A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Trig / Pre Cal/ identities Am I on the right path? Please guide me and do not give me the answers \[ (\tan x + \sin x)(1-cos x) = \sin^2 x \tan x \] Do I multiply like with foil starting with tan or do I turn tan into \( \frac{sin x}{cox x} \) and then use foil?

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

    foil! :-) i guess that will be great

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

    Just with tan \[ \tan x -(\tan x)(\cos x) + \sin x -(\sin x)(\cos x) \] correct so far?

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

    yep looks right :-)

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

    Ok, how do I approch it from here? do I turn all the tan into sin / cos? and work it out? \[ \tan x -(\tan x)(\cos x) + \sin x -(\sin x)(\cos x) \]

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

    yes convert 2nd tan to sin /cos and let me find notebook and a pencil :3 hehe

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

    make sure question is right :3 :-)

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

    Yes it is right :-)

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

    is it equal to tanxsinx or tan^2xsinx ??

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

    I have \[ \tan x -\sin x + \sin x -(\sin x)(\cos x) = \sin^2 x \tan x \] \[ \tan x -(\sin x)(\cos x) = \sin^2 x \tan x \]

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

    yeah so for the final answer i got sinx tanx so that's why ..hm are u sure it's sin square ??

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

    next step is to change tan to sin/cos

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

    Yes, it is sin^2 x tan x

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

    :( alright

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

    alright got it!! :-)

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

    hahah i'm soo happy :P

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

    \( \tan x -(\sin x)(\cos x) = \sin^2 x \tan x \) \( \frac{sin x}{cos x} -(\sin x)(\cos x) = \sin^2 x \tan x \) ????

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

    \( \huge\rm \frac{sin x}{cos x} -(\sin x)(\cos x) = \) ???? find common denominator :-)

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

    Got ya :-)

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

    one sec

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

    alright :-)

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

    I am using cos as the common denominator. that is correct?

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

    yes right

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

    Ok I got \[ \frac{sin x}{cos x} -(\sin x)(\cos x) = \sin^2 x \tan x \] \[ \frac{(sin x)(cos x)}{\cos x} -\frac{((\sin x))}{\cos x} \frac{\cos^2}{cos} = \sin^2 x \tan x \] correct???

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

    \[ \frac{(sin x)(\cancel{cos x})}{\cos x} -\frac{((\sin x))}{\cos x} \frac{\cos^2}{cos} = \sin^2 x \tan x \] multiply cosx/1 so when you multiply first fraction by cos both cos will cancel each other out

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

    so there isn't supposed to be any cosx \[ \frac{(sin x)}{\cos x} -\frac{((\sin x cos^2x))}{cos x} = \sin^2 x \tan x \] like this

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

    Oh right!!! Made a boo boo

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

    One sec, now we should have

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

    \[ \frac{(sin x)}{\cos x} -\frac{((\sin x cos^2x))}{cos x} = \sin^2 x \tan x \] \[ \frac{(sin x)}{\cos x} -\frac{((\sin x (1-\sin^2 x)2x))}{cos x} = \sin^2 x \tan x \] Ok I am lost

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

    \[ \frac{(sin x)}{\cos x} -\frac{((\sin x (1-cos^2 x))}{cos x} = \sin^2 x \tan x \]

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

    \[\cos x \times \frac{ \sin x }{ \cos x} - \frac{ sinx \cos x }{ \cos } \times \cos \] multiply |dw:1434231940649:dw|

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

    \[\cos x \times \frac{ \sin x }{ \cos x} - \frac{ sinx \cos x }{ \cos } \times \cos \] \[\huge\rm \frac{ \sin x - sinx cosx (cosx) }{ cosx }\]you will get this let me if that doesn't make sense

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

    Yes that makes sence

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

    alright now multiply cos x times cos x = ?? \[\huge\rm \frac{ \sin x - sinx \color{Red}{cosx (cosx)} }{ cosx }\]you will get this let me if that doesn't make sense

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

    Was this originally \[\frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \] \[\cos x \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ 1 } \times \cos \] \[\cos x \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos x }{ 1 } \times \cos \] Is that what we did there?

