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#