Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

kryton1212

  • one year ago

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

  • This Question is Closed
  1. gerryliyana
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sinΘ-2cosΘ = (sin Θ/ cos Θ) -(2cos Θ/cosΘ) = tan Θ - 2 = -4/3 - 2 = ???

  2. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why we need to divided by cosΘ??

  3. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you can't just divide by cos theta, you need to multiply also.

  4. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i have learnt sec yet, i have learnt sin cos tan only so far

  6. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1365910549853:dw|

  7. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    there is a picture of an angle whose tangent is \(\frac{4}{3}\)

  8. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you need the third side, which you get via pythagoras or by remembering the 3 - 4 - 5 right triangle

  9. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1365910630356:dw|

  10. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and?

  11. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    rather easy if you draw a triangle otherwise a pita

  14. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2?

  15. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah,2

  17. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank you both of you

  18. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and the next questions is : [1/cosθ + tanθ](1-sinθ)=?

  19. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    write tan = sin/cos.

  20. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    how you got -cos theta ?

  22. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    more specifically, how you got that - *minus*

  23. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[(\frac{1}{a}+\frac{b}{a})(1-b)=\frac{1-b^2}{a}\]

  24. Azteck
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large \tan \theta=-\frac{4}{3}\] \[\large \theta =126`52'\] \[\sin \theta -2cos\theta=\sin 126`52' -2\cos126`52'\] \[\large =2\]

  25. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1365911248263:dw|

  27. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ummm, may be i got something wrong

  28. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the algebra is easier if you put \(\cos(\theta)=a\) and \(\sin(\theta)=b\)

  29. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[(\frac{1}{\cos(\theta)}+\tan(\theta))(1-\sin(\theta))\] becomes the more simple to compute \[(\frac{1}{a}+\frac{b}{a})(1-b)\]

  31. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    or if you prefer \[(\frac{1+b}{a})(1-b)\]

  32. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    now you get \(\frac{1-b^2}{a}\) almost instantly

  33. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohhh, i write sin^(2)θ-1 ....

  34. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    actually \(1-\sin^2(\theta)\) for the numerator

  35. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i want to ask (1+sinθ)(1-sinθ)=1-sin^(2)θ??

  36. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes^

  37. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \((a+b)(a-b)=a^2-b^2\)

  38. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohhhh...then the answer is cosθ

  39. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, absolutely.

  40. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and the next question: If cosθ=1/k and 0°≤θ≤90°, then -tan(90°-θ)=?

  41. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1365911889934:dw| find the unknown side by using pythagoras theorem ?

  42. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    k^(2)-1

  43. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes.

  44. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1365911994361:dw| now find (tan 90-theta) = opposite side / adjacent side

  45. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry, that should read \(\large k^2-1\)

  46. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    tan(90-theta)=-1/tan theta?

  47. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you can use that identity , its just another way to solve the same problem.

  48. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you'll get the same answer using any of the 2 methods.

  49. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[-\frac{ 1 }{ \sqrt{k ^{2}}-1 }\] ??

  50. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  51. kryton1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks

  52. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    welcome ^_^

  53. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.