# Maths

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

thanks ... :)

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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?

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yeah that's right, can u go on?

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

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Error: sinA=-3/5

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

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well, i got:

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

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

therefore:

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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

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

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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

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