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anonymous

  • one year ago

can someone explain this to me???

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  1. anonymous
    • one year ago
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    simplify

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  2. kropot72
    • one year ago
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    The first step in simplifying is to multiply out the numerator: \[\large (1-\cos \theta)(1+\cos \theta)=?\] Can you do that and post the result?

  3. anonymous
    • one year ago
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    \[1+\cos \theta -\cos \theta + \cos^2 ?\]

  4. DecentNabeel
    • one year ago
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    (1−cosθ)(1+cosθ)=1-cos^2x \[\sqrt{\frac{1-\cos^2x}{\cos^2x}}\] \[\mathrm{Use\:the\:following\:identity}:\quad \:1-\cos ^2\left(x\right)=\sin ^2\left(x\right)\] \[=\sqrt{\frac{\sin ^2\left(x\right)}{\cos ^2\left(x\right)}}\]

  5. anonymous
    • one year ago
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    ok. so im left with sin(x) over cos(x)?

  6. DecentNabeel
    • one year ago
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    yes

  7. kropot72
    • one year ago
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    Nearly correct! However the sign in front of the cos^2 (theta) term should be negative. Therefore we get: \[\large \sqrt{\frac{(1-\cos^{2} \theta)}{\cos^{2} \theta}}\] Does that make sense?

  8. anonymous
    • one year ago
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    yes. but cant i break that down further?

  9. DecentNabeel
    • one year ago
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    are you understand @kapukawai

  10. kropot72
    • one year ago
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    You need to use the relationship: \[\large \sin^{2}\theta+\cos^{2}\theta=1\] From which you get: \[\large 1-\cos^{2}\theta=\sin^{2}\theta\] and finally by substitution we can write: \[\large \sqrt{\frac{\sin^{2}\theta}{\cos^{2}\theta}}\]

  11. kropot72
    • one year ago
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    Next you can use the relationship \[\large \frac{\sin\theta}{\cos\theta}=\tan\theta\]

  12. anonymous
    • one year ago
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    so tan theta is the simplified form of ...... ok that makes sence. i just have to memorise the relation ships?

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  13. kropot72
    • one year ago
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    Yes, those relationships are some of the basic ones in trigonometry.

  14. anonymous
    • one year ago
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    thank you!

  15. kropot72
    • one year ago
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    You're welcome :)

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