How can I solve the mentioned problem?Please,help.

Question

How can I solve the mentioned problem?Please,help.

If, $cos A + cos B + cos C = 0$ $cosA+cosB+cosC=0$ ;then prove that , $cos 3 A + cos 3 B + cos 3 C = 12 cos A cos B cos C$ $cos3A+cos3B+cos3C=12cosAcosBcosC$

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Answer 1 · 2017-12-20T14:46:56

Given

$cos A + cos B + cos C = 0$ $cosA+cosB+cosC=0$

$\Rightarrow cos A + cos B = - cos C$ $=>cosA+cosB=-cosC$

$\Rightarrow {(cos A + cos B)}^{3} = - {cos}^{3} C$ $=>(cosA+cosB)^3=-cos^3C$

$\Rightarrow {cos}^{3} A + {cos}^{3} B + 3 cos A cos B (cos A + cos B) = - {cos}^{3} C$ $=>cos^3A+cos^3B+3cosAcosB(cosA+cosB)=-cos^3C$

$\Rightarrow {cos}^{3} A + {cos}^{3} B + 3 cos A cos B (- cos C) = - {cos}^{3} C$ $=>cos^3A+cos^3B+3cosAcosB(-cosC)=-cos^3C$

$\Rightarrow {cos}^{3} A + {cos}^{3} B + {cos}^{3} C = 3 cos A cos B cos C$ $=>cos^3A+cos^3B+cos^3C=3cosAcosBcosC$

$\Rightarrow 4 {cos}^{3} A + 4 {cos}^{3} B + 4 {cos}^{3} C = 12 cos A cos B cos C$ $=>4cos^3A+4cos^3B+4cos^3C=12cosAcosBcosC$
Now
$cos 3 A + cos 3 B + cos 3 B$ $cos3A+cos3B+cos3B$

$= 4 {cos}^{3} A - 3 cos A + 4 {cos}^{3} B - 3 cos B + 4 {cos}^{3} C - 3 cos C$ $=4cos^3A-3cosA+4cos^3B-3cosB+4cos^3C-3cosC$

$= 4 {cos}^{3} A + 4 {cos}^{3} B + 4 {cos}^{3} C - 3 (cos A + cos B + cos C)$ $=4cos^3A+4cos^3B+4cos^3C-3(cosA+cosB+cosC)$

$= 12 cos A cos B cos C + 3 \cdot 0$ $=12cosAcosBcosC+3*0$

$= 12 cos A cos B cos C$ $=12cosAcosBcosC$

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How can I solve the mentioned problem?Please,help.

If, $cos A + cos B + cos C = 0$ $cosA+cosB+cosC=0$ ;then prove that , $cos 3 A + cos 3 B + cos 3 C = 12 cos A cos B cos C$ $cos3A+cos3B+cos3C=12cosAcosBcosC$

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If, $cos A + cos B + cos C = 0$ $cosA+cosB+cosC=0$ ;then prove that , $cos 3 A + cos 3 B + cos 3 C = 12 cos A cos B cos C$ $cos3A+cos3B+cos3C=12cosAcosBcosC$