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
Verified623.4k+ views
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 ${e^{ - i\gamma }} = \cos \left( { - \gamma } \right) + i\sin \left( { - \gamma } \right)$.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Master Class 11 Business Studies: Engaging Questions & Answers for Success
Master Class 11 Computer Science: Engaging Questions & Answers for Success
Master Class 11 Economics: Engaging Questions & Answers for Success
Master Class 11 Social Science: Engaging Questions & Answers for Success
Master Class 11 English: Engaging Questions & Answers for Success
Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Explain zero factorial class 11 maths CBSE
State and prove Bernoullis theorem class 11 physics CBSE
What steps did the French revolutionaries take to create class 11 social science CBSE
The transition element that has lowest enthalpy of class 11 chemistry CBSE
Can anyone list 10 advantages and disadvantages of friction