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anonymous

  • 5 years ago

simple trig identity! prove sinx/1+cosx= 1/sinx - 1/tan x

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  1. anonymous
    • 5 years ago
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    so i think you write not right equal

  2. anonymous
    • 5 years ago
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    you should do this on your own. it will help you learn about manipulation of trigonometric identities.

  3. anonymous
    • 5 years ago
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    honestly, i would if i knew how. i have 2 assignments of an online functions course left to do and once i'm done, i never have to do math again, so all i really need to do is finish them. i've done what i can on my own and now i'm asking for help.

  4. anonymous
    • 5 years ago
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    Note \[ 1-\cos^2x = \sin^2x \] And think about what you can multiply the top and bottom of the LHS (or RHS) to use this fact.

  5. anonymous
    • 5 years ago
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    tan(x)= sin(x)/cos(x)

  6. anonymous
    • 5 years ago
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    My method is more interesting ¬_¬

  7. anonymous
    • 5 years ago
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    \[a^2 - b^2 = (a+b)(a-b) \] May also be useful.

  8. anonymous
    • 5 years ago
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    I like Newton's method

  9. anonymous
    • 5 years ago
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    this is all just confusing me more. it's so much easier to understand when i see a solution.

  10. anonymous
    • 5 years ago
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    Sorry, I can't do your problem for you. Ask one of the other people.

  11. anonymous
    • 5 years ago
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    so far i took the right side and got 1/sinx-cosx/sinx which = 1-cosx/sinx but i dont know if that is correct, or what to do next

  12. anonymous
    • 5 years ago
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    That is correct, and useful. How can you get from that to the LHS?

  13. anonymous
    • 5 years ago
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    that's what i don't know.

  14. anonymous
    • 5 years ago
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    i some how need to flip it and chance the - to a +

  15. anonymous
    • 5 years ago
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    Not technically true. Because on the LHS, the sin(X) is on the top, but on the RHS, the sin(x) is on the bottom. See my above advice.

  16. anonymous
    • 5 years ago
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    but there are no ^2 in this proof

  17. anonymous
    • 5 years ago
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    OK. So if you take your LHS and RHS in the new form you have. \[\frac{\sin x}{1+\cos x} = \frac{1-\cos x}{\sin x} \] Multiplying both sides by (1+cosx)(sinx) gives a result that is clearly true. (s^2 - 1-c^2) HOWEVER it is best to go directly from LHS -> RHS (or vice versa) and not assume the result and work from that. So can you see how something similar could get from one side to the otehr directly?

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