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Hannah_Ahn Group Title

Prove the identity. \[1 \over 1+\sin \theta\]=\[\sec ^{2} \theta\] -\[ \tan \theta \over \cos \theta\]

  • 2 years ago
  • 2 years ago

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  1. nikvist Group Title
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    \[\frac{1}{1+\sin\theta}=\frac{1-\sin\theta}{1-\sin^2\theta}=\frac{1-\sin\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}-\frac{\sin\theta}{\cos^2\theta}=\sec^2\theta-\frac{\tan\theta}{\cos\theta}\]

    • 2 years ago
  2. meverett04 Group Title
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    Let us know if you need more help with this ...

    • 2 years ago
  3. myininaya Group Title
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    \[\frac{1}{1+\sin(x)} \cdot \frac{1-\sin(x)}{1-\sin(x)}=\frac{1-\sin(x)}{1-\sin^2(x)}=\frac{1-\sin(x)}{\cos^2(x)}\] \[=\frac{1}{\cos^2(x)}-\frac{\sin(x)}{\cos^2(x)}=\sec^2(x)-\frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)}\] \[=\sec^2(x)-\tan(x) \cdot \frac{1}{\cos(x)}=\sec^2(x)-\frac{\tan(x)}{\cos(x)}\] but nikvist already did this i might in a little extra steps if you wanted to see those

    • 2 years ago
  4. Hannah_Ahn Group Title
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    why did you muliply 1 over cos theta?

    • 2 years ago
  5. Hannah_Ahn Group Title
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    it's so interesting how we can do that :p

    • 2 years ago
  6. myininaya Group Title
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    \[\cos^2(x)=\cos(x) \cdot \cos(x)\]

    • 2 years ago
  7. Hannah_Ahn Group Title
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    oh i missed that. thanks guys!!!! :)

    • 2 years ago
  8. meverett04 Group Title
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    I don't think they multiplied 1 over cos theta .... I think they used the Pythagorean Identity sin^2 x + cos^2 x = 1 or 1 - sin^2 x = cos^2 x You will use this formula many many many times in trig

    • 2 years ago
  9. Hannah_Ahn Group Title
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    do you guys memorize these formulas?

    • 2 years ago
  10. myininaya Group Title
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    Yes!

    • 2 years ago
  11. Hannah_Ahn Group Title
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    I guess I have to do lots of these questions to master this chapter..

    • 2 years ago
  12. Hannah_Ahn Group Title
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    tehehehehe aywy thanks so much lifesaverss :)

    • 2 years ago
  13. meverett04 Group Title
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    I only memorize the ones I see over and over again. For trig, there are just a few that I memorize sin^2 x + cos^2 x = 1 is one of them, but I have used it so many times, it is no longer memorized, I know it like my name meverett lol

    • 2 years ago
  14. Hannah_Ahn Group Title
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    I am having troubles with every single questions.. wowo.... how do i prove this one.? cot theta over csc theta - 1 equals csc theta + 1 over cot theta ???

    • 2 years ago
  15. Hannah_Ahn Group Title
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    haha I don't even see the first step...

    • 2 years ago
  16. meverett04 Group Title
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    For cot x and csc x I would change them into sines and cosines, can you do this?

    • 2 years ago
  17. Hannah_Ahn Group Title
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    ......?

    • 2 years ago
  18. Hannah_Ahn Group Title
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    for csc x 1 over sin x

    • 2 years ago
  19. meverett04 Group Title
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    No .... for cot theta over csc theta

    • 2 years ago
  20. TuringTest Group Title
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    is it\[{\cot\theta\over\csc\theta-1}={\csc\theta+1\over\cot\theta}\]??

    • 2 years ago
  21. Hannah_Ahn Group Title
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    yes

    • 2 years ago
  22. meverett04 Group Title
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    I would change cot (theta) = cos(theta) / sin (theta) csc(theta) = 1/sin(theta) for all cot and csc

    • 2 years ago
  23. Hannah_Ahn Group Title
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    yes I did that just now :)

    • 2 years ago
  24. TuringTest Group Title
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    Yes meverett is right, though this one is a little tricky. I had to change both sides.

    • 2 years ago
  25. Hannah_Ahn Group Title
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    and multiply each side by 1 over sin theta - 1?

