найти cos x , cos 2x , sin 2x , если Sin x = 4/5, и pi/2 < x
cosx -?
cos2x = cos²x - sin²x ?
sin2x = 2sinxcosx ?
Sinx = 4/5
cosx = -√(1-sin²x) = - √(1 - 16/25) = -√9/25 = -3/5
cos2x = cos²x - sin²x = 9/25 - 16/25 = -7/25
sin2x = 2sinxcosx = 2 * 4/5*(-3/5) = -24/25
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Answers & Comments
cosx -?
cos2x = cos²x - sin²x ?
sin2x = 2sinxcosx ?
Sinx = 4/5
cosx = -√(1-sin²x) = - √(1 - 16/25) = -√9/25 = -3/5
cos2x = cos²x - sin²x = 9/25 - 16/25 = -7/25
sin2x = 2sinxcosx = 2 * 4/5*(-3/5) = -24/25