更新1:
hey, look at my comment on the first answer, still stuck! :/ thanks everyone!!
HELPPPPP!!!!!! Find Sin (A+B) When SinA=(3/5) with 90 degrees < A < 270 degrees, and CosB=(8/17) with 0 degrees < B< 180 degrees!!!!!!?
回答 (2)
2016-05-12 11:24 pm
3^2 + 4^2 = 5^2 so from the 3-4-5 right triangle, if sinA = 3/5 then cosA = 4/5, except in the quadrants you mention cosA = -4/5.
8^2 + 15^2 = 17^2 so from the 8, 15, 17 right triangle, if cosB=8/17 then sinB=15/17, positive in the quadrants you mention.
Now use addition formula sin(A+B)=sin A cos B + cos A sin B.
8^2 + 15^2 = 17^2 so from the 8, 15, 17 right triangle, if cosB=8/17 then sinB=15/17, positive in the quadrants you mention.
Now use addition formula sin(A+B)=sin A cos B + cos A sin B.
2016-05-12 10:55 pm
sin(A+B)=sin A cos B + cos A sin B
cos2(theta)+sin2(theta)=1
cos2(theta)+sin2(theta)=1
收錄日期: 2021-04-18 14:54:32
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https://hk.answers.yahoo.com/question/index?qid=20160512145024AAsMKpl