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Question

Prove that sin(AB)=sinAcosBcosAsinB

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Solution

We can prove this by using euler’s identity : eix=cosx+i sinx

therefore we have eiA=cosA+i sinA,

and eiB=cosB+isinB

now

eiAeiB=ei(A+B)

(cosA+i sinA)(cosB+i sinB)=(cos(A+B)+i sin(A+B))

(cosAcosBsinAsinB)+i(sinAcosB+cosAsinB)=(cos(A+B)+i sin(A+B))

On equating the imaginary parts on both sides of the equation we get the required result

i.e.sin(A+B)=sinAcosB+cosAsinB


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