    • 2 years ago
  26. TuringTest Group Title
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    sorry no... let me fix that.

    • 2 years ago
  27. TuringTest Group Title
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    \[{{\cos x\over\sin x}\over{1\over\sin x}-1}={{{1\over\sin x}+1}\over{\cos x\over\sin x}}\]right?

    • 2 years ago
  28. Hannah_Ahn Group Title
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    uh... i don't get the 2nd part. you flip them?

    • 2 years ago
  29. TuringTest Group Title
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    no, that's just the right side of the formula with csc written as 1/sin, and cot written as cos/sin.

    • 2 years ago
  30. Hannah_Ahn Group Title
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    oh hahahaha gotcha

    • 2 years ago
  31. TuringTest Group Title
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    I changed both sides to their elementary functions is all. I think there are too many sines in this equation; makes my head hurt. But it looks like if we multiply each side by\[\frac{\sin x}{\sin x}\]everything will be a bit simpler, so try that. See if you can write what you get.

    • 2 years ago
  32. Hannah_Ahn Group Title
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    uhh I am sorry I don't want your head to hurt.. but I don't know how to do it.. :S

    • 2 years ago
  33. TuringTest Group Title
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    no prob, here it is...

    • 2 years ago
  34. TuringTest Group Title
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    \[\frac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x}-1}\frac{\sin x}{\sin x}=\frac{\frac{1}{\sin x}+1}{\frac{\cos x}{\sin x}}\frac{\sin x}{\sin x}\]\[\frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x}\]reasonable?

    • 2 years ago
  35. TuringTest Group Title
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    we just multiplied the top and bottom of each fraction by sinx

    • 2 years ago
  36. Hannah_Ahn Group Title
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    yes!

    • 2 years ago
  37. TuringTest Group Title
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    Ok, so now I'm going to say that my mission is to get the left to look like the right. I'm not going to change the right anymore, just try to get the left to look like it. So see if you can multiply the left by (1+sinx)/(1+sinx). That should make you see something.

    • 2 years ago
  38. TuringTest Group Title
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    it should change the denominator in a familiar way...

    • 2 years ago
  39. Hannah_Ahn Group Title
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    i got it 'till \[cosx + sinxcosx \over 1-\sin ^{2} x\]

    • 2 years ago
  40. TuringTest Group Title
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    That's good. You didn't need to multiply out the top, you could have left it as cosx(1+sinx). You'll see why in this step. Look at the denominator. What is that identity?

    • 2 years ago
  41. Hannah_Ahn Group Title
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    \[\cos ^{2}x\]

    • 2 years ago
  42. TuringTest Group Title
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    so we can divide out our fraction by a cosx to make it...?

    • 2 years ago
  43. Hannah_Ahn Group Title
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    yes!!!!!!!!!!!!!!!!!!!!!!!!!!

    • 2 years ago
  44. TuringTest Group Title
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    there ya go :)

    • 2 years ago
  45. Hannah_Ahn Group Title
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    THANKS! yayayayayay!

    • 2 years ago
  46. Hannah_Ahn Group Title
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    you are really qualified as an awesome tutor! so patient.. :') i wish you the bestest luck !

    • 2 years ago
  47. myininaya Group Title
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    Turing is awesome!

    • 2 years ago
  48. TuringTest Group Title
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    so what was the strategy? 1)convert all tan, cot, sec, and csc into terms of sin and cos. 2)Simplify if possible. Remember that mertsj was right, you cannot "divide both sides" or "multiply both sides" in identities. You need to work each side separately. 3)Look for ways make things like (1-sin^2x) so we can change it to cos^2x. That was all my strategy was, plus practice. Happy New Year!

    • 2 years ago
  49. Hannah_Ahn Group Title
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    so are you though, your every each step helped me alot :)

    • 2 years ago
  50. Hannah_Ahn Group Title
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    myinininaya :)

    • 2 years ago
  51. TuringTest Group Title
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    Yay myin!

    • 2 years ago
  52. Hannah_Ahn Group Title
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    LOVELY! happy NEW year :) bye ttyl

    • 2 years ago
  53. AravindG Group Title
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    nikvist can u hlp me??

    • 2 years ago
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