Solveeit Logo
Login

Question

Question: If \( \sin \alpha + \sin \beta + \sin \gamma = 0 = \cos \alpha + \cos \beta + \cos \gamma \) , then ...

If sin⁡α+sin⁡β+sin⁡γ=0=cos⁡α+cos⁡β+cos⁡γ\sin \alpha + \sin \beta + \sin \gamma = 0 = \cos \alpha + \cos \beta + \cos \gamma , then show that cos⁡(α−β)+cos⁡(β−γ)+cos⁡(γ−α)=−32\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) = \dfrac{{ - 3}}{2} .

Explanation

Solution

Hint : In this question we need to prove that cos⁡(α−β)+cos⁡(β−γ)+cos⁡(γ−α)=−32\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) = \dfrac{{ - 3}}{2} . Here we will consider sin⁡α+sin⁡β=−sin⁡γ\sin \alpha + \sin \beta = - \sin \gamma and cos⁡α+cos⁡β=−cos⁡γ\cos \alpha + \cos \beta = - \cos \gamma . Then square and add both the equations. Then use algebraic and trigonometric identities to evaluate it. Then similarly consider the equation with alternative terms and evaluate it. Finally we will get three values for each equation. And by adding the three equations we will get the required proof

** Complete step-by-step answer** :
We need to prove that cos⁡(α−β)+cos⁡(β−γ)+cos⁡(γ−α)=−32\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) = \dfrac{{ - 3}}{2} .
It is given that sin⁡α+sin⁡β+sin⁡γ=0=cos⁡α+cos⁡β+cos⁡γ\sin \alpha + \sin \beta + \sin \gamma = 0 = \cos \alpha + \cos \beta + \cos \gamma
Therefore, sin⁡α+sin⁡β+sin⁡γ=0\sin \alpha + \sin \beta + \sin \gamma = 0
Then, sin⁡α+sin⁡β=−sin⁡γ\sin \alpha + \sin \beta = - \sin \gamma →(1)\to \left( 1 \right)
And, cos⁡α+cos⁡β+cos⁡γ=0\cos \alpha + \cos \beta + \cos \gamma = 0
So, cos⁡α+cos⁡β=−cos⁡γ\cos \alpha + \cos \beta = - \cos \gamma →(2)\to \left( 2 \right)
Now, let us square and add the equations (1) and (2), we have,
(sin⁡α+sin⁡β)2+(cos⁡α+cos⁡β)2=sin⁡2γ+cos⁡2γ{\left( {\sin \alpha + \sin \beta } \right)^2} + {\left( {\cos \alpha + \cos \beta } \right)^2} = {\sin ^2}\gamma + {\cos ^2}\gamma
Now, we know the algebraic identity (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}
And also, the trigonometric identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Therefore, applying these identities, we have,
sin⁡2α+sin⁡2β+2sin⁡αsin⁡β+cos⁡2α+cos⁡2β+2cos⁡αcos⁡β=1{\sin ^2}\alpha + {\sin ^2}\beta + 2\sin \alpha \sin \beta + {\cos ^2}\alpha + {\cos ^2}\beta + 2\cos \alpha \cos \beta = 1
(sin⁡2α+cos⁡2α)+(sin⁡2β+cos⁡2β)+2sin⁡αsin⁡β+2cos⁡αcos⁡β=1\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + \left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right) + 2\sin \alpha \sin \beta + 2\cos \alpha \cos \beta = 1
1+1+2(sin⁡αsin⁡β+cos⁡αcos⁡β)=11 + 1 + 2\left( {\sin \alpha \sin \beta + \cos \alpha \cos \beta } \right) = 1
2+2(sin⁡αsin⁡β+cos⁡αcos⁡β)=12 + 2\left( {\sin \alpha \sin \beta + \cos \alpha \cos \beta } \right) = 1
We know that, cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b
Therefore, we have, 2+2(cos⁡(α−β))=12 + 2\left( {\cos \left( {\alpha - \beta } \right)} \right) = 1
2(cos⁡(α−β))=1−22\left( {\cos \left( {\alpha - \beta } \right)} \right) = 1 - 2
Hence, (cos⁡(α−β))=−12\left( {\cos \left( {\alpha - \beta } \right)} \right) = \dfrac{{ - 1}}{2} →(3)\to \left( 3 \right)
Similarly, now sin⁡β+sin⁡γ=−sin⁡α\sin \beta + \sin \gamma = - \sin \alpha →(4)\to \left( 4 \right)
And cos⁡β+cos⁡γ=−cos⁡α\cos \beta + \cos \gamma = - \cos \alpha →(5)\to \left( 5 \right)
