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Question
If p=cos2α+isin2α,q=cos2β+isin2β then the value of √pq−√qp=
If p=cos2α+isin2α,q=cos2β+isin2β then the value of √pq−√qp=
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Solution
The correct option is B 2isin(α−β)
In euler form, p=e2α,q=e2β∴√pq−√qp=√e2αe2β−√e2βe2α=√e2(α−β)−√e2(β−α)=e(α−β)−e(β−α)e(α−β)−e(β−α)=(cos(α−β)+isin(α−β))−(cos(β−α)+isin(β−α))Now, cos(−θ)=cos(θ) and sin(−θ)=−sin(θ)So, above will be, (cos(α−β)+isin(α−β))−(cos(α−β)−isin(α−β))=2isin(α−β)
In euler form, p=e2α,q=e2β
∴√pq−√qp=√e2αe2β−√e2βe2α
=√e2(α−β)−√e2(β−α)=e(α−β)−e(β−α)
e(α−β)−e(β−α)=(cos(α−β)+isin(α−β))−(cos(β−α)+isin(β−α))
Now, cos(−θ)=cos(θ) and sin(−θ)=−sin(θ)
So, above will be, (cos(α−β)+isin(α−β))−(cos(α−β)−isin(α−β))=2isin(α−β)
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