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Question: If \[a = \cos \alpha + i\sin \alpha \] , \[b = \cos \beta + i\sin \beta \] , \[c = \cos \gamma + i\s...

If a=cos⁡α+isin⁡αa = \cos \alpha + i\sin \alpha , b=cos⁡β+isin⁡βb = \cos \beta + i\sin \beta , c=cos⁡γ+isin⁡γc = \cos \gamma + i\sin \gamma and bc+ca+ab=1\dfrac{b}{c} + \dfrac{c}{a} + \dfrac{a}{b} = 1 then cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)\cos (\beta - \gamma ) + \cos \left( {\gamma - \alpha } \right) + \cos (\alpha - \beta ) is equal to
A.00
B.11
C.−1 - 1
D.None of these

Explanation

Solution

Hint : A complex number is a number that can be expressed in the form x+iyx + iy where xx and yy are real numbers and ii s a symbol called the imaginary unit , and satisfying the equation i2=−1{i^2} = - 1 . Because no "real" number satisfies this equation ii was called an imaginary number . for a complex number x+iyx + iy , xx is called the real part and yy is called the imaginary part.

Complete step-by-step answer :
Complex numbers have a similar definition of equality to real numbers; two complex numbers are equal if and only if both their real and imaginary parts are equal. Nonzero complex numbers written in polar form are equal if and only if they have the same magnitude and their arguments differ by an integer multiple of 2π. complex numbers are naturally thought of as existing on a two-dimensional plane.
Formulas used in the solution part :
cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta
sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin (\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta
sin⁡(−θ)=−sin⁡θ\sin ( - \theta ) = - \sin \theta
We are given that a=cos⁡α+isin⁡αa = \cos \alpha + i\sin \alpha , b=cos⁡β+isin⁡βb = \cos \beta + i\sin \beta , c=cos⁡γ+isin⁡γc = \cos \gamma + i\sin \gamma
Now , bc=cos⁡β+isin⁡βcos⁡γ+isin⁡γ×cos⁡γ−isin⁡γcos⁡γ−isin⁡γ\dfrac{b}{c} = \dfrac{{\cos \beta + i\sin \beta }}{{\cos \gamma + i\sin \gamma }} \times \dfrac{{\cos \gamma - i\sin \gamma }}{{\cos \gamma - i\sin \gamma }} ( by rationalizing)
Which simplifies to
=cos⁡αcos⁡γ−icos⁡βsin⁡γ+isin⁡βcos⁡γ−i2sin⁡βsin⁡γcos⁡2γ+sin⁡2γ= \dfrac{{\cos \alpha \cos \gamma - i\cos \beta \sin \gamma + i\sin \beta \cos \gamma - {i^2}\sin \beta \sin \gamma }}{{{{\cos }^2}\gamma + {{\sin }^2}\gamma }}
Which further simplifies to
=cos⁡βcos⁡γ+sin⁡βsin⁡γ−i(cos⁡βsin⁡γ−sin⁡βcos⁡γ)= \cos \beta \cos \gamma + \sin \beta \sin \gamma - i\left( {\cos \beta \sin \gamma - \sin \beta \cos \gamma } \right)
On applying identities we get
=cos⁡(β−γ)−isin⁡(γ−β)= \cos (\beta - \gamma ) - i\sin (\gamma - \beta )
Since we know that sin⁡(−θ)=−sin⁡θ\sin ( - \theta ) = - \sin \theta
Therefore we get
bc=cos⁡(β−γ)+isin⁡(β−γ)\dfrac{b}{c} = \cos (\beta - \gamma ) + i\sin (\beta - \gamma )
Now ca=cos⁡γ+isin⁡γcos⁡α+isin⁡α×cos⁡α−isin⁡αcos⁡α−isin⁡α\dfrac{c}{a} = \dfrac{{\cos \gamma + i\sin \gamma }}{{\cos \alpha + i\sin \alpha }} \times \dfrac{{\cos \alpha - i\sin \alpha }}{{\cos \alpha - i\sin \alpha }}( by rationalizing)
Which simplifies to
