SOLUTION: How to prove that cos3x= 4cos^3x - 3cosx ?
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Question 586529: How to prove that cos3x= 4cos^3x - 3cosx ?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Use the Cosine of the Sum and the Cosine and Sine of Double Angles.
\ =\ \cos\left(\phi\right)\cos\left(\varphi\right)\ -\ \sin\left(\phi\right)\sin\left(\varphi\right))
\ =\ \cos^2\left(\varphi\right)\ -\ \sin^2\left(\varphi\right))
\ =\ 2\sin\left(\varphi\right)\cos\left(\varphi\right))
Then
\ =\ \cos\left(2x\ +\ x\right))
So
\ =\ \cos\left(2x)\cos\left(x\right)\ -\ \sin\left(2x\right)\sin\left(x\right))
\cos\left(x\right)\ -\ \sin\left(2x\right)\sin\left(x\right)\ =\ \left[\cos^2(x)\ -\ \sin^2{x}\right]\cos(x)\ -\ \left[2\sin(x)\cos{x}\right]\sin(x))
\ -\ \sin^2{x}\cos(x)\ -\ 2\sin^2(x)\cos{x})
\ -\ 3\sin^2{x}\cos(x))
Now use the Pythagorean Identity:
\ -\ 3\sin^2{x}\cos(x)\ =\ \cos^3(x)\ -\ 3\left(1\ -\ \cos^2{x}\right)\cos(x))
Distribute and collect terms
\ -\ 3\left(1\ -\ \cos^2{x}\right)\cos(x)\ =\ \underline{4\cos^3(x)\ -\ 3\cos(x)})
John
My calculator said it, I believe it, that settles it
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