A is a QII angle. We draw a 3-4-5 right triangle in QII.
, ,
2. Sin (4A).
Edwin
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website! A is an obtuse angle such that cos(A)=-4/5.
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The terminal point is in Q2.
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Find:
1. Cos(4A)
cos(2A) = 2cos^2(A) - 1 = 32/25 - 1 = 7/25
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cos(4A) = 2*(7/25)^2 - 1 = 98/625 - 1
= -527/625
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2. Sin (4A)
sin(4A) = sqrt(1 - cos^2(4A))
= sqrt(390625/390625 - 277729/390625)
= sqrt(112896/390625)
= -336/625