precal- PLEASE HELP!

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Given:cosA= 4/5 0(less than or equal to)A(less than of equal to)90*

find sinA, sin2A, cos2A

  • precal- PLEASE HELP! -

    cosA = 4/5 or opposite = 4, hypotenuse = 5
    you should recognize the 3-4-5 triangle,

    then sinA = 3/5

    sin 2A = 2sinAcosA = 2(3/5)(4/5) = 24/25

    cos 2A = cos^2 A - sin^2 A
    = 16/25 - 9/25 = 7/25

  • precal- PLEASE HELP! -

    thank you.

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