Given that sin(a)=2/3 and cos(b)=-1/5, with a and b both in the interval of pi - pi/2

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Hi Lauren.

This can be done with trig addition. Here are the formulas:

sin(a+b)=sin(a)cos(b) + sin(b)cos(a)

sin(a-b)=sin(a)cos(b) - sin(b)cos(a)

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)

cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

So... with your values of sin(a) = 2/3 and cos(b) = -1/5, we need to find cos(a) and sin(b).

Again, using a triangle, and putting the angle the 2nd quadrant, the angle a is in the bottom right vertex (angle from neg x-axis and the ray completing the angle), opposite side is vertical side = 2. Hypotenuse = 3. So the horizontal leg is √5. Since it is in the 2nd quadrant, it will be negative. That makes cos(a)=(-√5/3).

Do it again for the angle b. Make an angle starting at the origin and up to the left into the 2nd quadrant. The angle b is the bottom right vertex, between the neg x-axis and the ray from the origin. This time the adjacent side, the horizontal side is -1 and the hypotenuse is 5. Using the Pythag Thm find that the vertical side is √24 = 2√6. That makes the sin(b)=(2√6)/5.

Now just plug these values into the formulas:

sin(a+b)=sin(a)cos(b) + sin(b)cos(a) = (2/3)(-1/5) + (2√6/5)(-√5/3) = -2/15 - 2√30/15

cos(a-b)=cos(a)cos(b) + sin(a)sin(b) = (-√5/3)(-1/5) + (2/3)(2√6/5) = √5/15 + 4√6/15

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