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

    we know that common denominator is cos so we multiplied numerator by cos x

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

    if the original fraction already have a same denominator as common denominator so u just have to carrry it down|dw:1434232700973:dw|

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

    If we did that we have \[ \cos \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \times \cos \] \[ \sin x - (\sin x)(\cos^2 x) \] ??

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

    first of all we know that cosx is common denominator so that should be at the denominator \[\frac{ ?????? }{ \cos x }\]

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

    after that we have to multiply both fraction by cos If we did that we have \[ \cos \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \times \cos \] \[\large\rm \frac{?????????}{cosx }\] ??

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

    wait a sec that's not cos^2 yet

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

    let me fix it :-)

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

    \(\color{blue}{\text{Originally Posted by}}\) @Nixy If we did that we have \[ \cos \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \times \cos \] \[ \sin x - (\sin x)(\cos^2 x) \] ?? \(\color{blue}{\text{End of Quote}}\) we know common denominator is cosx now we have to find numerator \[\frac{ ?????? }{ \cos x }\] original equation is \[\huge\rm \frac{ sinx }{ cosx } - \frac{ sinx \cos x }{ 1 }\] to find out numerator we should multiply both fraction by cosx(which is common denominator ) \[\large\rm cosx \times \frac{ sinx }{ cosx } - \frac{ sinx \cos x }{ 1 } \times cosx\]

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

    and that's how \[\large\rm \cancel{cosx} \times \frac{ sinx }{\cancel{ cosx} } - \frac{ sinx \cos^2 x }{ 1 } \] that's the numerator part

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

    Ok, I think we are on the same page now WOW this stuff gets confusing lol Ok I am where you are at

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

    what ??

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

    do you got that denominator thingy ?

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

    So now we have \[ \large\rm\frac{ sinx }{{ 1} } - \frac{ sinx \cos^2 x }{ 1 } \]

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

    that's the numerator what's our common denominator ?

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

    From what section?? I am going from here \[ \large\rm \cancel{cosx} \times \frac{ sinx }{\cancel{ cosx} } - \frac{ sinx \cos^2 x }{ 1 } \]

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

    sinx/1 = sinx so you can write it as \[\huge\rm \frac{ sinx - sinxcos^2x }{ cosx }\]

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

    Yes correct

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

    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha sinx/1 = sinx so you can write it as \[\huge\rm \frac{ sinx - sinxcos^2x }{ cosx }\] \(\color{blue}{\text{End of Quote}}\) do you understand this step ??

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

    want to make sure!!

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

    Yes I understand Next ? \[ \huge\rm \frac{ sinx - sinx(1-sin^2) }{ cosx } \]

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

    nope next step is to take out common term

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

    \[ \huge\rm \frac{\color{ReD}{ sinx} -\color{ReD}{ sinx}cos^2x}{ cosx } \] factor the numerator :-)

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

    let sinx = a cosx = b so you can write it as \[\large\rm (a-ab^2)\]

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

    So this \( (a-ab^2) = (ab^2-a) = (ab+a)(ab-a) \) correct?

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

    I cannot believe this one problem is like this. It has been an hour and 30 minutes lol

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

    nope what is common in these both terms a - ab^2 ??

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

    almost done .-.

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

    so a(1-b^2)

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

    yes right!!! so now change it back to sin cos

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

    \[ \huge\rm \frac{\color{ReD}{ sinx}( 1-cos^2x)}{ cosx } \]

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

    and that is the answer :-)

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

    separate it \[ \huge\rm \frac{\color{ReD}{ sinx} }{ cosx } \times (1-cos^2x) \]

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

    WOW, I cannot believe we got a problem like this. Some of these things take a long time.

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

    yay! done!

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

    practice more and more! then you will be able to finish it just in 5 minutes :3

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

    You work them one way then come to find out it is wrong then work them another and repeat and finally you get it lol This was the worst one yet!!! thank you for your help!!!!

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

    I do not have time to practice we have a week to fit a lot in and then we move on. Next week we probably will not even mess with these. idk

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

    ohh but still try do 2 question every night :3 = nightmares haha jk

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

    gO_Od luck!!!

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

    Thank you again!!!

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.