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Question: If \[\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma \], then pr...

If cos⁡α+cos⁡β+cos⁡γ=0=sin⁡α+sin⁡β+sin⁡γ\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma , then prove that (cos⁡α)2+(cos⁡β)2+(cos⁡γ)2=32=(sin⁡α)2+(sin⁡β)2+(sin⁡γ)2{\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}.

Explanation

Solution

Hint: Here, we will proceed by using the general representation for any complex number given by z=eiθ=cos⁡θ+isin⁡θz = {e^{i\theta }} = \cos \theta + i\sin \theta in order to find the values of (eiα+eiβ+eiγ)\left( {{e^{i\alpha }} + {e^{i\beta }} + {e^{i\gamma }}} \right) and (e−iα+e−iβ+e−iγ)\left( {{e^{ - i\alpha }} + {e^{ - i\beta }} + {e^{ - i\gamma }}} \right). Then, apply the formula (a+b+c)2=a2+b2+c2+2(ab+bc+ac){\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right) where a=eiα,b=eiβ,c=eiγa = {e^{i\alpha }},b = {e^{i\beta }},c = {e^{i\gamma }}.

Complete step-by-step answer:
Given, cos⁡α+cos⁡β+cos⁡γ=0=sin⁡α+sin⁡β+sin⁡γ →(1)\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma {\text{ }} \to {\text{(1)}}
To prove: (cos⁡α)2+(cos⁡β)2+(cos⁡γ)2=32=(sin⁡α)2+(sin⁡β)2+(sin⁡γ)2{\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=eiθ=cos⁡θ+isin⁡θ →(2)z = {e^{i\theta }} = \cos \theta + i\sin \theta {\text{ }} \to {\text{(2)}}
Using the formula given by equation (2), we can write
eiα=cos⁡α+isin⁡α →(3) eiβ=cos⁡β+isin⁡β →(4) eiγ=cos⁡γ+isin⁡γ →(5)  {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
⇒eiα+eiβ+eiγ=cos⁡α+isin⁡α+cos⁡β+isin⁡β+cos⁡γ+isin⁡γ ⇒eiα+eiβ+eiγ=cos⁡α+cos⁡β+cos⁡γ+i(sin⁡α+sin⁡β+sin⁡γ)  \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
⇒eiα+eiβ+eiγ=0+i(0) ⇒eiα+eiβ+eiγ=0 →(6)  \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−iα=cos⁡(−α)+isin⁡(−α) →(7) e−iβ=cos⁡(−β)+isin⁡(−β) →(8) e−iγ=cos⁡(−γ)+isin⁡(−γ) →(9)  {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
⇒e−iα+e−iβ+e−iγ=cos⁡(−α)+isin⁡(−α)+cos⁡(−β)+isin⁡(−β)+cos⁡(−γ)+isin⁡(−γ)\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⁡(−θ)=cos⁡θ\cos \left( { - \theta } \right) = \cos \theta and sin⁡(−θ)=−sin⁡θ\sin \left( { - \theta } \right) = - \sin \theta in the above equation, we get

⇒e−iα+e−iβ+e−iγ=cos⁡α−isin⁡α+cos⁡β−isin⁡β+cos⁡γ−isin⁡γ ⇒e−iα+e−iβ+e−iγ=cos⁡α+cos⁡β+cos⁡γ−i(sin⁡α+sin⁡β+sin⁡γ)  \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
⇒e−iα+e−iβ+e−iγ=0−i(0) ⇒e−iα+e−iβ+e−iγ=0 ⇒1eiα+1eiβ+1eiγ=0 ⇒eiβeiγ+eiαeiγ+eiαeiβeiαeiβeiγ=0 ⇒eiβeiγ+eiαeiγ+eiαeiβ=0 ⇒eiαeiβ+eiβeiγ+eiαeiγ=0 →(10)  \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 (a+b+c)2=a2+b2+c2+2(ab+bc+ac){\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right) and taking a=eiα,b=eiβ,c=eiγa = {e^{i\alpha }},b = {e^{i\beta }},c = {e^{i\gamma }}, we get

⇒(eiα+eiβ+eiγ)2=(eiα)2+(eiβ)2+(eiγ)2+2(eiαeiβ+eiβeiγ+eiαeiγ) ⇒(eiα+eiβ+eiγ)2=e2iα+e2iβ+e2iγ+2(eiαeiβ+eiβeiγ+eiαeiγ)  \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
⇒(0)2=e2iα+e2iβ+e2iγ+2(0) ⇒0=e2iα+e2iβ+e2iγ+0 ⇒e2iα+e2iβ+e2iγ=0 ⇒ei(2α)+ei(2β)+ei(2γ)=0  \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
⇒cos⁡(2α)+isin⁡(2α)+cos⁡(2β)+isin⁡(2β)+cos⁡(2γ)+isin⁡(2γ)=0 ⇒cos⁡(2α)+cos⁡(2β)+cos⁡(2γ)+i[sin⁡(2α)+sin⁡(2β)+sin⁡(2γ)]=0 →(11)  \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 →(12)a + ib = 0{\text{ }} \to {\text{(12)}}, a = 0 and b = 0
By comparing equations (11) and (12), we have
a=cos⁡(2α)+cos⁡(2β)+cos⁡(2γ)=0 ⇒cos⁡(2α)+cos⁡(2β)+cos⁡(2γ)=0 →(13)  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⁡(2α)+sin⁡(2β)+sin⁡(2γ)=0b = \sin \left( {2\alpha } \right) + \sin \left( {2\beta } \right) + \sin \left( {2\gamma } \right) = 0
Using the trigonometric formula cos⁡(2θ)=(cos⁡θ)2−(sin⁡θ)2\cos \left( {2\theta } \right) = {\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2} in equation (13), we get
⇒[(cos⁡α)2−(sin⁡α)2]+[(cos⁡β)2−(sin⁡β)2]+[(cos⁡γ)2−(sin⁡γ)2]=0 ⇒(cos⁡α)2+(cos⁡β)2+(cos⁡γ)2=(sin⁡α)2+(sin⁡β)2+(sin⁡γ)2 →(14)  \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)}} \\

UsingthetrigonometricidentityUsing 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} \\

\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 ${e^{ - i\gamma }} = \cos \left( { - \gamma } \right) + i\sin \left( { - \gamma } \right)$.