How does this inital equation:
sinx+1/3(sin3x)
equal this equation?:
2sinx-(4/3)(sinx)(sinx)(sinx)
or
2sinx-(4/3)sin^3x
A explanation for how the rewritten equation was produced would be great. Please provide a detailed and full explanation please, don't provide a layman's explanation.
Verify the following identity:
sin(x)+(sin(3 x))/(3) = 2 sin(x)-(4 sin(x)^3)/(3)
Put sin(x)+1/3 sin(3 x) over the common denominator 3: sin(x)+1/3 sin(3 x) = (3 sin(x)+sin(3 x))/3:
(3 sin(x)+sin(3 x))/3 = ^?2 sin(x)-(4 sin(x)^3)/(3)
Put 2 sin(x)-(4 sin(x)^3)/3 over the common denominator 3: 2 sin(x)-(4 sin(x)^3)/3 = (6 sin(x)-4 sin(x)^3)/3:
(3 sin(x)+sin(3 x))/3 = ^?(6 sin(x)-4 sin(x)^3)/3
Multiply both sides by 3:
3 sin(x)+sin(3 x) = ^?6 sin(x)-4 sin(x)^3
sin(x)^3 = 1/4 (3 sin(x)-sin(3 x)):
3 sin(x)+sin(3 x) = ^?6 sin(x)-4(3 sin(x)-sin(3 x))/(4)
-(3 sin(x)-sin(3 x)) = sin(3 x)-3 sin(x):
3 sin(x)+sin(3 x) = ^?6 sin(x)+sin(3 x)-3 sin(x)
6 sin(x)-3 sin(x)+sin(3 x) = 3 sin(x)+sin(3 x):
3 sin(x)+sin(3 x) = ^?3 sin(x)+sin(3 x)
The left hand side and right hand side are identical:
Answer: |(identity has been verified)
