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anonymous

  • 4 years ago

Prove that (1 + secx)(1 - cosx) = sinxtanx??

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  1. amistre64
    • 4 years ago
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    write everything in terms of sin and cos to see things better

  2. anonymous
    • 4 years ago
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    so 1 - cosx + secx - cosxsecx = secx - cosx. Now what??

  3. anonymous
    • 4 years ago
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    sec=cos/sin

  4. anonymous
    • 4 years ago
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    no, sec is 1/cos

  5. anonymous
    • 4 years ago
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    hhaha

  6. amistre64
    • 4 years ago
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    i was thinking more of: \[1-cos+sec-cos.sec=sin.tan\] \[1-cos+sec-cos/cos=sin.sin/cos\] \[1-cos+sec-1=sin^2/cos\] \[-cos+1/cos=sin^2/cos\] for starters

  7. anonymous
    • 4 years ago
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    but we don't have to do that. we have to work from one side, and expand that to get the other side. It has to be proven.

  8. amistre64
    • 4 years ago
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    this is working one side; I havent changed the left at all except to see it better

  9. amistre64
    • 4 years ago
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    and by left i mean right lol

  10. amistre64
    • 4 years ago
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    \[-cos+1/cos=sin^2/cos\] \[(-cos^2+1)/cos=sin^2/cos\] \[(1-cos^2)/cos=sin^2/cos\] \[sin^2/cos=sin^2/cos\]

  11. anonymous
    • 4 years ago
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    oh silly me it is a product!! i thought it was a fraction...

  12. anonymous
    • 4 years ago
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    but we are not supposed to even supposed to do ANYTHING with the other side. It's like we're not supposed to know the other side.

  13. anonymous
    • 4 years ago
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    ^that's right :P

  14. amistre64
    • 4 years ago
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    your requirements are absurd

  15. amistre64
    • 4 years ago
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    your not spose to know tha tan = sin/cos? your not spose to know that sin.sin = sin^2? your not spose to know what; any trig relations and identities? but somehow transform them into each other?

  16. anonymous
    • 4 years ago
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    not that, yaar. Imagine they give you a question, saying expand ....., leaving it terms of sin/cos/tan. SO we just do that. we expand. but how??

  17. anonymous
    • 4 years ago
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    \[(1+\frac{1}{a})(1-a)=\frac{1}{a}-a=\frac{1-a^2}{a}\] now replace a by cosine and you should get it

  18. amistre64
    • 4 years ago
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    since the right side isnt changed; just revert it back into its origanl form and and leave it there; but the process is still the same ...

  19. anonymous
    • 4 years ago
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    it is much easier to do the algebra without the sines and cosines. then put them in at the end.

  20. anonymous
    • 4 years ago
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    but how does that give us sinxtanx??

  21. anonymous
    • 4 years ago
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    but what amistre said is important. it in not a matter of leaving one side untouched. if you can change one side, you can change it back. if you have to hand it in in a certain form, fine, but if you can alter one side you can presumably go backwards

  22. amistre64
    • 4 years ago
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    what? how does sin.sin/cos = sin tan?

  23. amistre64
    • 4 years ago
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    by definition of what sin and tan ARE lol

  24. anonymous
    • 4 years ago
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    ok we did the algebra part, now comes the trig part. replace a by cosine and get \[\frac{1-\cos^2(x)}{\cos(x)}\] now come the only trig step, writing \[1-\cos^2(x)=\sin^2(x)\] that is the one step that is trig

  25. amistre64
    • 4 years ago
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    can know that 1-cos^2 = sin^2 sat; its forbidden knowledge ;)

  26. anonymous
    • 4 years ago
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    I GOT IT. I love you guys :D

  27. anonymous
    • 4 years ago
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    so you get \[\frac{1-\cos^2(x)}{\cos(x)}=\frac{\sin^2(x)}{\cos(x)}=\frac{\sin(x)}{\cos(x)}\times \sin(x)\]

  28. amistre64
    • 4 years ago
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    :) good luck

  29. anonymous
    • 4 years ago
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    and we love you!

  30. anonymous
    • 4 years ago
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    okayy, another question. given that tanx = p, find an expression, in terms of p, for cosec^2 x. GO.

  31. amistre64
    • 4 years ago
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    gone!! ;)

  32. anonymous
    • 4 years ago
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    ??

  33. anonymous
    • 4 years ago
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    oh i see what they want

  34. anonymous
    • 4 years ago
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    so is it 1 + 1/p^2 ??

  35. anonymous
    • 4 years ago
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    i don't think so let me draw a triangle. you should draw one too

  36. anonymous
    • 4 years ago
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    |dw:1327848673010:dw|

  37. anonymous
    • 4 years ago
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    there is my triangle, the triangle where tan(x) = p

  38. anonymous
    • 4 years ago
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    now you need the hypotenuse which you find by pythagoras

  39. anonymous
    • 4 years ago
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    |dw:1327848731602:dw|

  40. anonymous
    • 4 years ago
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    hypotenuse is \[\sqrt{p^2+1}\] you are looking for \[\csc(x)\] and that is \[\frac{\sqrt{p^2+1}}{p}\]

  41. anonymous
    • 4 years ago
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    nonononon. See, cosec^2x = 1 + cot^2x = 1 + 1/tan^2x = 1 + 1/p^2, right??

  42. anonymous
    • 4 years ago
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    oh sorry you want cosecant squared, so it is the square of what i wrote, which is \[\frac{p^2+1}{p^2}=1+\frac{1}{p^2}\] yes you have it!

  43. anonymous
    • 4 years ago
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    i forgot about the square, sorry

  44. anonymous
    • 4 years ago
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    hahahahah, studd. Since you love me, you have to do me a favour. You have to join my group.

  45. anonymous
    • 4 years ago
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    group of what?

  46. anonymous
    • 4 years ago
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    my openstudy group. I created it B) It's called, wait for it, Herp-Derp.

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