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Question Number 51933 by Tawa1 last updated on 01/Jan/19

If p = cos θ + i sin θ and q = cos φ + i sin φ Show that (((p + q)(pq − 1))/((p − q)(pq + 1))) = ((sin θ + sin φ)/(sin θ − sin φ))

Ifp=cosθ+isinθandq=cosϕ+isinϕShowthat(p+q)(pq−1)(p−q)(pq+1)=sinθ+sinϕsinθ−sinϕ

Commented by peter frank last updated on 01/Jan/19

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jan/19

p=e^(iθ) q=e^(i∅) LHS (((p+q)(pq−1))/((p−q)((pq+1))) ((((p/q)+1))/(pq+1))×((pq−1)/((p/q)−1)) =((e^(i(θ−∅)) +1)/(e^(i(θ+∅)) +1))×((e^(i(θ+∅)) −1)/(e^(i(θ−∅)) −1)) =((cos(θ−∅)+isin(θ−∅)+1)/(cos(θ+∅)+isin(θ+∅)+1))×((cos(θ+∅)+isin(θ+∅)−1)/(cos(θ−∅)+isin(θ−∅)−1)) =((2cos^2 ((θ−∅)/2)+i2sin((θ−∅)/2)cos((θ−∅)/2))/(2cos^2 ((θ+∅)/2)+i2sin((θ+∅)/2)cos((θ+φ)/2)))×((1−cos(θ+∅)−isin(θ+φ))/(1−cos(θ−∅)−isin(θ−∅))) =((2cos((θ−∅)/2)[cos((θ−∅)/2)+isin((θ−∅)/2)])/(2cos((θ+∅)/2)[cos((θ+∅)/2)+isin((θ+∅)/2)]))×((2sin^2 ((θ+∅)/2)−2isin((θ+φ)/2)cos((θ+∅)/2))/(2sin^2 ((θ−φ)/2)−2isin((θ−∅)/2)cos((θ−φ)/2))) =((cos((θ−∅)/2))/(cos((θ+∅)/2)))×e^(i(((θ−∅)/2))−i(((θ+∅)/2))) ×((2sin((θ+∅)/2))/(2sin((θ−φ)/2)))×(([sin((θ+∅)/2)−icos((θ+∅)/2)])/([sin((θ−∅)/2)−icos((θ−∅)/2)])) ((tan((θ+φ)/2))/(tan((θ−φ)/2)))×e^(i(((θ−φ−θ−φ)/2))) ×(e^(−i[(π/2)−(((θ+φ)/2))]) /e^(−i[(π/2)−(((θ−∅)/2))]) ) =((tan((θ+φ)/2))/(tan((θ−∅)/2)))×e^(−iφ) ×e^(−((iπ)/2)+i(((θ+φ))/2)+((iπ)/2)−i(((θ−∅)/2))) =((tan((θ+∅)/2))/(tan((θ−φ)/2)))×e^(−i∅) ×e^(i(((θ+∅−θ+φ)/2))) =((tan((θ+∅)/2))/(tan((θ−∅)/2)))×e^(−i∅+i∅) =((2sin((θ+∅)/2)cos((θ−∅)/2))/(2sin((θ−∅)/2)cos((θ+∅)/2))) =((sinθ+sin∅)/(sinθ−sin∅))

p=eiθq=ei∅LHS(p+q)(pq−1)(p−q)((pq+1)(pq+1)pq+1×pq−1pq−1=ei(θ−∅)+1ei(θ+∅)+1×ei(θ+∅)−1ei(θ−∅)−1=cos(θ−∅)+isin(θ−∅)+1cos(θ+∅)+isin(θ+∅)+1×cos(θ+∅)+isin(θ+∅)−1cos(θ−∅)+isin(θ−∅)−1=2cos2θ−∅2+i2sinθ−∅2cosθ−∅22cos2θ+∅2+i2sinθ+∅2cosθ+ϕ2×1−cos(θ+∅)−isin(θ+ϕ)1−cos(θ−∅)−isin(θ−∅)=2cosθ−∅2[cosθ−∅2+isinθ−∅2]2cosθ+∅2[cosθ+∅2+isinθ+∅2]×2sin2θ+∅2−2isinθ+ϕ2cosθ+∅22sin2θ−ϕ2−2isinθ−∅2cosθ−ϕ2=cosθ−∅2cosθ+∅2×ei(θ−∅2)−i(θ+∅2)×2sinθ+∅22sinθ−ϕ2×[sinθ+∅2−icosθ+∅2][sinθ−∅2−icosθ−∅2]tanθ+ϕ2tanθ−ϕ2×ei(θ−ϕ−θ−ϕ2)×e−i[π2−(θ+ϕ2)]e−i[π2−(θ−∅2)]=tanθ+ϕ2tanθ−∅2×e−iϕ×e−iπ2+i(θ+ϕ)2+iπ2−i(θ−∅2)=tanθ+∅2tanθ−ϕ2×e−i∅×ei(θ+∅−θ+ϕ2)=tanθ+∅2tanθ−∅2×e−i∅+i∅=2sinθ+∅2cosθ−∅22sinθ−∅2cosθ+∅2=sinθ+sin∅sinθ−sin∅

Commented by Tawa1 last updated on 01/Jan/19

God bless you sir. I really appreciate your time.

Godblessyousir.Ireallyappreciateyourtime.

Answered by $@ty@m last updated on 01/Jan/19

see my solution to Q. No. 51905

seemysolutiontoQ.No.51905

Commented by Tawa1 last updated on 01/Jan/19

God bless you sir. Thanks for your time.

Godblessyousir.Thanksforyourtime.

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