if cosA = 4/5; sin <0; cosB = 12/13; 0<B<90, determine sin(A-B)

thanks ... :)

The first pairing says that sinA is -4/5

If cosB is 12/13, then sinB is 7/13

Sin(A-B)=CosASinB-SinACosB, right?

yeah that's right, can u go on?

I could, but I wont. However, I will be happy to critique your work or thinking

Error: sinA=-3/5

in all cases, sin^2 Theta + cos^2 Theta=1

well, i got:

(-4/5)(12/13) - (4/5)(7/13)
therefore

(-48/65) - (28/65)

therefore:

answer = (-76/65)

from if cosA = 4/5; sin <0 we know that A is in the fourth quadrant and sinA = -3/5

from cosB = 12/13; 0<B<90 we know sinB = 5/13

sin(A-B) = sinAcosB - cosAsinB
= (-3/5)(12/13) - (4/5)(5/13)
= -56/65

change the Sin A in the first term, and it works.

i knew that was wrong, coz my answer should be in the 50's haha..thanks though, great help

can i ask how you got sinA and sinB with that information :)

no point in me just copying it i would like to know how you got it

In any angle, the sum of the squares of the cosines and sines is equal to one. Use that to solve.

To determine sin(A-B), we can use the trigonometric identity that states sin(A-B) = sinA cosB - cosA sinB.

Given that:
cosA = 4/5 (which means A is an acute angle)
sinθ < 0 (which means θ is in the 3rd or 4th quadrant)
cosB = 12/13 (which means B is an acute angle and 0° < B < 90°)

First, let's find sinA using the Pythagorean identity sin^2A + cos^2A = 1:
sin^2A = 1 - cos^2A
sin^2A = 1 - (4/5)^2
sin^2A = 1 - 16/25
sin^2A = 9/25
sinA = √(9/25)
sinA = 3/5
Note: Since sinA must be positive, we use the positive square root.

Next, we find sinB using the Pythagorean identity:
sin^2B = 1 - cos^2B
sin^2B = 1 - (12/13)^2
sin^2B = 1 - 144/169
sin^2B = 25/169
sinB = √(25/169)
sinB = 5/13

Now we can calculate sin(A-B) using the formula sin(A-B) = sinA cosB - cosA sinB:
sin(A-B) = (3/5)(12/13) - (4/5)(5/13)
sin(A-B) = 36/65 - 20/65
sin(A-B) = 16/65

Therefore, sin(A-B) = 16/65.