hand, would be in the 1st quadrant because it is between 0 and pi, and tan beta is positive.

For us to determine sin(theta+beta), we have to use the expanded notation.

sin(theta+beta) = sin theta * cos beta + cos theta * sin beta

We have to figure out what sin theta, cos beta and sin beta are (since we already know what cos theta is).

Since cos theta = -1/2, theta being in the third quadrant, sin theta = - sqrt(3)/2 (using special right angles).

since sec beta = sqrt( tan^2 beta + 1), sec beta = sqrt( (1/3)^2 + 1), sec beta = sqrt(10)/3. Since beta is in 1st quadrant, sec beta = + sqrt(10)/3. cos beta = 3/sqrt(10) = 3*sqrt(10)/10. sin beta = sqrt(1 - cos^2 beta), thus, sin beta = sqrt(1 - (3*sqrt(10)/10)^2) = 1/sqrt(10) = sqrt(10)/10. Thus,

sin(theta+beta) = sin theta * cos beta + cos theta * sin beta
sin(theta+beta) = - sqrt(3)/2 * 3*sqrt(10)/10 + -1/2 *sqrt(10)/10

sin(theta+beta) = -3 sqrt(30)/20 - sqrt(10)/20

Hope that helps!!!

Answer 3

sin(theta+beta)
= sin(theta)cos(beta) + cos(theta)sin(beta)
= (-√3/2)(3/√10) + (-1/2)(1/√10)
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Ideas:theta is in the third quadrant, and beta is in the first quadrant.