Find cos(A+B).

cos A=1/3 and sin B=-1/2, with A in quadrant I and B in quadrant IV.

cos A = 1 / 3

sin A = + OR - sqrt ( 1 - cos A ^ 2 )

sin A = + OR - sqrt ( 1 - ( 1 / 3 ) ^ 2 )

sin A = + OR - sqrt ( 1 - 1 / 9 )

sin A = + OR - sqrt ( 9 / 9 - 1 / 9 )

sin A = + OR - sqrt ( 8 / 9 )

sin A = + OR - sqrt ( 4 * 2 / 9 )

sin A = + OR - sqrt ( 4 ) * sqrt ( 2 ) / sqrt ( 9 )

sin A = + OR - 2 * sqrt ( 2 ) / 3

sin A = + OR - ( 2 / 3 ) * sqrt ( 2 )

In quadrant I sine are positive so :

sin A = ( 2 / 3 ) * sqrt ( 2 )

sin B = - 1 / 2

sin B = + OR - sqrt ( 1- cos B ^ 2 )

cos B = + OR - sqrt ( 1 - ( - 1 / 2 ) ^ 2 )

cos B = + OR - sqrt ( 1 - 1 / 4 )

cos B = + OR - sqrt ( 4 / 4 - 1 / 4 )

cos B = + OR - sqrt ( 3 / 4 )

cos B = + OR - sqrt ( 3 ) / 2

In quadrant IV cosine are positive so :

cos B = sqrt ( 3 ) / 2

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

cos ( A + B ) = ( 1 / 3 ) * sqrt ( 3 ) / 2 - ( 2 / 3 ) * sqrt ( 2 ) * ( - 1 / 2 )

cos ( A + B ) = ( 1 / 6 )sqrt ( 3 ) + ( 2 / 6 ) * sqrt ( 2 )

cos ( A + B ) = ( 1 / 6 ) [ sqrt ( 3 ) + 2 sqrt ( 2 ) ]

cos ( A + B ) = [ sqrt ( 3 ) + 2 sqrt ( 2 ) ] / 6

what's all this work?

A is in QI, so sinA = √8/3
B is in QIV so cosB = √3/2

and then as done in the final paragraph

To find cos(A + B), we can use the trigonometric identity:

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

Given that cos A = 1/3 and sin B = -1/2, we need to find cos B and sin A.

Since A is in quadrant I, cos A is positive. So we can use the Pythagorean identity to find sin A:

sin A = √(1 - cos^2 A)
= √(1 - (1/3)^2)
= √(1 - 1/9)
= √(8/9)
= √8 / 3

Since B is in quadrant IV, sin B is negative. So we can use the Pythagorean identity to find cos B:

cos B = √(1 - sin^2 B)
= √(1 - (-1/2)^2)
= √(1 - 1/4)
= √(3/4)
= √3 / 2

Now we can substitute these values into the trigonometric identity to find cos(A + B):

cos(A + B) = cos A * cos B - sin A * sin B
= (1/3) * (√3 / 2) - (√8 / 3) * (-1/2)
= (√3 / 6) + (√8 / 6)
= (√3 + √8) / 6

Therefore, cos(A + B) = (√3 + √8) / 6.

To find cos(A+B), we can use the cosine addition formula:

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

Given that cos A = 1/3 and sin B = -1/2, we need to find cos B and sin A to substitute into the formula.

Since A is in quadrant I, cos A is positive, and we can use the Pythagorean identity to find sin A:

cos² A + sin² A = 1
(1/3)² + sin² A = 1
1/9 + sin² A = 1
sin² A = 1 - 1/9
sin² A = 8/9

Taking the square root of both sides, we get:

sin A = √(8/9) = √8/√9 = √8/3

Now, let's find cos B. Since B is in quadrant IV, both cos B and sin B are positive.

Using the Pythagorean identity:

cos² B + sin² B = 1
cos² B + (-1/2)² = 1
cos² B + 1/4 = 1
cos² B = 1 - 1/4
cos² B = 3/4

Taking the square root of both sides, we get:

cos B = √(3/4) = √3/√4 = √3/2

Finally, we can substitute the values of cos A, sin A, cos B, and sin B into the cosine addition formula:

cos(A+B) = (1/3)(√3/2) - (√8/3)(-1/2)
= (√3/6) + (√8/6)
= (√3 + √8)/6

Therefore, cos(A+B) = (√3 + √8)/6.