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natasha.aries

  • 3 years ago

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!

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

  3. natasha.aries
    • 3 years ago
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    what?

  4. natasha.aries
    • 3 years ago
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    does 15 d have an actual answer?

  5. natasha.aries
    • 3 years ago
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    i mean a lol

  6. TweT226
    • 3 years ago
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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)

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

  8. natasha.aries
    • 3 years ago
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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?

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

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

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

  12. natasha.aries
    • 3 years ago
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    haha yes youre right. lol thank you!

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

  14. TweT226
    • 3 years ago
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    mine or hero?

  15. natasha.aries
    • 3 years ago
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    yours

  16. cesckay
    • 3 years ago
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    14d: work from the right hand side

  17. cesckay
    • 3 years ago
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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

  18. natasha.aries
    • 3 years ago
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    i just figured out 15a! thanks!

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

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

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

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

  23. cesckay
    • 3 years ago
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    the identity of sin2x = 1-cos2x

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

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

  26. cesckay
    • 3 years ago
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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??

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

  28. natasha.aries
    • 3 years ago
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    okay! hold on!

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

  30. cesckay
    • 3 years ago
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    alright, i will finish the solution

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

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

  33. cesckay
    • 3 years ago
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    sure you can! lemme check

  34. natasha.aries
    • 3 years ago
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    okay cool! awh thanks!

  35. natasha.aries
    • 3 years ago
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    hold onn!

  36. natasha.aries
    • 3 years ago
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  37. cesckay
    • 3 years ago
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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

  38. natasha.aries
    • 3 years ago
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    okay thank youuu

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

  40. natasha.aries
    • 3 years ago
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    why is it sin2x−cosxsin2xcos2x

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

  42. natasha.aries
    • 3 years ago
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    yeah so shouldnt it be cos2x

  43. cesckay
    • 3 years ago
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    hang on, I'll upload the solution

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

  46. natasha.aries
    • 3 years ago
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    loll yeah ill try haha

  47. natasha.aries
    • 3 years ago
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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?

  48. cesckay
    • 3 years ago
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    ook yeah try this

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

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

  51. cesckay
    • 3 years ago
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    Oh.. ok lets try that again:

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

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

  54. cesckay
    • 3 years ago
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    where exactly?

  55. natasha.aries
    • 3 years ago
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    at the sin2x - sin2xcos2x / cos2x

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

  57. cesckay
    • 3 years ago
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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

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

  59. cesckay
    • 3 years ago
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    Ok i will post the solution from that point

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

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

  62. cesckay
    • 3 years ago
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    solve that..

  63. natasha.aries
    • 3 years ago
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    thanks @Hero !!!

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

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

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

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

  68. natasha.aries
    • 3 years ago
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    yeah i agree im just confused

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

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