If \[\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma \], then prove that \[{\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = \dfrac{3}{2} = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2}\].
Answer
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Hint: Here, we will proceed by using the general representation for any complex number given by $z = {e^{i\theta }} = \cos \theta + i\sin \theta $ in order to find the values of $\left( {{e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }}} \right)$ and \[\left( {{e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }}} \right)\]. Then, apply the formula ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$ where $a = {e^{i\alpha }},b = {e^{i\beta }},c = {e^{i\gamma }}$.
Complete step-by-step answer:
Given, \[\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma {\text{ }} \to {\text{(1)}}\]
To prove: \[{\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = \dfrac{3}{2} = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2}\]
Since, any complex number can be represented as $z = {e^{i\theta }} = \cos \theta + i\sin \theta {\text{ }} \to {\text{(2)}}$
Using the formula given by equation (2), we can write
$
{e^{i\alpha }} = \cos \alpha + i\sin \alpha {\text{ }} \to {\text{(3)}} \\
{e^{i\beta }} = \cos \beta + i\sin \beta {\text{ }} \to {\text{(4)}} \\
{e^{i\gamma }} = \cos \gamma + i\sin \gamma {\text{ }} \to {\text{(5)}} \\
$
By adding equations (3), (4) and (5), we get
$
\Rightarrow {e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }} = \cos \alpha + i\sin \alpha + \cos \beta + i\sin \beta + \cos \gamma + i\sin \gamma \\
\Rightarrow {e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }} = \cos \alpha + \cos \beta + \cos \gamma + i\left( {\sin \alpha + \sin \beta + \sin \gamma } \right) \\
$
Using equation (1) in the above equation, we get
$
\Rightarrow {e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }} = 0 + i\left( 0 \right) \\
\Rightarrow {e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }} = 0{\text{ }} \to {\text{(6)}} \\
$
Similarly, using the formula given by equation (2), we can write
$
{e^{ - i\alpha }} = \cos \left( { - \alpha } \right) + i\sin \left( { - \alpha } \right){\text{ }} \to {\text{(7)}} \\
{e^{ - i\beta }} = \cos \left( { - \beta } \right) + i\sin \left( { - \beta } \right){\text{ }} \to {\text{(8)}} \\
{e^{ - i\gamma }} = \cos \left( { - \gamma } \right) + i\sin \left( { - \gamma } \right){\text{ }} \to {\text{(9)}} \\
$
By adding equations (7), (8) and (9), we get
\[ \Rightarrow {e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }} = \cos \left( { - \alpha } \right) + i\sin \left( { - \alpha } \right) + \cos \left( { - \beta } \right) + i\sin \left( { - \beta } \right) + \cos \left( { - \gamma } \right) + i\sin \left( { - \gamma } \right)\]
Using the formulas $\cos \left( { - \theta } \right) = \cos \theta $ and $\sin \left( { - \theta } \right) = - \sin \theta $ in the above equation, we get
\[
\Rightarrow {e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }} = \cos \alpha - i\sin \alpha + \cos \beta - i\sin \beta + \cos \gamma - i\sin \gamma \\
\Rightarrow {e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }} = \cos \alpha + \cos \beta + \cos \gamma - i\left( {\sin \alpha + \sin \beta + \sin \gamma } \right) \\
\]
Using equation (1) in the above equation, we get
$
\Rightarrow {e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }} = 0 - i\left( 0 \right) \\
\Rightarrow {e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }} = 0 \\
\Rightarrow \dfrac{1}{{{e^{i\alpha }}}} + \dfrac{1}{{{e^{i\beta }}}} + \dfrac{1}{{{e^{i\gamma }}}} = 0 \\
