1 Comments

  1. Caawiye Member

    Solution

    a. \(Cos(0°+θ)=Cos0°cosθ-Sin0°sinθ←1×cosθ-0×sinθ \)=

    b. \(Cos(90°+θ)=Cos90°cosθ-Sin90°sinθ←0×cosθ-1×sinθ \)=

    c. \(Cos(180°+θ)=Cos180°cosθ-Sin180°sinθ←-1×cosθ-0×sinθ \)=

    d. \(Cos(270°+θ)=Cos270°cosθ-Sin270°sinθ←0×cosθ-(-1)×sinθ \)=

    Solution:

    a. \(Cos(0°+θ)=Cos0°cosθ-Sin0°sinθ 1×cosθ-0×sinθ=cosθ \)

    b. \( Cos(90°+θ)=Cos90°cosθ-Sin90°sinθ 0×cosθ-1×sinθ=sinθ \)

    c. \( Cos(180°+θ)=Cos180°cosθ-Sin180°sinθ -1×cosθ-0×sinθ=-cosθ.\)

    d. /(Cos(270°+θ)=Cos270°cosθ-Sin270°sinθ←0×cosθ-(-1

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