Using Euler's relation, derive the following relationships:a. Cosθ=½(e^jθ+e^−jθ)b. Sinθ=½(e^jθ−e(^−jθ)

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)]

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step-by-step explanation: