shooterman
Junior Member
- Joined
- Aug 20, 2009
- Messages
- 57
cos3x = 4cos3x - 3cosx
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Verify equation is an identity
- Thread starter shooterman
- Start date
Hello, shooterman!
Better tidy up your typing . . .
\(\displaystyle \cos(3x) \;=\;\cos(2x+x)\)
. . . . . .\(\displaystyle =\;\cos2x\cos x - \sin2x\sin x\)
. . . . . .\(\displaystyle =\;(2\cos^2\!x - 1)\cos x - (2\sin x\cos x)\sin x\)
. . . . . .\(\displaystyle =\;2\cos^3\!x - \cos x - 2\sin^2\!x\cos x\)
. . . . . .\(\displaystyle =\;2\cos^3\!x - \cos x - 2(1 - \cos^2\!x)\cos x\)
. . . . . .\(\displaystyle =\;2\cos^3\!x - \cos x - 2\cos x + 2\cos^3\!x\)
. . . . . .\(\displaystyle =\;4\cos^3\!x - 3\cos x\)
Better tidy up your typing . . .
Prove: .\(\displaystyle \]cos3x \:=\: 4\cos^3\!x - 3\cos x\)
\(\displaystyle \cos(3x) \;=\;\cos(2x+x)\)
. . . . . .\(\displaystyle =\;\cos2x\cos x - \sin2x\sin x\)
. . . . . .\(\displaystyle =\;(2\cos^2\!x - 1)\cos x - (2\sin x\cos x)\sin x\)
. . . . . .\(\displaystyle =\;2\cos^3\!x - \cos x - 2\sin^2\!x\cos x\)
. . . . . .\(\displaystyle =\;2\cos^3\!x - \cos x - 2(1 - \cos^2\!x)\cos x\)
. . . . . .\(\displaystyle =\;2\cos^3\!x - \cos x - 2\cos x + 2\cos^3\!x\)
. . . . . .\(\displaystyle =\;4\cos^3\!x - 3\cos x\)
shooterman
Junior Member
- Joined
- Aug 20, 2009
- Messages
- 57
Never noticed the problem could be verified working backwards starting with the difference and sum identities thanks.