SOLUTION: Prove the identity: {{{sin(3x) = 3*sin(x) – 4*(sin(x))^3}}} (x is theta)
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Question 138821: Prove the identity: (x is theta)
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Note: I'm only manipulating the left side. I'm not touching the right side. I'm only showing it for comparison.
 = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Start with the given equation
 = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Break up 
to get 
\cos\left(\theta\right)+\cos\left(2\theta\right)\sin\left(\theta\right) = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Expand using the sum-difference formulas
\cos\left(\theta\right)\cos\left(\theta\right)+\left(1-2\sin^2\left(\theta\right)\right)\sin\left(\theta\right) = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Replace )
with \cos\left(\theta\right))
. Replace )
with )
.
\left(2\cos^2\left(\theta\right)+1-2\sin^2\left(\theta\right)\right) = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Factor out )
\left(2\cos^2\left(\theta\right)-2\sin^2\left(\theta\right)+1\right) = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Rearrange the terms
\left(2\left(\cos^2\left(\theta\right)-\sin^2\left(\theta\right)\right)+1\right) = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Factor out 2
\left(2\cos\left(2\theta\right)+1\right) = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Replace -\sin^2\left(\theta\right))
with
\left(2\left(1-2\sin^2\left(\theta\right)\right)+1\right) = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Replace with
\left(2-4\sin^2\left(\theta\right)+1\right) = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Distribute
\left(3-4\sin^2\left(\theta\right)\right) = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Combine like terms
-4\sin^3\left(\theta\right) = 3\sin\left(\theta\right) - 4\sin^3\left(\theta\right))
Distribute
So we've just proven that
is an identity.
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