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Question: Let A and B denote the statements: A: \(\cos \alpha + \cos \beta + \cos \gamma = 0\) B: \(\sin \...

Let A and B denote the statements:
A: cos⁡α+cos⁡β+cos⁡γ=0\cos \alpha + \cos \beta + \cos \gamma = 0
B: sin⁡α+sin⁡β+sin⁡γ=0\sin \alpha + \sin \beta + \sin \gamma = 0
If cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−32\cos (\beta - \gamma ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) = - \dfrac{3}{2}
Then:
A) A is correct and B is incorrect
B) A is incorrect and B is correct
C) Both A and B are correct
D) Both A and B are incorrect

Explanation

Solution

We are given two statements we have to find whether these statements hold when we are given the equation:
cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−32\cos (\beta - \gamma ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) = - \dfrac{3}{2}
We will first move the denominator of the fraction given on the right side to the left side and then square the whole equation and remember we will express 33 as :
3=sin2α + cos2α + sin2β + cos2β + sin2γ + cos2γ 3 = si{n^2}\alpha {\text{ }} + {\text{ }}co{s^2}\alpha {\text{ }} + {\text{ }}si{n^2}\beta {\text{ }} + {\text{ }}co{s^2}\beta {\text{ }} + {\text{ }}si{n^2}\gamma {\text{ }} + {\text{ }}co{s^2}\gamma {\text{ }} and then we apply the pythagorean trigonometric identity   sin2A + cos2A = 1\;si{n^2}A{\text{ }} + {\text{ }}co{s^2}A{\text{ }} = {\text{ }}1 to get desired answer.

Complete step by step answer:
We are given the two statements A and B and also the equation:
cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−32\cos (\beta - \gamma ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta ) = - \dfrac{3}{2}
2 cos(β − γ) + 2 cos(γ − α) + 2 cos(α − β) = −32{\text{ }}cos\left( {\beta {\text{ }} - {\text{ }}\gamma } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\gamma {\text{ }} - {\text{ }}\alpha } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\alpha {\text{ }} - {\text{ }}\beta } \right){\text{ }} = {\text{ }} - 3
Moving 33 to the left hand side of the equation.
2 cos(β − γ) + 2 cos(γ − α) + 2 cos(α − β) + 3 = 02{\text{ }}cos\left( {\beta {\text{ }} - {\text{ }}\gamma } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\gamma {\text{ }} - {\text{ }}\alpha } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\alpha {\text{ }} - {\text{ }}\beta } \right){\text{ }} + {\text{ }}3{\text{ }} = {\text{ }}0

2 cos(β − γ) + 2 cos(γ − α) + 2 cos(α − β) + sin2α + cos2α + sin2β + cos2β +  sin2γ + cos2γ = 0  2{\text{ }}cos\left( {\beta {\text{ }} - {\text{ }}\gamma } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\gamma {\text{ }} - {\text{ }}\alpha } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\alpha {\text{ }} - {\text{ }}\beta } \right){\text{ }} + {\text{ }}si{n^2}\alpha {\text{ }} + {\text{ }}co{s^2}\alpha {\text{ }} + {\text{ }}si{n^2}\beta {\text{ }} + {\text{ }}co{s^2}\beta {\text{ }} + \\\ {\text{ }}si{n^2}\gamma {\text{ }} + {\text{ }}co{s^2}\gamma {\text{ }} = {\text{ }}0 \\\

We will now solve the terms 2 cos(β − γ) + 2 cos(γ − α) + 2 cos(α − β) 2{\text{ }}cos\left( {\beta {\text{ }} - {\text{ }}\gamma } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\gamma {\text{ }} - {\text{ }}\alpha } \right){\text{ }} + {\text{ }}2{\text{ }}cos\left( {\alpha {\text{ }} - {\text{ }}\beta } \right){\text{ }}
We know the identity:
cos(A − B) = cos A cos B + sin A sin Bcos\left( {A{\text{ }} - {\text{ }}B} \right){\text{ }} = {\text{ }}cos{\text{ }}A{\text{ }}cos{\text{ }}B{\text{ }} + {\text{ }}sin{\text{ }}A{\text{ }}sin{\text{ }}B
This will result in first term being written as : 2 cos(β − γ) = 2 cosβcosγ+2 sinβsinγ2{\text{ }}cos\left( {\beta {\text{ }} - {\text{ }}\gamma } \right){\text{ = }}2{\text{ }}cos\beta cos\gamma + 2{\text{ }}sin\beta sin\gamma
Other terms will also be similar and thus will result in the equation becoming

sin2α + cos2α + sin2β + cos2β + sin2γ + cos2γ = 0  si{n^2}\alpha {\text{ }} + {\text{ }}co{s^2}\alpha {\text{ }} + {\text{ }}si{n^2}\beta {\text{ }} + {\text{ }}co{s^2}\beta {\text{ }} + {\text{ }}si{n^2}\gamma {\text{ }} + {\text{ }}co{s^2}\gamma {\text{ }} = {\text{ }}0 \\\

We can now see that the left hand side is a perfect square identity of the form;
  (a + b + c)2  = a2  + b2  + c2  + 2(ab + bc + ac)\;{\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)^2}\; = {\text{ }}{a^2}\; + {\text{ }}{b^2}\; + {\text{ }}{c^2}\; + {\text{ }}2\left( {ab{\text{ }} + {\text{ }}bc{\text{ }} + {\text{ }}ac} \right)
(cos α + cos β + cos γ)2  + (sin α + sin β + sin γ)2  = 0{\left( {cos{\text{ }}\alpha {\text{ }} + {\text{ }}cos{\text{ }}\beta {\text{ }} + {\text{ }}cos{\text{ }}\gamma } \right)^2}\; + {\text{ }}{\left( {sin{\text{ }}\alpha {\text{ }} + {\text{ }}sin{\text{ }}\beta {\text{ }} + {\text{ }}sin{\text{ }}\gamma } \right)^2}\; = {\text{ }}0
Since for this statement to be true both the expressions must be equal to zero and hence both the statements are true. The option C is hence true.

Note:
While applying the pythagorean trigonometric identity we should be careful regarding the angle. That is both the sine and cosine function angle should be the same. If we have sin⁡2x+cos⁡2y{\sin ^2}x + {\cos ^2}y then it is not equal to 1, both the angles should be xx or it should be yy. We used the same concept above.