- #1
kirab
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Homework Statement
Prove that
[tex]\frac{cos 3x}{cos x} = 2cos (2x) - 1[/tex]
Homework Equations
The ones I used:
[tex] cos 2x = cos^2 x - sin^2 x[/tex]
[tex]sin^2 x + cos^2 x = 1
[/tex]
The Attempt at a Solution
I *think* that the left hand side cannot be manipulated so I only fooled around with the right hand side...
[tex] 2cos 2x - 1 = 2 (cos^2 x - sin^2 x) - 1 = 2 (cos^2 x - sin^2 x) - (sin^2 x + cos^2 x)
= 2cos^2 x - 2sin^2 x - sin^2 x - cos^2 x = cos^2 x - 3sin^2 x [/tex]
And I'm not sure what do to from there, so I did another approach (from the original right hand side):
[tex] 2cos 2x - 1 = 2 (cos2x - 1/2) = 2 (cos 2x - ((sin^2 x + cos^2 x)/2)) = 2 ((2cos2x - sin^2 x - cos^2 x)/2) = 2cos2x - sin^2 x - cos^2 x. [/tex]
And I'm stuck at this point. Any suggestions?
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