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natasha.aries Group Title

Can someone PLEASE help me with 14d. and 15a. I have my pre-cal final, and I still have no idea how to do some ?s!

  • 2 years ago
  • 2 years ago

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  1. natasha.aries Group Title
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    • 2 years ago
  2. caesar27 Group Title
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    14. C 15.B

    • 2 years ago
  3. natasha.aries Group Title
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    what?

    • 2 years ago
  4. natasha.aries Group Title
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    does 15 d have an actual answer?

    • 2 years ago
  5. natasha.aries Group Title
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    i mean a lol

    • 2 years ago
  6. TweT226 Group Title
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    tan^2(x) - sin^2(x) sin^2(x)/cos^2(x) - sin^2(x) [sin^2(x) - cos^2(x)sin^2(x)]/cos^2(x) sin^2(x)(1 - cos^2(x))/cos^2(x) sin^2(x)sin^2(x)/cos^2(x) tan^2(x)sin^2(x)

    • 2 years ago
  7. Hero Group Title
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    \[\tan^2x - \sin^2x = \tan^2x\sin^2x\] \[\frac{\tan^2x - \sin^2x}{\sin^2x} = \tan^2x\] \[\frac{\tan^2x}{\sin^2x} - \frac{\sin^2x}{\sin^2x} = \tan^2x\] \[\frac{\tan^2x}{\sin^2x} - 1 = \tan^2x\] \[\frac{\tan^2x}{\sin^2x} = \tan^2x + 1\] \[\frac{\tan^2x}{\sin^2x} = \frac{\sin^2x}{\cos^2x} + \frac{\cos^2x}{\cos^2x}\] \[\frac{\tan^2x}{\sin^2x} = \frac{\sin^2x+\cos^2x}{\cos^2x}\] \[\frac{\tan^2x}{\sin^2x} = \frac{1}{\cos^2x}\] \[\tan^2x = \frac{\sin^2x}{\cos^2x}\] \[\tan^2x = \tan^2x\]

    • 2 years ago
  8. natasha.aries Group Title
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    but in this kind of problem both sides have to equal to one another, so in this casem @TweT226 is right. but why did you subtract it by 1?

    • 2 years ago
  9. natasha.aries Group Title
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    but thank you for all the work @Hero your work was beautiful!

    • 2 years ago
  10. natasha.aries Group Title
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    but the answer has to be equal to tan^2xsin^2x

    • 2 years ago
  11. natasha.aries Group Title
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    but that is the answer, tan^xsin^2x

    • 2 years ago
  12. natasha.aries Group Title
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    haha yes youre right. lol thank you!

    • 2 years ago
  13. natasha.aries Group Title
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    do u know how in his way why he subtracted by one though?

    • 2 years ago
  14. TweT226 Group Title
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    mine or hero?

    • 2 years ago
  15. natasha.aries Group Title
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    yours

    • 2 years ago
  16. cesckay Group Title
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    14d: work from the right hand side

    • 2 years ago
  17. cesckay Group Title
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    15a. \[\sin ( x - \Theta) = sinx \cos \theta - \sin \theta cosx \] then solve and put theta = pi \[\cos \Pi = -1 \] \[\sin \Pi = 0\] Put that into the equation and obtain -sinx

    • 2 years ago
  18. natasha.aries Group Title
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    i just figured out 15a! thanks!

    • 2 years ago
  19. natasha.aries Group Title
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    do u know how to change something in polar form? like sqrt(3) + i?

    • 2 years ago
  20. cesckay Group Title
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    14d. i will solve this from the right hand side of the equation: \[\tan ^2 x \sin^2x = \frac{ \sin^2x }{ \cos^2x } \times \left( \sin^2x \right)\] \[= \frac{ \sin^2x }{ \cos^2x } \times ( 1 - \cos^2x)\]

    • 2 years ago
  21. cesckay Group Title
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    solve the equation and you will obtain the expression for the RHS

    • 2 years ago
  22. natasha.aries Group Title
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    i dont understand why you did (1−cos2x)

    • 2 years ago
  23. cesckay Group Title
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    the identity of sin2x = 1-cos2x

    • 2 years ago
  24. natasha.aries Group Title
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    so whenever you see that u change it to (1−cos2x)?

    • 2 years ago
  25. natasha.aries Group Title
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    cause i thought that only applied when you added sin2x + cos2x

    • 2 years ago
  26. cesckay Group Title
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    well you can determine what you will need to convert depending on whats on the other side of the equation. right??

    • 2 years ago
  27. cesckay Group Title
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    solve and let me know how it goes. OK :)

    • 2 years ago
  28. natasha.aries Group Title
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    okay! hold on!