Now, let us square and add the equations (4) and (5), we have,
(sin⁡β+sin⁡γ)2+(cos⁡β+cos⁡γ)2=sin⁡2α+cos⁡2α{\left( {\sin \beta + \sin \gamma } \right)^2} + {\left( {\cos \beta + \cos \gamma } \right)^2} = {\sin ^2}\alpha + {\cos ^2}\alpha
Then, by applying the identities, we have,
sin⁡2β+sin⁡2γ+2sin⁡βsin⁡γ+cos⁡2β+cos⁡2γ+2cos⁡βcos⁡γ=1{\sin ^2}\beta + {\sin ^2}\gamma + 2\sin \beta \sin \gamma + {\cos ^2}\beta + {\cos ^2}\gamma + 2\cos \beta \cos \gamma = 1
(sin⁡2β+cos⁡2β)+(sin⁡2γ+cos⁡2γ)+2sin⁡βsin⁡γ+2cos⁡βcos⁡γ=1\left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right) + \left( {{{\sin }^2}\gamma + {{\cos }^2}\gamma } \right) + 2\sin \beta \sin \gamma + 2\cos \beta \cos \gamma = 1
2+2(sin⁡βsin⁡γ+cos⁡βcos⁡γ)=12 + 2\left( {\sin \beta \sin \gamma + \cos \beta \cos \gamma } \right) = 1
Now by applying identity, we have,
2+2(cos⁡(β−γ))=12 + 2\left( {\cos \left( {\beta - \gamma } \right)} \right) = 1
2(cos⁡(β−γ))=1−22\left( {\cos \left( {\beta - \gamma } \right)} \right) = 1 - 2
Hence, (cos⁡(β−γ))=−12\left( {\cos \left( {\beta - \gamma } \right)} \right) = \dfrac{{ - 1}}{2} →(6)\to \left( 6 \right)
Similarly, now sin⁡γ+sin⁡α=−sin⁡β\sin \gamma + \sin \alpha = - \sin \beta →(7)\to \left( 7 \right)
And cos⁡γ+cos⁡α=−cos⁡β\cos \gamma + \cos \alpha = - \cos \beta →(8)\to \left( 8 \right)
Now, let us square and add the equations (7) and (8), we have,
(sin⁡γ+sin⁡α)2+(cos⁡γ+cos⁡α)2=sin⁡2β+cos⁡2β{\left( {\sin \gamma + \sin \alpha } \right)^2} + {\left( {\cos \gamma + \cos \alpha } \right)^2} = {\sin ^2}\beta + {\cos ^2}\beta
Then, by applying the identities, we have,
sin⁡2γ+sin⁡2α+2sin⁡γsin⁡α+cos⁡2γ+cos⁡2α+2cos⁡γcos⁡α=1{\sin ^2}\gamma + {\sin ^2}\alpha + 2\sin \gamma \sin \alpha + {\cos ^2}\gamma + {\cos ^2}\alpha + 2\cos \gamma \cos \alpha = 1
(sin⁡2γ+cos⁡2γ)+(sin⁡2α+cos⁡2α)+2sin⁡γsin⁡α+2cos⁡γcos⁡α=1\left( {{{\sin }^2}\gamma + {{\cos }^2}\gamma } \right) + \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + 2\sin \gamma \sin \alpha + 2\cos \gamma \cos \alpha = 1
2+2(sin⁡γsin⁡α+cos⁡γcos⁡α)=12 + 2\left( {\sin \gamma \sin \alpha + \cos \gamma \cos \alpha } \right) = 1
Now by applying identity, we have,
2+2(cos⁡(γ−α))=12 + 2\left( {\cos \left( {\gamma - \alpha } \right)} \right) = 1
2(cos⁡(γ−α))=1−22\left( {\cos \left( {\gamma - \alpha } \right)} \right) = 1 - 2
Hence, (cos⁡(γ−α))=−12\left( {\cos \left( {\gamma - \alpha } \right)} \right) = \dfrac{{ - 1}}{2} →(9)\to \left( 9 \right)
Now, from equation (3), (6) and (9), we have,
cos⁡(α−β)+cos⁡(β−γ)+cos⁡(γ−α)=−12+−12+−12\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) = \dfrac{{ - 1}}{2} + \dfrac{{ - 1}}{2} + \dfrac{{ - 1}}{2}
cos⁡(α−β)+cos⁡(β−γ)+cos⁡(γ−α)=−32\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) = \dfrac{{ - 3}}{2}
Hence, cos⁡(α−β)+cos⁡(β−γ)+cos⁡(γ−α)=−32\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) = \dfrac{{ - 3}}{2} which is the required solution we need to prove

Note : In this question, it is important to note here that when we are crossing these kinds of problems, we need to think about how to determine the terms that we need to prove from given conditions. For this we need to be pretty good at the algebraic and trigonometric identities. Here, the place where we can make mistakes is while choosing the formulas to apply.