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kryton1212 Group Title

If tanΘ= -4/3 and Θ lies in the second quadrant, then sinΘ-2cosΘ=?

  • one year ago
  • one year ago

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  1. gerryliyana Group Title
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    sinΘ-2cosΘ = (sin Θ/ cos Θ) -(2cos Θ/cosΘ) = tan Θ - 2 = -4/3 - 2 = ???

    • one year ago
  2. kryton1212 Group Title
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    why we need to divided by cosΘ??

    • one year ago
  3. hartnn Group Title
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    you can't just divide by cos theta, you need to multiply also.

    • one year ago
  4. hartnn Group Title
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    i would suggest you find sec theta first, from the identity sec^2 theta = 1+ tan^2 theta and then, cos = 1/ sec, sin = tan * cos.

    • one year ago
  5. kryton1212 Group Title
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    i have learnt sec yet, i have learnt sin cos tan only so far

    • one year ago
  6. satellite73 Group Title
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    |dw:1365910549853:dw|

    • one year ago
  7. satellite73 Group Title
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    there is a picture of an angle whose tangent is \(\frac{4}{3}\)

    • one year ago
  8. satellite73 Group Title
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    you need the third side, which you get via pythagoras or by remembering the 3 - 4 - 5 right triangle

    • one year ago
  9. satellite73 Group Title
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    |dw:1365910630356:dw|

    • one year ago
  10. kryton1212 Group Title
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    and?

    • one year ago
  11. satellite73 Group Title
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    now you see that \(\sin(\theta)=\frac{4}{5}\) and \(\cos(\theta)=\frac{3}{5}\) except that since you are in quadrant II you have \(\cos(\theta)=-\frac{3}{5}\) because in quadrant II cosine is negative

    • one year ago
  12. satellite73 Group Title
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    you want \[\sin(\theta)-2\cos(\theta)\] and you know the numbers that you need \[\frac{4}{5}-2\times (-\frac{3}{5})=\frac{4}{5}+\frac{6}{5}\]

    • one year ago
  13. satellite73 Group Title
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    rather easy if you draw a triangle otherwise a pita

    • one year ago
  14. kryton1212 Group Title
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    2?

    • one year ago
  15. hartnn Group Title
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    Alternate way to look at the same thing, |dw:1365910821672:dw| so the angle inside triangle is \(\pi - \theta\) so, \(\cos(\pi-\theta)=3/5, -\cos \theta=3/5 , \cos \theta = -3/5\)

    • one year ago
  16. satellite73 Group Title
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    yeah,2

    • one year ago
  17. kryton1212 Group Title
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    thank you both of you

    • one year ago
  18. kryton1212 Group Title
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    and the next questions is : [1/cosθ + tanθ](1-sinθ)=?

    • one year ago
  19. hartnn Group Title
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    write tan = sin/cos.

    • one year ago
  20. kryton1212 Group Title
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    i got the answer -cosθ but there are only five choices: A. sinθ B. cosθ C. cos^(2)θ D. 1+sinθ E. sinθtanθ

    • one year ago
  21. hartnn Group Title
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    how you got -cos theta ?

    • one year ago
  22. hartnn Group Title
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    more specifically, how you got that - *minus*

    • one year ago
  23. satellite73 Group Title
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    \[(\frac{1}{a}+\frac{b}{a})(1-b)=\frac{1-b^2}{a}\]

    • one year ago
  24. Azteck Group Title
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    \[\large \tan \theta=-\frac{4}{3}\] \[\large \theta =126`52'\] \[\sin \theta -2cos\theta=\sin 126`52' -2\cos126`52'\] \[\large =2\]

    • one year ago
  25. satellite73 Group Title
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    in this case \(a=\cos(\theta)\) and \(b=\sin(\theta)\) so you get \[\frac{1-\sin^2(\theta)}{\cos(\theta)}\] since \(1-\sin^2(\theta)=\cos^2(\theta)\) you end up with \[\frac{\cos^2(\theta)}{\cos(\theta)}=\cos(\theta)\]

    • one year ago
  26. kryton1212 Group Title
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    |dw:1365911248263:dw|

    • one year ago
  27. kryton1212 Group Title
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    ummm, may be i got something wrong

    • one year ago
  28. satellite73 Group Title
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    the algebra is easier if you put \(\cos(\theta)=a\) and \(\sin(\theta)=b\)

    • one year ago
  29. satellite73 Group Title
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    then you won't get so confused when you multiply stuff out almost all the steps are algebra, there is very little trig here

    • one year ago
  30. satellite73 Group Title
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    \[(\frac{1}{\cos(\theta)}+\tan(\theta))(1-\sin(\theta))\] becomes the more simple to compute \[(\frac{1}{a}+\frac{b}{a})(1-b)\]

    • one year ago
  31. satellite73 Group Title
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    or if you prefer \[(\frac{1+b}{a})(1-b)\]

    • one year ago
  32. satellite73 Group Title
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    now you get \(\frac{1-b^2}{a}\) almost instantly

    • one year ago
  33. kryton1212 Group Title
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    ohhh, i write sin^(2)θ-1 ....

    • one year ago
  34. satellite73 Group Title
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    actually \(1-\sin^2(\theta)\) for the numerator

    • one year ago
  35. kryton1212 Group Title
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    i want to ask (1+sinθ)(1-sinθ)=1-sin^(2)θ??

    • one year ago
  36. hartnn Group Title
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    yes^

    • one year ago
  37. hartnn Group Title
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    \((a+b)(a-b)=a^2-b^2\)

    • one year ago
  38. kryton1212 Group Title
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    ohhhh...then the answer is cosθ

    • one year ago
  39. hartnn Group Title
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    yes, absolutely.

    • one year ago
  40. kryton1212 Group Title
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    and the next question: If cosθ=1/k and 0°≤θ≤90°, then -tan(90°-θ)=?

    • one year ago
  41. hartnn Group Title
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    |dw:1365911889934:dw| find the unknown side by using pythagoras theorem ?

    • one year ago
  42. kryton1212 Group Title
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    k^(2)-1

    • one year ago
  43. hartnn Group Title
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    yes.

    • one year ago
  44. hartnn Group Title
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    |dw:1365911994361:dw| now find (tan 90-theta) = opposite side / adjacent side

    • one year ago
  45. hartnn Group Title
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    sorry, that should read \(\large k^2-1\)

    • one year ago
  46. kryton1212 Group Title
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    tan(90-theta)=-1/tan theta?

    • one year ago
  47. hartnn Group Title
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    you can use that identity , its just another way to solve the same problem.

    • one year ago
  48. hartnn Group Title
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    you'll get the same answer using any of the 2 methods.

    • one year ago
  49. kryton1212 Group Title
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    \[-\frac{ 1 }{ \sqrt{k ^{2}}-1 }\] ??

    • one year ago
  50. hartnn Group Title
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    if you meant , \(\Large -\frac{ 1 }{ \sqrt{k ^{2}-1} }\) then you are correct.

    • one year ago
  51. kryton1212 Group Title
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    thanks

    • one year ago
  52. hartnn Group Title
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    welcome ^_^

    • one year ago
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