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\[sin(2~arc~cos{3\over5})\]i forgot how to do this...
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@artofspeed
@Hero
sin(2x)=2 * sinx * cos x
arccos(3/5) = x can be drawn as|dw:1358402188442:dw|
hence the third side is sqrt(5^2-3^2)=4
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sinx = 4/5 and cosx=3/5
sin(2x)=2 * sinx * cos x = 2*4/5*3/5 = 24/25
i dont get how sin2x=2sinxcosx
nevermind...
thanks @artofspeed ! :)
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nooo problem
lol :D
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