A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Verify the trigonometric equation cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x)?

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

    @Hero , @ganeshie8 @iambatman help please

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

    are you sure it is not tan^2x ?

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

    1 second please let me make sure

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

    oh, no nvm, it should be like that... but still make sure.

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

    it is how I initially wrote it

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

    I would start from multiplying times tan(x) on both sides.

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

    \(\large\color{black}{ \displaystyle \cot x\sec^4x=\cot x+2\tan x+\tan^3x }\) \(\large\color{black}{ \displaystyle \sec^4x=1+2\tan^2x+\tan^4x }\)

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

    I have to focus on one side though, I am trying to prove that they are the same, not solve.

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

    oh, okay, it would have been wonderful, but I guess not

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

    \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\) would you agree with this?

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

    I factored out of cot(x).

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

    I have seen that everywhere I have looked, but I dont understand where the tan^2 or the Tan^4 come from

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

    well, do you agree that: \(\cot(x) \times \tan(x)=1\) \(\cot(x) \times \tan^2(x)=\tan(x)\) \(\cot(x) \times \tan^3(x)=\tan^2(x)\) \(\cot(x) \times \tan^4(x)=\tan^3(x)\) and so on... \(\cot(x) \times \tan^{N}(x)=\tan^{N-1}(x)\)

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

    if you don't understand why these are true, then ask

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

    I didn't know those identities. I knew that sin^2x + cos^2x =1 ....

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

    well, do you understand why they are logically true?

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

    @SolomonZelman i want ask u how did u get sec^4(x) = 1 +2tan^2(x) +tan^4(x)

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

    Not really.

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

    I didn't get that yet, but we will get that.

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

    \(\large\color{black}{ \displaystyle (x+1)^2=x^2+2x+1 }\) right?

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

    we have only sec(x) = 1/cos(x)

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

    turmoilx2, do you agree that: \(\cot(x) \times \tan(x)=1\)

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

    I suppose... if you tell me that it's true

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

    @dinamix . Are you in FLVS?

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

    \(\cot(x) \times \tan(x)=1\) \(\dfrac{\cos(x)}{\sin(x)}\times \dfrac{\sin(x)}{\cos(x)}=1\)

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

    I understand that.

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

    turmolix, how about now?

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

    Yes!

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

    ok, my comp glitched:)

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

    \(\cot(x) \times \tan(x)=1\) So, now, we will be multiplying times tan(x) on both sides: \(\cot(x) \times \tan^2(x)=\tan(x)\) \(\cot(x) \times \tan^3(x)=\tan^2(x)\) \(\cot(x) \times \tan^4(x)=\tan^3(x)\) and so on... \(\cot(x) \times \tan^{N}(x)=\tan^{N-1}(x)\) -------------------------------- Is this better now, why the list of these identites is true?

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

    @turmoilx2 no mate

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

    turmoilx, when you read, reply me if you got what I elaborates so far in my last reply.

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

    I understand why the identities are true now

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

    Ok, so now back to our question;)

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

    \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\) Now, using these identies, I am factoring the left side out of \(\cot(x)\). \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\) Do you get how I am factoring?

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

    No, please explain.

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

    let me do an intermediate step in blue: \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\) \(\large\color{blue}{ \displaystyle \cot x \sec^4x=\cot x\cdot 1+2\cot x\tan^2x+\cot x\tan^4x }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\) is it better now?

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

    after the blue, I factor out of \(\cot(x)\). If you still have questions about how I factored, then please ask.

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

    Im still lost. Sorry if I'm being slow here, online math classes are not for me.

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

    (it is fine if you don't I can attempt to explain a little better than what I did. I am not a very good tutor after all).

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

    I dont understand how you can just multiply the tan by cot and I dont know how to get the exponents for the tangents still. I understand the identities you mentioned above but I dont know how they apply to this first step if that makes sense...

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

    oh wait...

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

    Is it allowed because Cotxtanx is the same as multiplying by 1 which wouldnt change the equation?

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

    yes

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

    I think I understand some more now...

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

    but I would still just in case post my explanation on that as throuhg as I can. I will try not to take too much time.

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

    Yes. That would be appreciated, thank you!

