Question

Если cosA=5/13,sinB=-0.6,A∈Ⅰчетв.,B∈Ⅲчетв,то найдите sin(A-B).

Answer 1

Решение в приложении.

Answer 2

Sin(A-B)=sinA*cosB-cosB*sinA

Answer 3

To find the value of sin(A-B), we can use trigonometric identities. First, let's find the values of sin(A) and cos(B) using the given information:

1. cos(A) = 5/13 Since A is in the first quadrant, both sine and cosine are positive in the first quadrant.

2. sin^2(A) + cos^2(A) = 1 sin^2(A) + (5/13)^2 = 1 sin^2(A) + 25/169 = 1 sin^2(A) = 1 - 25/169 sin^2(A) = 144/169 sin(A) = √(144/169) sin(A) = 12/13 (since sine is positive in the first quadrant)

3. sin(B) = -0.6 Since B is in the third quadrant, both sine and cosine are negative in the third quadrant.

4. cos^2(B) + sin^2(B) = 1 (-0.6)^2 + sin^2(B) = 1 0.36 + sin^2(B) = 1 sin^2(B) = 1 - 0.36 sin^2(B) = 0.64 sin(B) = √0.64 sin(B) = 0.8 (since sine is negative in the third quadrant)

Now that we have sin(A) and sin(B), we can find sin(A-B) using the trigonometric identity:

sin(A - B) = sin(A) * cos(B) - cos(A) * sin(B)

Substitute the values:

sin(A - B) = (12/13) * (-0.6) - (5/13) * (0.8)

sin(A - B) = (-12/13) - (4/13)

sin(A - B) = -16/13

So, sin(A - B) = -16/13.