Question

If $a = \cos \alpha + i\sin \alpha$, $b = \cos \beta + i\sin \beta$, $c = \cos$ $c = \cos \gamma + i\sin \gamma$and $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} = 1$. Then find $\cos (\alpha - \beta ) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right)$ :
A) $\dfrac{3}{2}$
B) $0$
C) $\dfrac{{ - 3}}{2}$
D) $1$

Answer 1

We have to find $\dfrac{a}{b}$, $\dfrac{b}{c}$and $\dfrac{c}{a}$then, the values we get are to be put in the equation given above in question. Then, we have to simplify and solve that equation using appropriate trigonometric formulas and after that we will reach the answer by solving the equation step by step.

Complete step by step answer:
According to the question,
Given: $a = \cos \alpha + i\sin \alpha$-----(1)
$b = \cos \beta + i\sin \beta$--------(2)
$c = \cos \gamma + i\sin \gamma$---------(3)
Now, we will find the value of $\dfrac{b}{c}$ , by putting in the value of b and c from equation (2) and (3)
$\dfrac{b}{c} = \dfrac{{\cos \beta + i\sin \beta }}{{\cos \gamma + i\sin \gamma }}$
Now we will rationalize this equation to make some trigonometric formula which will further help to reach the next step,
$\dfrac{b}{c} = \dfrac{{\cos \beta + i\sin \beta }}{{\cos \gamma + i\sin \gamma }} \times \dfrac{{\cos \gamma - i\sin \gamma }}{{\cos \gamma - i\sin \gamma }}$
$\dfrac{b}{c} = \dfrac{{\cos \beta \cos \gamma - {i^2}\sin \beta \sin \gamma + i\left( {\cos \beta \sin \gamma + \sin \beta \cos \gamma } \right)}}{{{{\cos }^2}\gamma - {i^2}{{\sin }^2}\gamma }}$
The value of $i = \sqrt { - 1}$so, ${i^2} = - 1$now we will replace ${i^2}$ by $- 1$.
$\dfrac{b}{c} = \dfrac{{\cos \beta \cos \gamma + \sin \beta \sin \gamma + i(\cos \beta \sin \gamma + \sin \beta \cos \gamma )}}{{{{\cos }^2}\gamma + {{\sin }^2}\gamma }}$
(now as we know ${\cos ^2}a + {\sin ^2}a = 1$ so, we will replace ${\cos ^2}\gamma + {\sin ^2}\gamma$ from $1$)
$\dfrac{b}{c} = \cos \beta \cos \gamma + \sin \beta \sin \gamma + i(\cos \beta \sin \gamma + \sin \beta \cos \gamma )$
(now as we know $\cos a\cos b - \sin a\sin b = \cos \left( {a - b} \right)$ and $\cos a\sin b + \sin a\cos b = \sin \left( {a - b} \right)$ so we will apply this in our above equation) and we get,
$\dfrac{b}{c} = \cos \left( {\beta - \gamma } \right) + i\sin \left( {\beta - \gamma } \right)$------(4)
Similarly, we will calculate the values of $\dfrac{a}{b}$ and $\dfrac{c}{a}$, we will get
$\dfrac{a}{b} = \cos \left( {\alpha - \beta } \right) + i\sin \left( {\alpha - \beta } \right)$-----(5) and
$\dfrac{c}{a} = \cos \left( {\gamma - \alpha } \right) + i\sin \left( {\gamma - \alpha } \right)$-----(6)
Now we will put the values from equation (4), (5) and (6) in this equation
$\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} = 1$
And we get,
$\cos \left( {\gamma - \alpha } \right) + i\sin \left( {\gamma - \alpha } \right) + \cos \left( {\beta - \gamma } \right) + i\sin \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + i\sin \left( {\gamma - \alpha } \right) = 1$
$\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) + i[\sin (\beta - \gamma ) + \sin \left( {\gamma - \alpha } \right) + \sin \left( {\alpha - \beta } \right)] = 1$
After equating real parts we get,
$\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = 1$
Therefore, option (D) is correct.

Note:
In order to solve such complex trigonometric problems with difference in angles in the given terms, we must identify the trigonometric entity connecting both the angles. In order to solve such problems all the trigonometric identities must be remembered to find the solution easily and we should manipulate different trigonometric identities to find the answer.