If sinA=4/5 and cosB= -5/13, where A belongs to QI and B belongs to QIII, then find sin(A+B). How do you write the second half of the formula where cos becomes A and sin is B and how do the quadrants affect these?

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Answer 1 · 2018-04-30T02:58:45

A is in QI, so $cos A = + \sqrt{1- (4/5)^2}= 3/5$ and B is in QIII, so $sin B = -sqrt{1- (-5/13)^2} = -12/13$ so

$sin(A+B) = sin A cos B + cos A sin B = -56/65$

Are we being played by Socratic here? It's says this was asked 13 minutes ago in the feed but 3 years ago in the comment.

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Everybody should immediately recognize the Pythagorean Triples underlying this problem: $3^2 + 4^2 = 5^2$ and $5^2 + 12^2=13^2.$

$cos^2 A + sin ^2 A = 1$

$cos A =\pm \sqrt{1 - sin ^2A}$

$sin A= 4/5$ and $A$ is in quadrant I, so the cosine is positive too.

$cos A= \sqrt{1 - (4/5)^2} = sqrt{9/25}=3/5$

$cos B = -5/13$ and $B$ is in quadrant III, so the sine is negative too.

$sin B = - \sqrt{1 - (-5/13)^2} = -12/13$

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

$= (4/5)(-5/13) + (3/5)(-12/13) = -56/65$