Given Tan(A) = 5 in Quadrant III and Sin(B) = ⅔ in Quadrant II, find Sin(A-B).

Draw your triangles and quickly see that

sinA = -5/√26
cosA = -1/√26

sinB = 2/3
cosB = -√5/3

now just use your sum-of-angles formula for sin(A-B)

To find Sin(A-B), we can use the identity Sin(A-B) = Sin(A) * Cos(B) - Cos(A) * Sin(B).

First, let's find the values of Cos(A) and Cos(B) using the information given.

Since Tan(A) = 5 in Quadrant III, we can use the definition of the tangent function: Tan(A) = Sin(A) / Cos(A).

Given that Tan(A) = 5, we have Sin(A) / Cos(A) = 5.

By rearranging the equation and using the fact that the square of Sin(A) + the square of Cos(A) equals 1, we can solve for Cos(A):

Sin(A) = 5 * Cos(A)

Squaring both sides:

Sin(A)^2 = 25 * Cos(A)^2

Using the property of the Pythagorean Identity: Sin^2(A) + Cos^2(A) = 1

25 * Cos(A)^2 + Cos(A)^2 = 1

25 * Cos(A)^2 + 1 * Cos(A)^2 = 1

26 * Cos(A)^2 = 1

Dividing both sides by 26:

Cos(A)^2 = 1/26

Taking the square root of both sides:

Cos(A) = ±sqrt(1/26)

Since A is in Quadrant III, which means Cos(A) < 0, we take the negative square root:

Cos(A) = -sqrt(1/26)

Now, let's find the value of Cos(B).

Given Sin(B) = 2/3 in Quadrant II, we can use the definition of the sine function: Sin(B) = Sin(B) / Cos(B).

Given that Sin(B) = 2/3, we have 2/3 = Sin(B) / Cos(B).

By rearranging the equation and using the fact that the square of Sin(B) + the square of Cos(B) equals 1, we can solve for Cos(B):

Sin(B) = (2/3) * Cos(B)

Squaring both sides:

Sin(B)^2 = (4/9) * Cos(B)^2

Using the property of the Pythagorean Identity: Sin^2(B) + Cos^2(B) = 1

(4/9) * Cos(B)^2 + Cos(B)^2 = 1

(4/9 + 1) * Cos(B)^2 = 1

(13/9) * Cos(B)^2 = 1

Dividing both sides by (13/9):

Cos(B)^2 = 9/13

Taking the square root of both sides:

Cos(B) = ±sqrt(9/13)

Since B is in Quadrant II, which means Cos(B) < 0, we take the negative square root:

Cos(B) = -sqrt(9/13)

Now that we have the values of Cos(A) and Cos(B), we can find Sin(A-B):

Sin(A-B) = Sin(A) * Cos(B) - Cos(A) * Sin(B)

Substituting the values we found:

Sin(A-B) = (5) * (-sqrt(9/13)) - (-sqrt(1/26)) * (2/3)

Simplifying:

Sin(A-B) = -5sqrt(9/13) + sqrt(1/26) * (2/3)

At this point, the calculation becomes a numerical evaluation. Using a calculator, we can find the decimal approximation for Sin(A-B):

Sin(A-B) ≈ -0.569