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Trig Identity or something

  1. Feb 21, 2013 #1
    cosθ + sinθ = √2 cos(θ-∏/4)

    what are the steps in between?
     
  2. jcsd
  3. Feb 21, 2013 #2

    ehild

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    Expand cos(θ-∏/4). What do you get?

    ehild
     
  4. Feb 21, 2013 #3
    But how would I get from cosθ + sinθ to √2 cos(θ-∏/4)
    I can expand and get the original expression but don't understand the other way around.
     
    Last edited: Feb 21, 2013
  5. Feb 21, 2013 #4
    Multiply and divide by ##\sqrt{2}##.
     
  6. Feb 21, 2013 #5

    NascentOxygen

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    Staff: Mentor

    Expand cos(θ-/4) means to apply the well-known :smile: trig expansion cos (A - B) = (cos A)(cos B) + (sin A)(sin B)

    This formula (along with a few others) should be well-known by the time you sit for your next closed-book exam. :wink:
     
  7. Feb 21, 2013 #6

    eumyang

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    Are you saying that you can get from
    [itex]\sqrt{2} \cos \left( \theta - \frac{\pi}{4} \right)[/itex]
    to
    [itex]\cos \theta + \sin \theta[/itex]
    but not the other way around? Just take the steps you get from the RHS to the LHS and go backwards.
     
  8. Feb 21, 2013 #7
    Write [tex]Acos(\theta)+Bsin(\theta)=\sqrt{A^2+B^2}(\frac{A}{\sqrt{A^2+B^2}}cos(\theta)+\frac{B}{\sqrt{A^2+B^2}}sin(\theta))[/tex]

    Then let [itex]\phi=\arctan{\frac{B}{A}}[/itex]
    Then,[tex]Acos(\theta)+Bsin(\theta)=\sqrt{A^2+B^2}(cos(\phi)cos(\theta)+sin(\phi)sin(\theta))=\sqrt{A^2+B^2}cos(\theta-\phi)[/tex]
     
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