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Prove that `cos^3A+cos^3(120^0+A)+cos^3(240^0+A)=3/4cos3A` |
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Answer» `cosA + cos(120^@+A) +cos(240^@+A)` `= cosA + (2cos((120+A+240+A)/2)cos((120+A-240-A)/2))` `= cosA + (2cos(180+A)cos(-60^@))` `= cosA + (2cos(180+A)cos60^@)` `= cosA + (-2cosA(1/2))` `=cosA - cosA = 0` We know, when, `a+b+c = 0`, `a^3+b^3+c^3 = 3abc` `:. L.H.S. = cos^3A + cos^3(120^@+A) +cos^3(240^@+A) = 3cosAcos(120^@+A)cos(240^@+A)` `=3cosAcos(180-(60-A))cos(180+(60+A))` `=3cosAcos(-(60-A))cos(-(60+A))` `=3cosAcos(60-A)cos(60+A)` Using `cos(C-D)cos(C+D) = cos^2C - sin^2D` `=3cosA[cos^2 60^@-sin^2A]` `=3cosA[1/4-(1-cos^2A)]` `=3cosA[-3/4+cos^2A]` `=3/4(4cos^3A - 3cosA)` `=3/4cos3A = R.H.S.` |
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