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Chapter1 (Q8) Find each of the following a)\(Cosθ, given Cos2θ=1/2, and θ terminates in quadrant 2.\)b). \(Cosx, given Cos2x=-5/12, With π/2
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1 Answer

  1. Solution
    \(COS2θ=2Sin^2 θ-1
    ∴Cosθ=√((Cos2θ+1)/2) ←√((1/2+1)/2) √((3/2)/2) √(3/4 \)) =

    Solution
    \(COS2θ=2Sin^2 θ-1
    ∴Sinx=√((Cos2x+1)/2) ←√((-5/12+1)/2) √((7/12)/2) =- √(7/24) .
    c.\( Cosx, given Cos2x=2/3, With π<x<3π/2 \)
    Solution
    \(COS2θ=2Sin^2 θ-1
    ∴Cosx=√((1-cos2x)/2) ←√((1-2/3)/2) √((1/3)/2 \))=

    Solution
    a. /(COS2θ=2Sin^2 θ-1
    ∴Cosθ=√((Cos2θ+1)/2) ←√((1/2+1)/2) √((3/2)/2) √(3/4) =- √3/2. /)
    Solution
    b. /( COS2θ=2Sin^2 θ-1
    ∴Sinx=√((Cos2x+1)/2) ←√((-5/12+1)/2) √((7/12)/2) =- √(7/24) ./)

    Solution
    c. /(COS2θ=2Sin^2 θ-1
    ∴Cosx=√(

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