\Rightarrow \dfrac{{{e^{i\beta }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\beta }}}}{{{e^{i\alpha }}{e^{i\beta }}{e^{i\gamma }}}} = 0 \\
\Rightarrow {e^{i\beta }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\beta }} = 0 \\
\Rightarrow {e^{i\alpha }}{e^{i\beta }} + {e^{i\beta }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\gamma }} = 0{\text{ }} \to {\text{(10)}} \\
$
Using the formula ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$ and taking $a = {e^{i\alpha }},b = {e^{i\beta }},c = {e^{i\gamma }}$, we get
\[
\Rightarrow {\left( {{e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }}} \right)^2} = {\left( {{e^{i\alpha }}} \right)^2} + {\left( {{e^{i\beta }}} \right)^2} + {\left( {{e^{i\gamma }}} \right)^2} + 2\left( {{e^{i\alpha }}{e^{i\beta }} + {e^{i\beta }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\gamma }}} \right) \\
\Rightarrow {\left( {{e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }}} \right)^2} = {e^{2i\alpha }} + {e^{2i\beta }} + {e^{2i\gamma }} + 2\left( {{e^{i\alpha }}{e^{i\beta }} + {e^{i\beta }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\gamma }}} \right) \\
\]
By substituting equation (6) and equation (10) in the above equation, we get
$
\Rightarrow {\left( 0 \right)^2} = {e^{2i\alpha }} + {e^{2i\beta }} + {e^{2i\gamma }} + 2\left( 0 \right) \\
\Rightarrow 0 = {e^{2i\alpha }} + {e^{2i\beta }} + {e^{2i\gamma }} + 0 \\
\Rightarrow {e^{2i\alpha }} + {e^{2i\beta }} + {e^{2i\gamma }} = 0 \\
\Rightarrow {e^{i\left( {2\alpha } \right)}} + {e^{i\left( {2\beta } \right)}} + {e^{i\left( {2\gamma } \right)}} = 0 \\
$
Using the formula given by equation (2) in the above equation, we have
$
\Rightarrow \cos \left( {2\alpha } \right) + i\sin \left( {2\alpha } \right) + \cos \left( {2\beta } \right) + i\sin \left( {2\beta } \right) + \cos \left( {2\gamma } \right) + i\sin \left( {2\gamma } \right) = 0 \\
\Rightarrow \cos \left( {2\alpha } \right) + \cos \left( {2\beta } \right) + \cos \left( {2\gamma } \right) + i\left[ {\sin \left( {2\alpha } \right) + \sin \left( {2\beta } \right) + \sin \left( {2\gamma } \right)} \right] = 0{\text{ }} \to {\text{(11)}} \\
$
For any complex number z, $a + ib = 0{\text{ }} \to {\text{(12)}}$, a = 0 and b = 0
By comparing equations (11) and (12), we have
$
a = \cos \left( {2\alpha } \right) + \cos \left( {2\beta } \right) + \cos \left( {2\gamma } \right) = 0 \\
\Rightarrow \cos \left( {2\alpha } \right) + \cos \left( {2\beta } \right) + \cos \left( {2\gamma } \right) = 0{\text{ }} \to {\text{(13)}} \\
$ and $b = \sin \left( {2\alpha } \right) + \sin \left( {2\beta } \right) + \sin \left( {2\gamma } \right) = 0$
Using the trigonometric formula $\cos \left( {2\theta } \right) = {\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2}$ in equation (13), we get
$
\Rightarrow \left[ {{{\left( {\cos \alpha } \right)}^2} - {{\left( {\sin \alpha } \right)}^2}} \right] + \left[ {{{\left( {\cos \beta } \right)}^2} - {{\left( {\sin \beta } \right)}^2}} \right] + \left[ {{{\left( {\cos \gamma } \right)}^2} - {{\left( {\sin \gamma } \right)}^2}} \right] = 0 \\
\Rightarrow {\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2}{\text{ }} \to {\text{(14)}} \\
$
Using the trigonometric identity \[
{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1 \\
\Rightarrow {\left( {\sin \theta } \right)^2} = 1 - {\left( {\cos \theta } \right)^2} \\
\] in equation (14), we get
\[
\Rightarrow {\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = 1 - {\left( {\cos \alpha } \right)^2} + 1 - {\left( {\cos \beta } \right)^2} + 1 - {\left( {\cos \gamma } \right)^2} \\
