First slide
Multiple and sub- multiple Angles
Question

 If cosA+cosB+cosC=0 and cos3A+cos3B+cos3C=λcosAcosBcosC then λ is 

Moderate
Solution

 If cosA+cosB+cosC=0 then cos3A+cos3B+cos3C=3cosAcosBcosC Given cos3A+cos3B+cos3c=λcosAcosBcosC4cos3A3cosA+4cos3B3cosB+4cos3C3cosC=λcosAcosBcosC=4cos3A+cos3B+cos3C3(cosA+cosB+cosC)=4(3cosAcosBcosC)3(0)=12cosAcosBcosC

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