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

    \(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\tan x}+\color{red}{\tan^3x} }\) `-----------------------------------------------`\({\\[0.8em]}\) \(\large\color{blue}{ \displaystyle \cot x =\cot x }\) (We know that)\({\\[0.8em]}\) `-----------------------------------------------`\({\\[0.8em]}\) \(\large\color{green}{ \displaystyle \tan x =\cot x\tan^2x }\)\({\\[0.8em]}\) (As we showed before in the list of those identites)\({\\[0.8em]}\) \(\large\color{green}{ \displaystyle \tan x =\cot x\tan^2x }\)\({\\[0.8em]}\) And then multiply times 2 on both sides\({\\[0.8em]}\) \(\large\color{green}{ \displaystyle 2\tan x =2\cot x\tan^2x }\) \({\\[0.8em]}\) `-----------------------------------------------`\({\\[0.8em]}\) \(\large\color{red}{ \displaystyle \tan^3x =\cot x\tan^4x }\)\({\\[0.8em]}\) (Also, as we showed before in the list of those identites)\({\\[0.8em]}\) `-----------------------------------------------` So it comes out that: \(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\cot x\tan^2 x}+\color{red}{\cot x\tan^4x} }\)

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

    take your time.

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

    When you are done reading, tell me if you understand how I got the following equation below: \(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\cot x\tan^2 x}+\color{red}{\cot x\tan^4x} }\)

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

    Here is what I have so far: \[CotxSec^4x=Cotx+2Tanx+Tan^3x\] \[CotxSec^4x=Cotx+2(CotxTan^2x)+(CotxTan^4x)\] \[CotxSec^4x=Cotx+Cot(2Tan^2x+tan^4x) Correct???

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

    oops \[CotxSec^4x=Cotx+Cot(2Tan^2x+tan^4x)\]

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

    \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}\) Rather, take out \(\cot(x)\) from the entire right side. \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\)

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

    did you get everything I have done so far/

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

    \[cotxsec4x=cotx((((1))))+2\tan2x+\tan4x)\] So Im assuming this "1" is from removing the cotagents... but how?

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

    well when I factor cot(x) out of cot(x), then I get 1.

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

    \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}\) I am factoring out of \(\cot(x)\). And I factor every term on the left side. \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\) still questions about that factoring out?

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

    I think what I dont get is why Cotx+2CotxTan^2x doesnt become 1+2Tanx instead of 1+ 2Tan^2x

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

    2tan^2x is what I got, didn't I?

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

    oh, because you have 2cot(x)tan²(x), and thus when you take out the cot(x) part, you remain with 2tan²(x).

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

    I'll just go with it. Cot can be factored from itself to equal 1.

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

    yes, true

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

    \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\) then, you know that: \(\color{red}{({\rm w}+1)^2={\rm w}^2+2{\rm w}+1}\) So, then look at the parenthesis on the right side and tell me how THAT can be factors.

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

    \(\large (\)\(\color{red}{\rm w}\) in our case is \(\tan^2x\)\(\large )\)

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

    \[(w+1)^2=w^2+1^2\] I know that this much is true.

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

    no, that is not true

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

    w²+1 would be w²+1², BUT (w+1)² is different:

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

    \((w+1)^2=(w+1)(w+1)=w(w+1)+1(w+1) \\ =w^2+w+w+1=w^2+2w+1\)

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

    so, therefore we know that: \((w+1)^2=w^2+2w+1\)

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

    ok, tell me if you get my very last post...

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

    I get it now. Mind blown

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

    You are very patient. Thanks

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

    Ok, so now if you look at, and understand the fact that: \((w+1)^2=w^2+2w+1\) Then you should also know that: \((\tan^2x+1)^2=\left(\tan^2x\right)^2+2\left(\tan^2x\right)+1\) `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` And then take a look at your problem: \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\) and it should follow that: \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+\tan^2 x\right)^2}\)

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

    By the way, it is not a problem.

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

    and then (1+tan^2x)^2 = (sec^2x)^2 which then equals sec^4x

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

    Yes, that is correct!

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

    THANK YOU SO MUCH!

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

    And then, we have verified the original equation to be an identity.

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

    You are welcome:)

  80. 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.