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Chapter1 (Q8) Find each of the following a)\(Cosθ, given Cos2θ=1/2, and θ terminates in quadrant 2.\)b). \(Cosx, given Cos2x=-5/12, With π/2<x<π .\)c). \( Cosx, given Cos2x=2/3, With π<x<3π/2 \)
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1 Answers
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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=√(