math

posted by isaiah

if cosØ=0.6 and 0°<Ø<90°, find the exact value of 2Ø

  1. Reiny

    remember cos^2 Ø + sin^2 Ø = 1
    .36 + sin^2 Ø = 1
    sin^2 Ø = .64
    sinØ = .8

    sin 2Ø = 2sinØcosØ
    = 2(.8)(.6)
    = .96

    2Ø = sin^-1 (.96) or arcsin(.96)

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