if 270°<A<360°,90°<B<180°,cosA=5/13,tanB=(-15/8)then sin(A+B)=?
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if 270°<A<360°,90°<B<180°,cosA=5/13,tanB=(-15/8)then sin(A+B)=?
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Step-by-step explanation:
cosA=5/13
sinA=-√(1-cos^2A) (since theta is in 4th quadrant)
i.e. sinA=-√144/169
=-12/13
tanB=-15/8
sinB=+(15)/√{(-15)^2+(8)^2} (B is in Q2)
i.e. sinB=15/17
cosB=tanB/sinB
i.e. cosB=-15/8 * 17/15
cosB=17/8
now
sin(A+B)=sinAcosB+cosAsinB
sin(A+B)=(-12/13)(17/8)+(5/13)(15/17)
={1/13}(-51/2 + 75/17)
=(1/13)(-710/34)
=-710/442