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amydh Group Title

sin^2 theta - cos^2 theta = -1

  • 3 months ago
  • 3 months ago

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  1. ParthKohli Group Title
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    \[-\left(\cos^2 \theta - \sin^2 \theta\right) = -1\]\[\implies -\cos(2\theta) = -1\]\[\implies \cos(2\theta) = 1\]Or use either of the two identities:\[\cos^2 \theta = 1 - \sin^2 \theta\]\[\sin^2 \theta = 1- \cos^2 \theta\]

    • 3 months ago
  2. myininaya Group Title
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    or ... ANOTHER WAY! \[\sin^2(\theta)-\cos^2(\theta)=-1 \] Recall \[\sin^2(\theta)+\cos^2(\theta)=1\] So add \[\cos^2(\theta)\] on both sides to use the identity that is being recalled \[\sin^2(\theta)+\cos^2(\theta)-\cos^2(\theta)=-1+\cos^2(\theta) \] Now use the identity \[1-\cos^2(\theta)=-1+\cos^2(\theta)\] \[2=2\cos^2(\theta)\] \[1=\cos^2(\theta) \] which results into two equations

    • 3 months ago
  3. ParthKohli Group Title
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    Just add those two equations to get \(2\sin^2 \theta = 0\).

    • 3 months ago
  4. myininaya Group Title
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    oh yeah that is pretty too

    • 3 months ago
  5. mathslover Group Title
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    This problem can be done in several ways. I was going to write some of them here, but I guess, they will all be actually similar to each other. But @myininaya and @ParthKohli have done well to find 2 of them.

    • 3 months ago
  6. myininaya Group Title
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    3 of them actually

    • 3 months ago
  7. myininaya Group Title
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    parth did one way at first then he found another way off what i was doing

    • 3 months ago
  8. ParthKohli Group Title
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    why are you typing like satellite does

    • 3 months ago
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