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anonymous

  • 5 years ago

can someone help with verifying a trig function?

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  1. anonymous
    • 5 years ago
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    try us? lols won't know till you post the question

  2. anonymous
    • 5 years ago
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    \[(\sin \theta-\cos \theta)^{2}=1-2\sin^{2} \theta \cot \theta , Verify?\]

  3. anonymous
    • 5 years ago
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    i'm going to skip the theta sign so here goes sin^2 - 2sin*cos + cos^2 = 1 - 2sin^2(cos/sin) sin^2 - cos^2 = 1 1 - 2 sin*cos = 1 - 2sin(cos they are equal make sense?

  4. anonymous
    • 5 years ago
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    thank you but just give me a second to make sure i got it

  5. anonymous
    • 5 years ago
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    sure thing

  6. anonymous
    • 5 years ago
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    how did you get to sin^2 - 2sin*cos + cos^2 = 1 - 2sin^2(cos/sin)

  7. anonymous
    • 5 years ago
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    where did the 2sin come from specifically

  8. anonymous
    • 5 years ago
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    i foiled the left side of the equation and tangent is equal to sin/cos so cot is equal to cos/sin

  9. anonymous
    • 5 years ago
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    (a - b)^2 a^2 -2ab + b^2 it's foil

  10. anonymous
    • 5 years ago
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    does that make sense?

  11. anonymous
    • 5 years ago
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    i think im having trouble with the foiling

  12. anonymous
    • 5 years ago
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    umm do you know how to foil? (a - b)(a - b) a^2 - ab - ab + b^2 (sin - cos)(sin - cos) sin^2 - sin*cos - sin*cos + cos^2 sin^2 - 2sin*cos + cos^2

  13. anonymous
    • 5 years ago
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    okay thank you for yor help i see what i was doing wrong, i was foiling wrong, little rusty on the algebra

  14. anonymous
    • 5 years ago
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    no problem

  15. anonymous
    • 5 years ago
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    hey you have time for another?

  16. anonymous
    • 5 years ago
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    sure

  17. anonymous
    • 5 years ago
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    Sec^4-tan^4, what trig function is it equal to?

  18. anonymous
    • 5 years ago
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    sec is 1/cos

  19. anonymous
    • 5 years ago
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    1/cos^4 - sin^4/ cos^4 (1 - sin^4) / cos^4 cos^2 + sin^2 = 1 so cos^2 = 1 - sin^2 [(1 - sin^2)(1 + sin^2)]/ cos^4 (1 + sin^2)/ cos^2

  20. anonymous
    • 5 years ago
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    i'm not sure if you can get more simplified than that..

  21. anonymous
    • 5 years ago
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    sin^2 = 1 - cos^2 (1 + 1 - cos^2)/ cos^2 (2 - cos^2) / cos^2 2sec^2 - 1 there that's one trig function

  22. anonymous
    • 5 years ago
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    the choices are A) Sec^2 B) csc C) tan D) Sec^2+tan^2 E) sin^2 F) sin*tan

  23. anonymous
    • 5 years ago
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    the answer is D where i stopped first (1 + sin^2)/ cos^2 if you divide that out you'll get sec^2 + tan^2

  24. anonymous
    • 5 years ago
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    ohh man thanks that one was killin me lol just for my own knowledge where do you start first when you see a problem like that? like what goes through your mind first?

  25. anonymous
    • 5 years ago
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    i hate doing trig functions all i can say is something that you've probably heard 100s of times practice but seriously, the more you practice, the easier it is key hints cos^2 + sin^2 = 1 <-- super big and useful tan = sin/cos there are also the rarer ones half angle formulas

  26. anonymous
    • 5 years ago
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    just move the functions around until you see something familiar and you'll see things to cancel out your object to make it as simple as possible

  27. anonymous
    • 5 years ago
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    ok cool keep those in mind i have one question, how did you go from [(1 - sin^2)(1 + sin^2)]/ cos^4 to (1 + sin^2)/ cos^2

  28. anonymous
    • 5 years ago
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    i wrote before that line that cos^2 = 1 - sin^2 which i got from sin^2 + cos^2 = 1 see? very useful since those are equal, i canceled from top and bottom does that make sense?

  29. anonymous
    • 5 years ago
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    yes i see it now, and i see how you were left with the answer. your helping me out big time lol

  30. anonymous
    • 5 years ago
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    lol awesome ^_^

  31. anonymous
    • 5 years ago
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    alright ive got another one for you ive been trying it for a while but keep hittin dead ends, got time?

  32. anonymous
    • 5 years ago
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    go for it

  33. anonymous
    • 5 years ago
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    1/2(1+sin/cos+cos/1+sin) , what trig function is equal? Sin cos tan csc sec cot

  34. anonymous
    • 5 years ago
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    cos/1? you sure, caz that's just cos

  35. anonymous
    • 5 years ago
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    yup cos/(1+sin)

  36. anonymous
    • 5 years ago
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    oh i see you need to use parenthesis rocco, otherwise there's a lot of miss understandings

  37. anonymous
    • 5 years ago
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    ohh im sorry

  38. anonymous
    • 5 years ago
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    oh and the first part is (1+sin)/cos

  39. anonymous
    • 5 years ago
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    1/2(1+tan+cos+cot))

  40. anonymous
    • 5 years ago
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    T-T

  41. anonymous
    • 5 years ago
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    i feel quite useless....

  42. anonymous
    • 5 years ago
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    give me some time, i'm going to write this out, sorry

  43. anonymous
    • 5 years ago
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    no problem, and ure not useless dont worry about that lol

  44. anonymous
    • 5 years ago
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    rocco i have no luck for this one :x sorry! repost this one

  45. anonymous
    • 5 years ago
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    ok no prob, found another one, wanna take a crack at it, i thnk i got half of it?

  46. anonymous
    • 5 years ago
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    rofl sure

  47. anonymous
    • 5 years ago
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    ok its csc - cot = 1/ (csc + cot) verify

  48. anonymous
    • 5 years ago
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    (csc - cot)(csc + cot) = 1 csc^2 - cot^2 = 1 ^-- that's another rule you should remember

  49. anonymous
    • 5 years ago
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    ohh but can you cross the = sign when verifying?

  50. anonymous
    • 5 years ago
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    i'm not sure i understand what you mean?

  51. anonymous
    • 5 years ago
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    somewhere i heard that when you verify the = becomes like a wall that you cant cross and only work on either side to make them equal, maybe im just understanding your work wrong

  52. anonymous
    • 5 years ago
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    i don't think i've heard that rule 2 + 4 = 8 - 2 if i add 2 on to both sides (crossing?) i get 6 = 6 do you mean crossing like this?

  53. anonymous
    • 5 years ago
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    yea like that

  54. anonymous
    • 5 years ago
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    it's perfectly fine as i just shown up there LOL 6 is equal to 6 so it's fine cheers ^_^

  55. anonymous
    • 5 years ago
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    ohh i see, haha thanks youve been a big help!!!

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