Asked by isaiah
if cosØ=0.6 and 0°<Ø<90°, find the exact value of 2Ø
Answers
Answered by
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)
.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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