
Yefim S. answered 02/20/21
Math Tutor with Experience
cscu = 5, sinu = 1/5
cos(2u) = 1 - 2sin2u = 1 - 2/25 = 23/25;
sin(2u) = 2sinucosu = 2·1/5·√1 - 1/25 = 2/5·2√6/5 = 4√6/25;tan(2u) = sin(2u)/cos(2u) = (4√6/25)/(23/25) = 4√6/23
Aerin H.
asked 02/20/21csc(u) = 5, 0 < u < 𝜋/2
sin(2u) =
cos(2u) =
tan(2u) =
Yefim S. answered 02/20/21
Math Tutor with Experience
cscu = 5, sinu = 1/5
cos(2u) = 1 - 2sin2u = 1 - 2/25 = 23/25;
sin(2u) = 2sinucosu = 2·1/5·√1 - 1/25 = 2/5·2√6/5 = 4√6/25;tan(2u) = sin(2u)/cos(2u) = (4√6/25)/(23/25) = 4√6/23
Mark M. answered 02/20/21
Mathematics Teacher - NCLB Highly Qualified
csc u = 5 ⇒ sin u = 1/5
sin u = 1/5 ⇒ cos u √3/2
Now substitute values in appropriate formula.
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