Question

If \[a = \cos \alpha + i\sin \alpha \] , \[b = \cos \beta + i\sin \beta \] , \[c = \cos \gamma + i\sin \gamma \] and \[\dfrac{b}{c} + \dfrac{c}{a} + \dfrac{a}{b} = 1\] then \[\cos (\beta - \gamma ) + \cos \left( {\gamma - \alpha } \right) + \cos (\alpha - \beta )\] is equal to
A.\[0\]
B.\[1\]
C.\[ - 1\]
D.None of these

Answer 1

Hint: A complex number is a number that can be expressed in the form \[x + iy\] where \[x\] and \[y\] are real numbers and \[i\] s a symbol called the imaginary unit , and satisfying the equation \[{i^2} = - 1\] . Because no "real" number satisfies this equation \[i\] was called an imaginary number . for a complex number \[x + iy\] , \[x\] is called the real part and \[y\] 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 :
\[\cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]
\[\sin (\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \]
\[\sin ( - \theta ) = - \sin \theta \]
We are given that \[a = \cos \alpha + i\sin \alpha \] , \[b = \cos \beta + i\sin \beta \] , \[c = \cos \gamma + i\sin \gamma \]
Now , \[\dfrac{b}{c} = \dfrac{{\cos \beta + i\sin \beta }}{{\cos \gamma + i\sin \gamma }} \times \dfrac{{\cos \gamma - i\sin \gamma }}{{\cos \gamma - i\sin \gamma }}\] ( by rationalizing)
Which simplifies to
\[ = \dfrac{{\cos \alpha \cos \gamma - i\cos \beta \sin \gamma + i\sin \beta \cos \gamma - {i^2}\sin \beta \sin \gamma }}{{{{\cos }^2}\gamma + {{\sin }^2}\gamma }}\]
Which further simplifies to
\[ = \cos \beta \cos \gamma + \sin \beta \sin \gamma - i\left( {\cos \beta \sin \gamma - \sin \beta \cos \gamma } \right)\]
On applying identities we get
\[ = \cos (\beta - \gamma ) - i\sin (\gamma - \beta )\]
Since we know that \[\sin ( - \theta ) = - \sin \theta \]
Therefore we get
\[\dfrac{b}{c} = \cos (\beta - \gamma ) + i\sin (\beta - \gamma )\]
Now \[\dfrac{c}{a} = \dfrac{{\cos \gamma + i\sin \gamma }}{{\cos \alpha + i\sin \alpha }} \times \dfrac{{\cos \alpha - i\sin \alpha }}{{\cos \alpha - i\sin \alpha }}\]( by rationalizing)
Which simplifies to
\[ = \dfrac{{\cos \gamma \cos \alpha - i\cos \gamma \sin \alpha + i\sin \gamma \cos \alpha - {i^2}\sin \alpha \sin \gamma }}{{{{\cos }^2}\alpha + {{\sin }^2}\alpha }}\]
Which further simplifies to
\[\cos \gamma \cos \alpha + \sin \alpha \sin \gamma - i\left( {\cos \gamma \sin \alpha - \sin \gamma \cos \alpha } \right)\]
On applying identities we get
\[ = \cos \left( {\gamma - \alpha } \right) - i\sin (\alpha - \gamma )\]
Since we know that \[\sin ( - \theta ) = - \sin \theta \]
Therefore we get
\[\dfrac{c}{a} = \cos \left( {\gamma - \alpha } \right) + i\sin (\gamma - \alpha )\]
Now \[\dfrac{a}{b} = \dfrac{{\cos \alpha + i\sin \alpha }}{{\cos \beta + i\sin \beta }} \times \dfrac{{\cos \beta - i\sin \beta }}{{\cos \beta - i\sin \beta }}\]( by rationalizing)
Which simplifies to
\[ = \dfrac{{\cos \alpha \cos \beta - i\sin \beta \cos \alpha + i\sin \alpha \cos \beta - {i^2}\sin \alpha \sin \beta }}{{{{\cos }^2}\beta + {{\sin }^2}\beta }}\]
Which further simplifies to
\[ = \cos \alpha \cos \beta + \sin \alpha \sin \beta - i\left( {\sin \beta \cos \alpha - \sin \alpha \cos \beta } \right)\]
On applying identities we get
\[ = \cos \left( {\alpha - \beta } \right) - i\sin \left( {\beta - \alpha } \right)\]
Since we know that \[\sin ( - \theta ) = - \sin \theta \]
Therefore we get
\[\dfrac{a}{b} = \cos \left( {\alpha - \beta } \right) + i\sin \left( {\alpha - \beta } \right)\]
Now adding all the three equations we get
\[\dfrac{b}{c} + \dfrac{c}{a} + \dfrac{a}{b} = \cos (\beta - \alpha ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) + i\left[ {\sin (\beta - \gamma ) + \sin (\gamma - \alpha ) + \sin (\alpha - \beta )} \right]\]
Also we are given that \[\dfrac{b}{c} + \dfrac{c}{a} + \dfrac{a}{b} = 1\]
Therefore on comparing the equations we get
\[1 = \cos (\beta - \alpha ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) + i\left[ {\sin (\beta - \gamma ) + \sin (\gamma - \alpha ) + \sin (\alpha - \beta )} \right]\]
On comparing real and imaginary parts we get
\[1 = \cos (\beta - \alpha ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta )\] and \[\sin (\beta - \gamma ) + \sin (\gamma - \alpha ) + \sin (\alpha - \beta ) = 0\]
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 \[x + iy\] where \[x\] and \[y\] are real numbers and \[i\] s a symbol called the imaginary unit. Rationalizing the fraction plays a very important role in simplifying the solution.