Find the exact value of sin(u/2), cos(u/2), and tan(u/2) using the half-angle formula, given that cscu = -5/3 and π < u < 3π/2

Answer 1

π < u < 3 π / 2

180° < u < 270°

u lies in Quadrant III

csc ( u ) = 1 / sin ( u ) = - 5 / 3

sin ( u ) = - 3 / 5

cos ( u ) = ±√ ( 1 - sin² u )

cos ( u ) = ±√ [ 1 - ( - 3 / 5 )² ]

cos ( u ) = ±√ ( 1 - 9 / 25 )

cos ( u ) = ±√ ( 25 / 25 - 9 / 25 )

cos ( u ) = ±√ ( 16 / 25 )

cos ( u ) = ± 4 / 5

in Quadrant III cosine is negative so:

cos ( u ) = - 4 / 5

π / 2 < u / 2 < 3 π / 4

90° < u / 2 < 135°

u / 2 lies in Quadrant II

sin ( u / 2 ) = ±√ [ ( 1 - cos u ) / 2 ]

sin ( u / 2 ) = ±√ [ ( 1 - ( - 4 / 5 ) ) / 2 ]

sin ( u / 2 ) = ±√ [ ( 1 + 4 / 5 ) / 2 ]

sin ( u / 2 ) = ±√ [ ( 5 / 5 + 4 / 5 ) / 2 ]

sin ( u / 2 ) = ±√ [ ( 9 / 5 ) / 2 ]

sin ( u / 2 ) = ±√ ( 9 / 10 )

sin ( u / 2 ) = ±√ 9 / √ 10

sin ( u / 2 ) = ± 3 / √10

in Quadrant II sine is postive so:

sin ( u / 2 ) = 3 / √10

cos ( u / 2 ) = ±√ [ ( 1 + cos u ) / 2 ]

cos ( u / 2 ) = ±√ [ ( 1 + ( - 4 / 5 ) ) / 2 ]

cos ( u / 2 ) = ±√ [ ( 1 - 4 / 5 ) / 2 ]

cos ( u / 2 ) = ±√ [ ( 5 / 5 - 4 / 5 ) / 2 ]

cos ( u / 2 ) = ±√ [ ( 1 / 5 ) / 2 ]

cos ( u / 2 ) = ±√ ( 1 / 10 )

cos ( u / 2 ) = ±√ 1 / √ 10

cos ( u / 2 ) = ± 1 / √10

in Quadrant II cosine is negative so:

cos ( u / 2 ) = - 1 / √10

tan ( u / 2 ) = ±√ [ ( 1 - cos u ) / ( 1 + cos u ) ]

tan ( u / 2 ) = ±√ [ ( 1 - ( - 4 / 5 ) ) / ( 1 + ( - 4 / 5 ) ]

tan ( u / 2 ) = ±√ [ ( 1 + 4 / 5 ) / ( 1 - 4 / 5 ) ]

tan ( u / 2 ) = ±√ [ ( 5 / 5 + 4 / 5 ) / ( 5 / 5 - 4 / 5 ) ]

tan ( u / 2 ) = ±√ [ ( 9 / 5 ) / ( 1 / 5 ) ]

tan ( u / 2 ) = ±√ 9

tan ( u / 2 ) = ± 3

in Quadrant II tangent is negative so:

tan ( u / 2 ) = - 3

Answer 2

Thank you very much