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Question
If a=cosα−isinα;b=cosβ−isinβ;c=cosγ−isinγ then (a2c2−b2abc) is
If a=cosα−isinα;b=cosβ−isinβ;c=cosγ−isinγ then (a2c2−b2abc) is
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Solution
The correct option is A −2isin(α−β+γ)
Applying Euler's form a=cosα−isinα=e−iαb=cosβ−isinβ=e−iβc=cosγ−isinγ=e−iγThereforea2c2=e−2i(α+γ)...(i)b2=e−2iβ ...(ii)abc=e−i(α+β+γ) ...(iii)Now a2c2−b2abc=a2c2abc−b2abc=e−2i(α+γ)e−i(α+β+γ)−e−2iβe−i(α+β+γ)=e−i(2α+2γ−α−β−γ)−e−i(2β−α−β−γ)=e−i(α+γ−β)−e−i(β−α−γ)=ei(β−α−γ)−e−i(β−α−γ)=[cos(β−α−γ)+isin(β−α−γ)]−[cos(β−α−γ)−isin(β−α−γ)]=2isin(β−α−γ)=−2isin(α−β+γ)
Applying Euler's form
a=cosα−isinα=e−iα
b=cosβ−isinβ=e−iβ
c=cosγ−isinγ=e−iγ
Therefore
a2c2=e−2i(α+γ)...(i)
b2=e−2iβ ...(ii)
abc=e−i(α+β+γ) ...(iii)
Now
a2c2−b2abc
=a2c2abc−b2abc
=e−2i(α+γ)e−i(α+β+γ)−e−2iβe−i(α+β+γ)
=e−i(2α+2γ−α−β−γ)−e−i(2β−α−β−γ)
=e−i(α+γ−β)−e−i(β−α−γ)
=ei(β−α−γ)−e−i(β−α−γ)
=[cos(β−α−γ)+isin(β−α−γ)]−[cos(β−α−γ)−isin(β−α−γ)]
=2isin(β−α−γ)
=−2isin(α−β+γ)
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