Can i have a solution on how i can derive on this answer 4cos^3x - 3 cos x.....please..
3 answers
after doing what ?
If you want to know the solutions to
4cos^3x - 3 cos x = 0,
cos x (4cos^2 x - 3) = 0
so,
cos x = 0
or
cos x = ±√3/2
You should recognize those values, as they come up often.
4cos^3x - 3 cos x = 0,
cos x (4cos^2 x - 3) = 0
so,
cos x = 0
or
cos x = ±√3/2
You should recognize those values, as they come up often.
Ah. If you just wanted to verify that
cos 3x = 4cos^3 x - 3cos x, then consider
cos 3x = cos(2x+x)
= cos(2x)cos(x) - sin(2x)sin(x)
= (2cos^2(x)-1)(cos x) - (2sin x cos x)(sin x)
= 2cos^3(x)-cos(x) - 2sin^2(x)cos(x)
= 2cos^3(x) - cos(x) - 2(1-cos^2(x))cos(x)
= 2cos^3x - cosx - 2cosx + 2cos^3x
= 4cos^3x - 3cosx
cos 3x = 4cos^3 x - 3cos x, then consider
cos 3x = cos(2x+x)
= cos(2x)cos(x) - sin(2x)sin(x)
= (2cos^2(x)-1)(cos x) - (2sin x cos x)(sin x)
= 2cos^3(x)-cos(x) - 2sin^2(x)cos(x)
= 2cos^3(x) - cos(x) - 2(1-cos^2(x))cos(x)
= 2cos^3x - cosx - 2cosx + 2cos^3x
= 4cos^3x - 3cosx