1 Comments

1. 

   August 1, 2022

   Caawiye Member

   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=√(

   Reply