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unicwaan

  • one year ago

Can someone help me verify the identity? sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x

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  1. BTaylor
    • one year ago
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    so, this? \[\frac{\sin x}{1-\cos{x}} + \frac{\sin x}{1-\cos{x}} = 2\csc x\]

  2. unicwaan
    • one year ago
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    yes except the it's 1- cos and 1 + cos sorry

  3. welshfella
    • one year ago
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    first convert the 2 fractions on left side to one fraction

  4. BTaylor
    • one year ago
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    You want to use the identity \(1-\cos^2 x = \sin^2 x\). Multiply the 1 - cos x by (1 + cos x) on top and bottom, and you will have sin^2 x on the bottom.

  5. unicwaan
    • one year ago
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    Okay I understand that step BTaylpr but I don't know how the make the other fraction the same like what welshfella wants me to do

  6. misty1212
    • one year ago
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    HI!!

  7. BTaylor
    • one year ago
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    \[\frac{\sin x}{1-\cos{x}} \cdot \frac{1+\cos x}{1+\cos x} + \frac{\sin x}{1+\cos{x}} \cdot \frac{1-\cos x}{1-\cos x} = 2\csc x\]

  8. misty1212
    • one year ago
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    you can also just add, you will get it that way too

  9. unicwaan
    • one year ago
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    Oh the conjugate! I forgot about that!

  10. misty1212
    • one year ago
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    \[\frac{b}{1+a}+\frac{b}{1-a}=\frac{b(1-a)+b(1+a)}{(1-a)(1+a)}\]

  11. misty1212
    • one year ago
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    you get \[\frac{2b}{1-a^2}\]

  12. unicwaan
    • one year ago
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    Thank you I justed needed that step to get me started, I can figure the rest out :)

  13. unicwaan
    • one year ago
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    just*

  14. BTaylor
    • one year ago
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    no problem

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