Question

Given w =StartRoot 2 EndRoot (cosine (StartFraction pi Over 4 EndFraction) + I sine (StartFraction pi Over 4 EndFraction) ) and z = 2 (cosine (StartFraction pi Over 2 EndFraction) + I sine (StartFraction pi Over 2 EndFraction) ) , what is w – z expressed in polar form?

StartRoot 2 EndRoot (cosine (StartFRaction pi Over 4 EndFraction) + I sine (StartFraction pi Over 4 EndFraction) )
StartRoot 2 EndRoot (cosine (StartFraction 3 pi Over 4 EndFraction) + I sine (StartFraction 3 pi Over 4 EndFraction) )
StartRoot 2 EndRoot (cosine (StartFraction 5 pi Over 4 EndFraction) + I sine (StartFraction 5 pi Over 4 EndFraction) )
StartRoot 2 EndRoot (cosine (StartFraction 7 pi Over 4 EndFraction) + I sine (StartFraction 7 pi Over 4 EndFraction) )Solve the following triangle using the Law of Sines.
(9 points)

Answer 1

d) √2∠(7π/4)

Explanation:

Sometimes the expression A(cos(α)+i·sin(α)) is written as A·cis(α). I prefer the notation A∠α, because it is more compact.

You are given ...

w = √2∠(π/4)

z = 2∠(π/2)

And you are asked to find the difference w-z in polar form.

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In rectangular form, ...

w = √2∠(π/4) = √2(1/√2 +i/√2) = 1 +i

z = 2∠(π/2) = 2(0 +i·1) = 2i

Then the difference is ...

w -z = (1 +i) -(2i) = 1 -i

You may notice this is the conjugate of w, so will have the opposite angle:

w - z = w* = √2∠(-π/4) = √2∠(7π/4)

In the terms used by the problem statement, ...

w -z = √2(cos(7π/4) +i·sin(7π/4)) . . . . matches choice D