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Hannah_Ahn

  • 2 years ago

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

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

  3. myininaya
    • 2 years ago
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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

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

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

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

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

  8. meverett04
    • 2 years ago
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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

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

  10. myininaya
    • 2 years ago
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    Yes!

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

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

  13. meverett04
    • 2 years ago
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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

  14. Hannah_Ahn
    • 2 years ago
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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 ???

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

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

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

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

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

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

  21. Hannah_Ahn
    • 2 years ago
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    yes

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

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

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

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

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

  27. TuringTest
    • 2 years ago
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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?

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

  29. TuringTest
    • 2 years ago
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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.

  30. Hannah_Ahn
    • 2 years ago
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    oh hahahaha gotcha

  31. TuringTest
    • 2 years ago
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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.

  32. Hannah_Ahn
    • 2 years ago
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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

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

  34. TuringTest
    • 2 years ago
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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?

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

  36. Hannah_Ahn
    • 2 years ago
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    yes!

  37. TuringTest
    • 2 years ago
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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.

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

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

  40. TuringTest
    • 2 years ago
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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?

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

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

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

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

  45. Hannah_Ahn
    • 2 years ago
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    THANKS! yayayayayay!

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

  47. myininaya
    • 2 years ago
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    Turing is awesome!

  48. TuringTest
    • 2 years ago
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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!

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

  50. Hannah_Ahn
    • 2 years ago
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    myinininaya :)

  51. TuringTest
    • 2 years ago
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    Yay myin!

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

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

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