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kryton1212

  • 2 years ago

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

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  1. gerryliyana
    • 2 years ago
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    sinΘ-2cosΘ = (sin Θ/ cos Θ) -(2cos Θ/cosΘ) = tan Θ - 2 = -4/3 - 2 = ???

  2. kryton1212
    • 2 years ago
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    why we need to divided by cosΘ??

  3. hartnn
    • 2 years ago
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    you can't just divide by cos theta, you need to multiply also.

  4. hartnn
    • 2 years ago
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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.

  5. kryton1212
    • 2 years ago
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    i have learnt sec yet, i have learnt sin cos tan only so far

  6. satellite73
    • 2 years ago
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    |dw:1365910549853:dw|

  7. satellite73
    • 2 years ago
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    there is a picture of an angle whose tangent is \(\frac{4}{3}\)

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

  9. satellite73
    • 2 years ago
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    |dw:1365910630356:dw|

  10. kryton1212
    • 2 years ago
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    and?

  11. satellite73
    • 2 years ago
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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

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

  13. satellite73
    • 2 years ago
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    rather easy if you draw a triangle otherwise a pita

  14. kryton1212
    • 2 years ago
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    2?

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

  16. satellite73
    • 2 years ago
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    yeah,2

  17. kryton1212
    • 2 years ago
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    thank you both of you

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

  19. hartnn
    • 2 years ago
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    write tan = sin/cos.

  20. kryton1212
    • 2 years ago
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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

  21. hartnn
    • 2 years ago
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    how you got -cos theta ?

  22. hartnn
    • 2 years ago
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    more specifically, how you got that - *minus*

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

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

  25. satellite73
    • 2 years ago
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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 \[\frac{\cos^2(\theta)}{\cos(\theta)}=\cos(\theta)\]

  26. kryton1212
    • 2 years ago
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    |dw:1365911248263:dw|

  27. kryton1212
    • 2 years ago
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    ummm, may be i got something wrong

  28. satellite73
    • 2 years ago
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    the algebra is easier if you put \(\cos(\theta)=a\) and \(\sin(\theta)=b\)

  29. satellite73
    • 2 years ago
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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

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

  31. satellite73
    • 2 years ago
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    or if you prefer \[(\frac{1+b}{a})(1-b)\]

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

  33. kryton1212
    • 2 years ago
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    ohhh, i write sin^(2)θ-1 ....

  34. satellite73
    • 2 years ago
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    actually \(1-\sin^2(\theta)\) for the numerator

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

  36. hartnn
    • 2 years ago
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    yes^

  37. hartnn
    • 2 years ago
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    \((a+b)(a-b)=a^2-b^2\)

  38. kryton1212
    • 2 years ago
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    ohhhh...then the answer is cosθ

  39. hartnn
    • 2 years ago
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    yes, absolutely.

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

  41. hartnn
    • 2 years ago
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    |dw:1365911889934:dw| find the unknown side by using pythagoras theorem ?

  42. kryton1212
    • 2 years ago
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    k^(2)-1

  43. hartnn
    • 2 years ago
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    yes.

  44. hartnn
    • 2 years ago
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    |dw:1365911994361:dw| now find (tan 90-theta) = opposite side / adjacent side

  45. hartnn
    • 2 years ago
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    sorry, that should read \(\large k^2-1\)

  46. kryton1212
    • 2 years ago
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    tan(90-theta)=-1/tan theta?

  47. hartnn
    • 2 years ago
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    you can use that identity , its just another way to solve the same problem.

  48. hartnn
    • 2 years ago
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    you'll get the same answer using any of the 2 methods.

  49. kryton1212
    • 2 years ago
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    \[-\frac{ 1 }{ \sqrt{k ^{2}}-1 }\] ??

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

  51. kryton1212
    • 2 years ago
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    thanks

  52. hartnn
    • 2 years ago
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    welcome ^_^

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