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kassandratb Group Title

find all solutions of sin^2theta=cos^2theta

  • one year ago
  • one year ago

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  1. mathstudent55 Group Title
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    |dw:1359396748498:dw|

    • one year ago
  2. kassandratb Group Title
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    what do i do one i've done this

    • one year ago
  3. JamesJ Group Title
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    Well, remember that cos(2x) = cos^2 x - sin^2 x So if cos^2 x = sin^2 x then cos^2 x - sin^2 x = 0 or cos(2x) = 0 Can you take it from here?

    • one year ago
  4. kassandratb Group Title
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    when I do this : cos^2x-sin^2x=0 cos(2x)=0 cosx=0 x=pi, 2pi, +2k ?

    • one year ago
  5. kassandratb Group Title
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    JamesJ i feel like i did this wrong

    • one year ago
  6. mathstudent55 Group Title
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    |dw:1359397774601:dw|

    • one year ago
  7. Mertsj Group Title
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    You need to make a substitution. Replace sin^2 theta with 1 - cos^2 theta or replace cos^2 theta with 1-sin^2 theta.

    • one year ago
  8. JamesJ Group Title
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    If cos(2x) = 0 then 2x = pi/2 + k.pi, for any integer k. Hence x = pi/4 + k.pi/2, for any integer k.

    • one year ago
  9. JamesJ Group Title
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    You can see those solutions on this graph as well.

    • one year ago
  10. Mertsj Group Title
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    \[\sin^2\theta =1-\sin ^2\theta \] \[2\sin ^2\theta =1\] \[\sin^2\theta =\frac{1}{2}\] \[\sin \theta = \pm \frac{\sqrt{2}}{2}\] \[\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} + 2\pi n\]

    • one year ago
  11. kassandratb Group Title
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    Thank you all it's starting to make a lot more since to me now so I just gotta know my identities?

    • one year ago
  12. Mertsj Group Title
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    Yes. That is key.

    • one year ago
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