\Rightarrow {\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} + {\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = 1 + 1 + 1 \\
\Rightarrow 2\left[ {{{\left( {\cos \alpha } \right)}^2} + {{\left( {\cos \beta } \right)}^2} + {{\left( {\cos \gamma } \right)}^2}} \right] = 3 \\
\Rightarrow {\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = \dfrac{3}{2}{\text{ }} \to {\text{(15)}} \\
\]
Using the trigonometric identity \[
{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1 \\
\Rightarrow {\left( {\cos \theta } \right)^2} = 1 - {\left( {\sin \theta } \right)^2} \\
\] in equation (14), we get
\[
\Rightarrow 1 - {\left( {\sin \alpha } \right)^2} + 1 - {\left( {\sin \beta } \right)^2} + 1 - {\left( {\sin \gamma } \right)^2} = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2} \\
\Rightarrow 1 + 1 + 1 = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2} + {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2} \\
\Rightarrow 3 = 2\left[ {{{\left( {\sin \alpha } \right)}^2} + {{\left( {\sin \beta } \right)}^2} + {{\left( {\sin \gamma } \right)}^2}} \right] \\
\Rightarrow {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2} = \dfrac{3}{2}{\text{ }} \to {\text{(16)}} \\
\]
By combining equations (15) and (16), we have
\[{\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = \dfrac{3}{2} = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2}\]
The above equation is the same equation which we needed to prove.
Note: In this particular problem, we have written ${e^{ - i\alpha }} = \cos \left( { - \alpha } \right) + i\sin \left( { - \alpha } \right)$ which is obtained by replacing the angle $\theta $ in the formula ${e^{i\theta }} = \cos \theta + i\sin \theta $ by $ - \alpha $. Similarly, $\theta $ in the formula ${e^{i\theta }} = \cos \theta + i\sin \theta $ is replaced by $ - \beta $ to get ${e^{ - i\beta }} = \cos \left( { - \beta } \right) + i\sin \left( { - \beta } \right)$. Also, $\theta $ in the formula ${e^{i\theta }} = \cos \theta + i\sin \theta $ is replaced by $ - \gamma $ to get ${e^{ - i\gamma }} = \cos \left( { - \gamma } \right) + i\sin \left( { - \gamma } \right)$.
Complete step-by-step answer:
Given, \[\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma {\text{ }} \to {\text{(1)}}\]
To prove: \[{\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = \dfrac{3}{2} = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2}\]
Since, any complex number can be represented as $z = {e^{i\theta }} = \cos \theta + i\sin \theta {\text{ }} \to {\text{(2)}}$
Using the formula given by equation (2), we can write
$
{e^{i\alpha }} = \cos \alpha + i\sin \alpha {\text{ }} \to {\text{(3)}} \\
{e^{i\beta }} = \cos \beta + i\sin \beta {\text{ }} \to {\text{(4)}} \\
{e^{i\gamma }} = \cos \gamma + i\sin \gamma {\text{ }} \to {\text{(5)}} \\
$
By adding equations (3), (4) and (5), we get
$
\Rightarrow {e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }} = \cos \alpha + i\sin \alpha + \cos \beta + i\sin \beta + \cos \gamma + i\sin \gamma \\
\Rightarrow {e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }} = \cos \alpha + \cos \beta + \cos \gamma + i\left( {\sin \alpha + \sin \beta + \sin \gamma } \right) \\
$
Using equation (1) in the above equation, we get
$
\Rightarrow {e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }} = 0 + i\left( 0 \right) \\
\Rightarrow {e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }} = 0{\text{ }} \to {\text{(6)}} \\
$
Similarly, using the formula given by equation (2), we can write
$
{e^{ - i\alpha }} = \cos \left( { - \alpha } \right) + i\sin \left( { - \alpha } \right){\text{ }} \to {\text{(7)}} \\
