ДОКАЖИТЕ ТОЖДЕСТВО
sin(a+π)/sin(a+3π/2) + cos(3π-a)/(cos(π/2+a)-1) = 1/cosa
sin(a+π)/sin(a+3π/2) + cos(3π-a)/(cos(π/2+a)-1) = 1/cosa
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Sin(A + π) = -sinAsin(A + 3π/2) = sin(A + π + π/2) = -sin(A + π/2) = -cosA
cos(3π - A) = cos(2π + π - A) = cos(π - A) = -cosA
cos(π/2 + A) = -sinA
-sinA/(-cosA) - cosA/(-sinA - 1) = sinA/cosA + cosA/(sinA + 1) =
= [sinA(sinA + 1) + cos²A)]/(sinAcosA + cosA) =
(sin²A + cos²A + sinA)/(sinAcosA + cosA) = (1 + sinA)/cos(1 + sinA) =
= 1/cosA, чтд