Question: Considering Section 7.1, suppose we began the analysis to find E = E 1 + E 2 with two cosine functions E 1 = E

Considering Section 7.1, suppose we began the analysis to find E = E1+ E2with two cosine functions E1= E01cos (ωt + α1) and E2= E02 cos (ωt + α2). To make things a little less complicated, let E01= E02 and α= 0. Add the two waves algebraically and make use of the familiar trigonometric identity cos θ + cos Φ = 2 cos 1/2 (θ + Φ) cos 1/2 (θ + Φ) in order to show that E = E0cos (ωt + α), where E0= 2E01cos α2/2 and α = α2/2. Now show that these same results follow from Eqs. (7.9) and (7.10).

E = Eổi + Eổ2 + 2E01E02 cos (œ2 – aj) (7.9)

E01 sin aj + E2 sina2 tan a (7.10) E01 cos aj + E02 cos a2

E = Ei + E2 + 2E01E02 cos (2 aj) (7.9) E01 sin aj + E2 sina2 tan a (7.10) E01 cos aj + E02 cos a2

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