{e^{ - i\beta }} = \cos \left( { - \beta } \right) + i\sin \left( { - \beta } \right){\text{ }} \to {\text{(8)}} \\
{e^{ - i\gamma }} = \cos \left( { - \gamma } \right) + i\sin \left( { - \gamma } \right){\text{ }} \to {\text{(9)}} \\
$
By adding equations (7), (8) and (9), we get
\[ \Rightarrow {e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }} = \cos \left( { - \alpha } \right) + i\sin \left( { - \alpha } \right) + \cos \left( { - \beta } \right) + i\sin \left( { - \beta } \right) + \cos \left( { - \gamma } \right) + i\sin \left( { - \gamma } \right)\]
Using the formulas $\cos \left( { - \theta } \right) = \cos \theta $ and $\sin \left( { - \theta } \right) = - \sin \theta $ in the above equation, we get
\[
\Rightarrow {e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }} = \cos \alpha - i\sin \alpha + \cos \beta - i\sin \beta + \cos \gamma - i\sin \gamma \\
\Rightarrow {e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }} = \cos \alpha + \cos \beta + \cos \gamma - i\left( {\sin \alpha + \sin \beta + \sin \gamma } \right) \\
\]
Using equation (1) in the above equation, we get
$
\Rightarrow {e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }} = 0 - i\left( 0 \right) \\
\Rightarrow {e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }} = 0 \\
\Rightarrow \dfrac{1}{{{e^{i\alpha }}}} + \dfrac{1}{{{e^{i\beta }}}} + \dfrac{1}{{{e^{i\gamma }}}} = 0 \\
\Rightarrow \dfrac{{{e^{i\beta }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\beta }}}}{{{e^{i\alpha }}{e^{i\beta }}{e^{i\gamma }}}} = 0 \\
\Rightarrow {e^{i\beta }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\beta }} = 0 \\
\Rightarrow {e^{i\alpha }}{e^{i\beta }} + {e^{i\beta }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\gamma }} = 0{\text{ }} \to {\text{(10)}} \\
$
Using the formula ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$ and taking $a = {e^{i\alpha }},b = {e^{i\beta }},c = {e^{i\gamma }}$, we get
\[
\Rightarrow {\left( {{e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }}} \right)^2} = {\left( {{e^{i\alpha }}} \right)^2} + {\left( {{e^{i\beta }}} \right)^2} + {\left( {{e^{i\gamma }}} \right)^2} + 2\left( {{e^{i\alpha }}{e^{i\beta }} + {e^{i\beta }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\gamma }}} \right) \\
\Rightarrow {\left( {{e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }}} \right)^2} = {e^{2i\alpha }} + {e^{2i\beta }} + {e^{2i\gamma }} + 2\left( {{e^{i\alpha }}{e^{i\beta }} + {e^{i\beta }}{e^{i\gamma }} + {e^{i\alpha }}{e^{i\gamma }}} \right) \\
\]
By substituting equation (6) and equation (10) in the above equation, we get
$
\Rightarrow {\left( 0 \right)^2} = {e^{2i\alpha }} + {e^{2i\beta }} + {e^{2i\gamma }} + 2\left( 0 \right) \\
\Rightarrow 0 = {e^{2i\alpha }} + {e^{2i\beta }} + {e^{2i\gamma }} + 0 \\
\Rightarrow {e^{2i\alpha }} + {e^{2i\beta }} + {e^{2i\gamma }} = 0 \\
\Rightarrow {e^{i\left( {2\alpha } \right)}} + {e^{i\left( {2\beta } \right)}} + {e^{i\left( {2\gamma } \right)}} = 0 \\
$
Using the formula given by equation (2) in the above equation, we have
$
\Rightarrow \cos \left( {2\alpha } \right) + i\sin \left( {2\alpha } \right) + \cos \left( {2\beta } \right) + i\sin \left( {2\beta } \right) + \cos \left( {2\gamma } \right) + i\sin \left( {2\gamma } \right) = 0 \\
\Rightarrow \cos \left( {2\alpha } \right) + \cos \left( {2\beta } \right) + \cos \left( {2\gamma } \right) + i\left[ {\sin \left( {2\alpha } \right) + \sin \left( {2\beta } \right) + \sin \left( {2\gamma } \right)} \right] = 0{\text{ }} \to {\text{(11)}} \\
$
For any complex number z, $a + ib = 0{\text{ }} \to {\text{(12)}}$, a = 0 and b = 0
By comparing equations (11) and (12), we have