=cos⁡γcos⁡α−icos⁡γsin⁡α+isin⁡γcos⁡α−i2sin⁡αsin⁡γcos⁡2α+sin⁡2α= \dfrac{{\cos \gamma \cos \alpha - i\cos \gamma \sin \alpha + i\sin \gamma \cos \alpha - {i^2}\sin \alpha \sin \gamma }}{{{{\cos }^2}\alpha + {{\sin }^2}\alpha }}
Which further simplifies to
cos⁡γcos⁡α+sin⁡αsin⁡γ−i(cos⁡γsin⁡α−sin⁡γcos⁡α)\cos \gamma \cos \alpha + \sin \alpha \sin \gamma - i\left( {\cos \gamma \sin \alpha - \sin \gamma \cos \alpha } \right)
On applying identities we get
=cos⁡(γ−α)−isin⁡(α−γ)= \cos \left( {\gamma - \alpha } \right) - i\sin (\alpha - \gamma )
Since we know that sin⁡(−θ)=−sin⁡θ\sin ( - \theta ) = - \sin \theta
Therefore we get
ca=cos⁡(γ−α)+isin⁡(γ−α)\dfrac{c}{a} = \cos \left( {\gamma - \alpha } \right) + i\sin (\gamma - \alpha )
Now ab=cos⁡α+isin⁡αcos⁡β+isin⁡β×cos⁡β−isin⁡βcos⁡β−isin⁡β\dfrac{a}{b} = \dfrac{{\cos \alpha + i\sin \alpha }}{{\cos \beta + i\sin \beta }} \times \dfrac{{\cos \beta - i\sin \beta }}{{\cos \beta - i\sin \beta }}( by rationalizing)
Which simplifies to
=cos⁡αcos⁡β−isin⁡βcos⁡α+isin⁡αcos⁡β−i2sin⁡αsin⁡βcos⁡2β+sin⁡2β= \dfrac{{\cos \alpha \cos \beta - i\sin \beta \cos \alpha + i\sin \alpha \cos \beta - {i^2}\sin \alpha \sin \beta }}{{{{\cos }^2}\beta + {{\sin }^2}\beta }}
Which further simplifies to
=cos⁡αcos⁡β+sin⁡αsin⁡β−i(sin⁡βcos⁡α−sin⁡αcos⁡β)= \cos \alpha \cos \beta + \sin \alpha \sin \beta - i\left( {\sin \beta \cos \alpha - \sin \alpha \cos \beta } \right)
On applying identities we get
=cos⁡(α−β)−isin⁡(β−α)= \cos \left( {\alpha - \beta } \right) - i\sin \left( {\beta - \alpha } \right)
Since we know that sin⁡(−θ)=−sin⁡θ\sin ( - \theta ) = - \sin \theta
Therefore we get
ab=cos⁡(α−β)+isin⁡(α−β)\dfrac{a}{b} = \cos \left( {\alpha - \beta } \right) + i\sin \left( {\alpha - \beta } \right)
Now adding all the three equations we get
bc+ca+ab=cos⁡(β−α)+cos⁡(γ−α)+cos⁡(α−β)+i[sin⁡(β−γ)+sin⁡(γ−α)+sin⁡(α−β)]\dfrac{b}{c} + \dfrac{c}{a} + \dfrac{a}{b} = \cos (\beta - \alpha ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) + i\left[ {\sin (\beta - \gamma ) + \sin (\gamma - \alpha ) + \sin (\alpha - \beta )} \right]
Also we are given that bc+ca+ab=1\dfrac{b}{c} + \dfrac{c}{a} + \dfrac{a}{b} = 1
Therefore on comparing the equations we get
1=cos⁡(β−α)+cos⁡(γ−α)+cos⁡(α−β)+i[sin⁡(β−γ)+sin⁡(γ−α)+sin⁡(α−β)]1 = \cos (\beta - \alpha ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) + i\left[ {\sin (\beta - \gamma ) + \sin (\gamma - \alpha ) + \sin (\alpha - \beta )} \right]
On comparing real and imaginary parts we get
1=cos⁡(β−α)+cos⁡(γ−α)+cos⁡(α−β)1 = \cos (\beta - \alpha ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) and sin⁡(β−γ)+sin⁡(γ−α)+sin⁡(α−β)=0\sin (\beta - \gamma ) + \sin (\gamma - \alpha ) + \sin (\alpha - \beta ) = 0
Therefore option (2) is the correct answer.
So, the correct answer is “Option 2”.

Note : A complex number is a number that can be expressed in the form x+iyx + iy where xx and yy are real numbers and ii s a symbol called the imaginary unit. Rationalizing the fraction plays a very important role in simplifying the solution.