Solve Sin2x=Cos2x I know: Sin2x=2SinxCosx I also know: Cos2x=cos²x-sin²x Cos2x=2cos²x-1 Cos2x=1-2sin²x How do I rearrange and solve? Thanks x
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Originally Posted by hymnseeker Solve Sin2x=Cos2x I know: Sin2x=2SinxCosx I also know: Cos2x=cos²x-sin²x Cos2x=2cos²x-1 Cos2x=1-2sin²x How do I rearrange and solve? Thanks x sin2x=2sincosx is the double angle formula, which ones do you wish to solve?
Sorry, the question was: Solve Sin2x=Cos2x
Originally Posted by hymnseeker Sorry, the question was: Solve Sin2x=Cos2x Ok. Try using $\displaystyle SinA/CosA = TanA$ Then $\displaystyle Tan2x=1$ and $\displaystyle 2x = 45 $degrees plus the solution in the third fifth and seventh quadrant Does this help atol?
Thanks man. Got it.
Originally Posted by hymnseeker Thanks man. Got it. No problem, just make sure for these types of questions that you know 1.sine is positive in quads 1 and 2 2.cosine is positive in 1 and 4 3.tangent is positive in 1 and 3 4.sin0=sin180=0 5.cos90=cos270=0
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