$
a = \cos \left( {2\alpha } \right) + \cos \left( {2\beta } \right) + \cos \left( {2\gamma } \right) = 0 \\
\Rightarrow \cos \left( {2\alpha } \right) + \cos \left( {2\beta } \right) + \cos \left( {2\gamma } \right) = 0{\text{ }} \to {\text{(13)}} \\
$ and $b = \sin \left( {2\alpha } \right) + \sin \left( {2\beta } \right) + \sin \left( {2\gamma } \right) = 0$
Using the trigonometric formula $\cos \left( {2\theta } \right) = {\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2}$ in equation (13), we get
$
\Rightarrow \left[ {{{\left( {\cos \alpha } \right)}^2} - {{\left( {\sin \alpha } \right)}^2}} \right] + \left[ {{{\left( {\cos \beta } \right)}^2} - {{\left( {\sin \beta } \right)}^2}} \right] + \left[ {{{\left( {\cos \gamma } \right)}^2} - {{\left( {\sin \gamma } \right)}^2}} \right] = 0 \\
\Rightarrow {\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2}{\text{ }} \to {\text{(14)}} \\
$
Using the trigonometric identity \[
{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1 \\
\Rightarrow {\left( {\sin \theta } \right)^2} = 1 - {\left( {\cos \theta } \right)^2} \\
\] in equation (14), we get
\[
\Rightarrow {\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = 1 - {\left( {\cos \alpha } \right)^2} + 1 - {\left( {\cos \beta } \right)^2} + 1 - {\left( {\cos \gamma } \right)^2} \\
\Rightarrow {\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} + {\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = 1 + 1 + 1 \\
\Rightarrow 2\left[ {{{\left( {\cos \alpha } \right)}^2} + {{\left( {\cos \beta } \right)}^2} + {{\left( {\cos \gamma } \right)}^2}} \right] = 3 \\
\Rightarrow {\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = \dfrac{3}{2}{\text{ }} \to {\text{(15)}} \\
\]
Using the trigonometric identity \[
{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1 \\
\Rightarrow {\left( {\cos \theta } \right)^2} = 1 - {\left( {\sin \theta } \right)^2} \\
\] in equation (14), we get
\[
\Rightarrow 1 - {\left( {\sin \alpha } \right)^2} + 1 - {\left( {\sin \beta } \right)^2} + 1 - {\left( {\sin \gamma } \right)^2} = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2} \\
\Rightarrow 1 + 1 + 1 = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2} + {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2} \\
\Rightarrow 3 = 2\left[ {{{\left( {\sin \alpha } \right)}^2} + {{\left( {\sin \beta } \right)}^2} + {{\left( {\sin \gamma } \right)}^2}} \right] \\
\Rightarrow {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2} = \dfrac{3}{2}{\text{ }} \to {\text{(16)}} \\
\]
By combining equations (15) and (16), we have
\[{\left( {\cos \alpha } \right)^2} + {\left( {\cos \beta } \right)^2} + {\left( {\cos \gamma } \right)^2} = \dfrac{3}{2} = {\left( {\sin \alpha } \right)^2} + {\left( {\sin \beta } \right)^2} + {\left( {\sin \gamma } \right)^2}\]
The above equation is the same equation which we needed to prove.
Note: In this particular problem, we have written ${e^{ - i\alpha }} = \cos \left( { - \alpha } \right) + i\sin \left( { - \alpha } \right)$ which is obtained by replacing the angle $\theta $ in the formula ${e^{i\theta }} = \cos \theta + i\sin \theta $ by $ - \alpha $. Similarly, $\theta $ in the formula ${e^{i\theta }} = \cos \theta + i\sin \theta $ is replaced by $ - \beta $ to get ${e^{ - i\beta }} = \cos \left( { - \beta } \right) + i\sin \left( { - \beta } \right)$. Also, $\theta $ in the formula ${e^{i\theta }} = \cos \theta + i\sin \theta $ is replaced by $ - \gamma $ to get ${e^{ - i\gamma }} = \cos \left( { - \gamma } \right) + i\sin \left( { - \gamma } \right)$.
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