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Rosy95

  • 3 years ago

PreCal Help Please..I am so stuck on what to do

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  1. Rosy95
    • 3 years ago
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    Use the equation below to find the value of sin x. \[\frac{ \tan x }{ \cot x } - \frac{ \sec x }{ \cos x }= \frac{ 2 }{ \csc x }\]

  2. hartnn
    • 3 years ago
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    write everything in terms of sin and cos

  3. Rosy95
    • 3 years ago
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    That is what I am not understanding. I have a really hard time writting out what it is supposed to be once you start changing everything.

  4. Hero
    • 3 years ago
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    Hint: Write it this way. Then simply change everything to sin and cos \[\tan(x) \div \cot(x) - \sec(x) \div \cos(x) = 2 \div \csc(x)\]

  5. Rosy95
    • 3 years ago
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    So how would you change it?

  6. Hero
    • 3 years ago
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    \[\frac{\sin(x)}{\cos(x)} \div \frac{\cos(x)}{\sin(x)} - \frac{1}{\cos(x)} \div \cos(x) = 2 \div \frac{1}{\sin(x)}\]

  7. Rosy95
    • 3 years ago
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    so could the first two fracts cancel out?

  8. Hero
    • 3 years ago
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    Not exactly

  9. Hero
    • 3 years ago
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    How do you divide: \[\frac{1}{4} \div \frac{7}{6}\]

  10. Rosy95
    • 3 years ago
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    Why not? BTW i missed the lessons over this stuff so i am really lost on almost all of it.

  11. Hero
    • 3 years ago
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    Answer the previous question. How do you divide that?

  12. Rosy95
    • 3 years ago
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    Oh wait I remember

  13. Rosy95
    • 3 years ago
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    you switch the #'s of the second fraction so it is now 6/7 and you change divide to multiply. you now have 1/4 X 6/7 which is 6/28 simplified to 3/14

  14. Hero
    • 3 years ago
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    Yes, so do that same thing with the trig equation

  15. Rosy95
    • 3 years ago
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    so it is now \[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos x }\div \cos x=2\div \frac{ 1 }{ \sin x }\]??

  16. Hero
    • 3 years ago
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    Yes, but do the rest of them. Change all of the divisions into multiplication.

  17. Rosy95
    • 3 years ago
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    so it looks like\[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ cosx }\times \frac{ cosx }{ 1 }=2\times \frac{ sinx }{ 1 }\]

  18. Rosy95
    • 3 years ago
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    Did I do that right?

  19. Rosy95
    • 3 years ago
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    If so, wouldn't 1/cosx times cosx/1 cancel out?

  20. Hero
    • 3 years ago
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    no, you did the cos(x)/1 incorrectly. cos(x) is already equal to cos(x)/1. You have to flip it.

  21. Rosy95
    • 3 years ago
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    oh duh ok so then it is \[\frac{ \sin^2 x}{ \cos^2x } -\frac{ 1 }{ \cos x }\times \frac{ 1 }{ \cos x }= 2\times \frac{ sinx }{ 1 }\]

  22. Hero
    • 3 years ago
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    Yes

  23. Rosy95
    • 3 years ago
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    so do the mult fractions combine to get -1/cos^2x

  24. Rosy95
    • 3 years ago
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    so then what do you do?

  25. Hero
    • 3 years ago
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    Yes, so what do you have after that?

  26. Rosy95
    • 3 years ago
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    \[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos^2x }= 2\times \frac{ sinx }{ 1 }\]

  27. Hero
    • 3 years ago
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    Okay, do you know how to combine fractions on the left side?

  28. Rosy95
    • 3 years ago
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    couldn't you turn 2x(sinx/1) into just 2sinx? and Im not sure what to do on the other side

  29. Rosy95
    • 3 years ago
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    What do I do next?

  30. Hero
    • 3 years ago
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    On the left side, you have to combine fractions if they have the same denominator.

  31. Hero
    • 3 years ago
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    For example: \[\frac{1}{4} + \frac{x}{4} = \frac{1 + x}{4}\]

  32. Hero
    • 3 years ago
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    The general rule for combining fractions with the same denominator is: \[\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}\]

  33. Rosy95
    • 3 years ago
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    so it is now \[\frac{ \sin^2x-1 }{ \cos^2x }=2sinx\]

  34. Rosy95
    • 3 years ago
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    or is the 2sin x supposed to be sin^2x

  35. Hero
    • 3 years ago
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    Very good

  36. Hero
    • 3 years ago
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    Now change the 1 to \(\sin^2x + \cos^2x\)

  37. Hero
    • 3 years ago
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    No, you have it right. 2sin(x)

  38. Rosy95
    • 3 years ago
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    wait, you put sin^2x + cos^2x instead of one? what is the new equation?

  39. Hero
    • 3 years ago
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    Yes, you put \(\sin^2x + \cos^2x\) in place of 1 because they are equal.

  40. Hero
    • 3 years ago
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    What do you get afterwards?

  41. Rosy95
    • 3 years ago
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    \[\frac{ \sin^2x-\sin^2x+\cos^2x }{ \cos^2x }=2sinx\] then you can subtract the sin^2x to get \[\frac{ \cos^2x }{ \cos^2x }=2sinx\] then if thats the case you would just have 2sinx

  42. Hero
    • 3 years ago
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    Actually, if you do it correctly you get: \[-\frac{ \cos^2x }{ \cos^2x }=2\sin x\]

  43. Hero
    • 3 years ago
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    Which simplifies to \[-1 = 2 \sin x\] Remember that we are still solving for x

  44. Rosy95
    • 3 years ago
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    so sinx is -2

  45. Hero
    • 3 years ago
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    You were supposed to get this after substituting the 1: \[\frac{ \sin^2x-(\sin^2x+\cos^2x) }{ \cos^2x }=2\sin x\] Then get : \[\frac{ \sin^2x-\sin^2x-cos^2x }{ \cos^2x }=2\sin x\]

  46. Rosy95
    • 3 years ago
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    or no sinx would be -1/2

  47. Hero
    • 3 years ago
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    Ultimately you end up isolating the sin(x) on the right side: \[-\frac{1}{2} = \sin x\]

  48. Hero
    • 3 years ago
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    Now all you have to do is take the inverse sin of both sides to get \[\sin^{-1}\left(-\frac{1}{2}\right) = x\]

  49. Hero
    • 3 years ago
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    So evaluate the left side using your calc. Let me know what you get. Also make sure it is in radian mode.

  50. Rosy95
    • 3 years ago
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    -.52

  51. Rosy95
    • 3 years ago
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    And thank you for your help :)

  52. Hero
    • 3 years ago
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    Well, in exact mode, you should have gotten \[x = -\frac{\pi}{6}\]

  53. Hero
    • 3 years ago
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    Nevertheless:\[-\frac{\pi}{6} \approx -.52\]

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