Question

Prove that cos^3a+cos^3(120+a)+cos^3(240+a)=3/4cos3a

Answer 1

cos^3 A+ cos^3 (120+A) +cos^3 (240+A)=3/4 cos 3A 

<=> cos^3A + ( -1/2.cosA - √3/2.sinA )^3 + ( -1/2.cosA + √3/2sinA )^3 = 3/4 cos3A 

<=> cos^3A + ( -1/2.cosA - √3/2.sinA - 1/2.cosA + √3/2sinA )[ ( -1/2.cosA - √3/2.sinA )² - ( -1/2.cosA - √3/2.sinA )( -1/2.cosA + √3/2sinA ) + ( -1/2.cosA + √3/2sinA )² ] = 3/4.cos3A 

<=> cos^3A + ( -cosA )[ 1/4.cos²A + √3/2.sinAcosA + 3/4.sin²A - (-1/2cosA)² + (√3/2.sinA )² +1/4.cos²A - √3/2.sinAcosA + 3/4.sin²A ] = 3/4.cos3A 

<=> cos^3A + ( -cosA )[ 1/2.cos²A + 3/2.sin²A - 1/4.cos²A + 3/4.sin²A ] = 3/4.cos3A 

<=> cos^3A + ( -cosA )[ 1/4.cos²A + 9/4.sin²A ] = 3/4.cos3A 

<=> cos^3A - cos^3A / 4 - 9/4.cosA.sin²A = 3/4.cos3A 

<=> 3cos^3A / 4 - 9/4.cosA.sin²A = 3/4.cos3A 

<=> cos^3A - 3cosAsin²A = cos3A 

<=> cos^3A - 3cosA(1 - cos²A ) = cos3A 

<=> cos^3A - 3cosA + 3cos^4A = cos3A 

<=> 4cos^3A - 3cosA = cos3A

Answer 2

we know cos3A=4 cos^3 A - 3 cosA

cos^3A = 1/4(cos3A +3 cosA)

=1/4(cos3A+3cosA) + 1/4[cos(360+3A) + 3 cos(120+A)] + 1/4[cos(360-3A) + 3 cos(120-A)]

=3/4 cos3A + 1/4 [ 3cosA + 3 .2(cos120 cosA)]

=3/4 cos3A + 1/4[3cosA-3cosA]

= 3/4 cos3A