    • 2 years ago
  29. natasha.aries Group Title
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    no it still didnt realy work out... here let me try again

    • 2 years ago
  30. cesckay Group Title
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    alright, i will finish the solution

    • 2 years ago
  31. natasha.aries Group Title
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    can i take a picture of what im doing so u can tell me what im doing wrong

    • 2 years ago
  32. cesckay Group Title
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    \[\frac{ \sin^2x-\cos^xsin^2x }{ \cos^2x }\] \[\frac{ \sin^2x }{ \cos^2x } - \frac{ \cos^2xsin^2x }{ \cos^2x }\] \[\frac{ \sin^2x }{ \cos^2x }= \tan^2x\] and : \[-\frac{ \cos^2xsin^2x }{ \cos^2x }= -\sin^2x\]

    • 2 years ago
  33. cesckay Group Title
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    sure you can! lemme check

    • 2 years ago
  34. natasha.aries Group Title
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    okay cool! awh thanks!

    • 2 years ago
  35. natasha.aries Group Title
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    hold onn!

    • 2 years ago
  36. natasha.aries Group Title
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    • 2 years ago
  37. cesckay Group Title
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    I worked on from the RHS.. its much easier to figure out from the RHS but I will try to do it from the LHS as well. it should be the same answer. In the meantime, you can check the solutions i put up

    • 2 years ago
  38. natasha.aries Group Title
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    okay thank youuu

    • 2 years ago
  39. natasha.aries Group Title
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    when u did sin2x−cosxsin2xcos2x why isnt it sin2x−cos2xsin2xcos2x?

    • 2 years ago
  40. natasha.aries Group Title
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    why is it sin2x−cosxsin2xcos2x

    • 2 years ago
  41. cesckay Group Title
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    from the top, we can see that its just a multiplication of sin2X and whats in the bracket

    • 2 years ago
  42. natasha.aries Group Title
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    yeah so shouldnt it be cos2x

    • 2 years ago
  43. cesckay Group Title
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    hang on, I'll upload the solution

    • 2 years ago
  44. cesckay Group Title
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    • 2 years ago
  45. cesckay Group Title
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    i know right.. :D got it from my computer.. mind reading it backwards?? :/

    • 2 years ago
  46. natasha.aries Group Title
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    loll yeah ill try haha

    • 2 years ago
  47. natasha.aries Group Title
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    or u could take a picture on ur phone or something and email it to yo self! and did u find a common denominator?

    • 2 years ago
  48. cesckay Group Title
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    ook yeah try this

    • 2 years ago
  49. cesckay Group Title
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    LaTex is horrible, i'm much of a "down on paper" breed//

    • 2 years ago
  50. natasha.aries Group Title
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    ugh ijust dont understand!!! im sorry :(

    • 2 years ago
  51. cesckay Group Title
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    Oh.. ok lets try that again:

    • 2 years ago
  52. cesckay Group Title
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    Solve from the Right Hand Side.. Did you get the solutions up to a certain point?

    • 2 years ago
  53. natasha.aries Group Title
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    yes, and each time i do it i stop at that point

    • 2 years ago
  54. cesckay Group Title
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    where exactly?

    • 2 years ago
  55. natasha.aries Group Title
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    at the sin2x - sin2xcos2x / cos2x

    • 2 years ago
  56. natasha.aries Group Title
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    idk what to d after that because i get tan2x- sin2xcos2x

    • 2 years ago
  57. cesckay Group Title
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    Oh there' the mistake right there.. you have to divide sin2x by Cos2x as well as divide by -sin2xcos2x by cos2x

    • 2 years ago
  58. natasha.aries Group Title
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    yeah but when i do it i dont have both of those cases thats all i have left

    • 2 years ago
  59. cesckay Group Title
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    Ok i will post the solution from that point

    • 2 years ago
  60. cesckay Group Title
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    \[\frac{ \sin^2x-\sin^2xcos^2x }{ \cos^2x }\]

    • 2 years ago
  61. cesckay Group Title
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    \[\frac{ \sin^2x }{ \cos^2x }-\frac{ \sin^2xcos^2x }{ \cos^2x }\]

    • 2 years ago
  62. cesckay Group Title
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    solve that..

    • 2 years ago
  63. natasha.aries Group Title
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    thanks @Hero !!!

    • 2 years ago
  64. natasha.aries Group Title
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    when i solve that yeah i get the answer but i dont see how u got that :(

    • 2 years ago
  65. natasha.aries Group Title
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    lol i agree with you but my teacher wont give me credit unless i do it that way

    • 2 years ago
  66. natasha.aries Group Title
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    she told us if we do it any other way we wont get credit

    • 2 years ago
  67. cesckay Group Title
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    You are supposed to get credits for either solving from LHS or RHS

    • 2 years ago
  68. natasha.aries Group Title
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    yeah i agree im just confused

    • 2 years ago
  69. natasha.aries Group Title
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    i mean i honestly dont mind this way, just this problem..

    • 2 years ago
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