Answer:
a. cosθ = ¹/₂[e^jθ + e^(-jθ)] b. sinθ = ¹/₂[e^jθ - e^(-jθ)]
Step-by-step explanation:
a.We know that
e^jθ = cosθ + jsinθ and
e^(-jθ) = cosθ - jsinθ
Adding both equations, we have
e^jθ = cosθ + jsinθ
+
e^(-jθ) = cosθ - jsinθ
e^jθ + e^(-jθ) = cosθ + cosθ + jsinθ - jsinθ
Simplifying, we have
e^jθ + e^(-jθ) = 2cosθ
dividing through by 2 we have
cosθ = ¹/₂[e^jθ + e^(-jθ)]
b. We know that
e^jθ = cosθ + jsinθ and
e^(-jθ) = cosθ - jsinθ
Subtracting both equations, we have
e^jθ = cosθ + jsinθ
-
e^(-jθ) = cosθ - jsinθ
e^jθ + e^(-jθ) = cosθ - cosθ + jsinθ - (-jsinθ)
Simplifying, we have
e^jθ - e^(-jθ) = 2jsinθ
dividing through by 2 we have
sinθ = ¹/₂[e^jθ - e^(-jθ)]
