SOLUTION: How to prove that {{{sinxcos^3x-cosxsin^3x=(1/4)(sin4x)}}}?

Algebra.Com
Question 1107683: How to prove that ?
Found 2 solutions by rothauserc, ikleyn:
Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
sin(x)cos^3(x) - cos(x)sin^3(x) = (1/4) * sin(4x)
:
multiply both sides of = by 4
:
4(sin(x)cos^3(x) - cos(x)sin^3(x)) = sin(4x)
:
apply the identity cos^2(x) = 1-sin^2(x) to cos^3(x)
:
4(cos(x)(1-sin^2(x))sin(x) - cos(x)sin^3(x)) = sin(4x)
:
4((cos(x)-cos(x)sin^2(x))sin(x) - cos(x)sin^3(x)) = sin(4x)
:
4(cos(x)sin(x)-cos(x)sin^3(x) - cos(x)sin^3(x)) = sin(4x)
:
4(cos(x)sin(x)-2cos(x)sin^3(x)) = sin(4x)
:
4cos(x)sin(x)-8cos(x)sin^3(x) = sin(4x)
:
use double angle identity on sin(4x)
:
4cos(x)sin(x)-8cos(x)sin^3(x) = 2cos(2x)sin(2x)
:
use double angle identity on cos(2x)
:
4cos(x)sin(x)-8cos(x)sin^3(x) = 2(1-2sin^2(x))sin(2x)
:
use double angle identity on sin(2x)
:
4cos(x)sin(x)-8cos(x)sin^3(x) = 2 * 2cos(x)sin(x) * (1-2sin^2(x))
:
4cos(x)sin(x)-8cos(x)sin^3(x) = 4cos(x)sin(x) - 8cos(x)sin^3(x)
:
both sides are the same, the identity is correct
:

Answer by ikleyn(52846)   (Show Source): You can put this solution on YOUR website!
.
   =  = 


     Use the basic trigonometric formulas  sin(x)*cos(x) = ,   cos^2(x)-sin^2(x) = cos(2x)  to get


=  =  //use again  sin(2x)*cos(2x) =   to get //   = .


That's all.


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