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sabika13 Group Title

How do I solve this trig identity: cos2x/(1+sinx^2)=tan(pie/4 -x)

  • one year ago
  • one year ago

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  1. matricked Group Title
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    i think ur question is cos2x/(1+sin2x) now cos2x =cos^2x -sin^2x=(cosx+sinx)(cosx-sinx) and 1+sin2x = cos^2x + sin^2x +2sinxcosx =(cosx+sinx)^2 thus cos2x/(1+sinx)^2= (cosx+sinx)(cosx-sinx)/(cosx+sinx)^2 =(cosx- sinx)/((cosx+sinx) divide both num and denom by cosx and then we have =(1-sinx/cosx)/(1+sinx/cosx) =(1-tanx)/(1+tanx) = (tan(pi/4) -tanx)/(1+tan(pi/4)tanx)=tan(pie/4 -x)

    • one year ago
  2. sabika13 Group Title
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    for the second last step.. where did the square roots go? isnt it suppose to be: (1-sinx/cosx)(1-sinx/cos)/(1+sinx/cos)(1+sinx/cosx)

    • one year ago
  3. matricked Group Title
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    actually the factors (cosx+sinx) are there both in num and denom hence they cancel each other

    • one year ago
  4. sabika13 Group Title
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    but its (cosx-sinx) in the num :s

    • one year ago
  5. matricked Group Title
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    (cosx+sinx)(cosx-sinx) / (cosx+sinx)^2 =(cosx-sinx) / (cosx+sinx)

    • one year ago
  6. matricked Group Title
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    i hope its clear now

    • one year ago
  7. matricked Group Title
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    why 13 u hv included in ur id ? if u don't mind

    • one year ago
  8. sabika13 Group Title
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    if its (cosx+sinx)^2, it has to equal (cosx+sinx)(cosx+sinx)

    • one year ago
  9. sabika13 Group Title
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    idk random.

    • one year ago
  10. matricked Group Title
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    (cosx+sinx)^2= (cosx+sinx)(cosx+sinx) (correct ) that is why one of the (cosx+sinx)

    • one year ago
  11. matricked Group Title
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    cancel each other

    • one year ago
  12. sabika13 Group Title
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    how does cos^2x -sin^2x=(cosx+sinx)(cosx-sinx)

    • one year ago
  13. matricked Group Title
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    USE ..A*A -B*B =(A)^2 -(B)^2 = (A+B)(A-B)...

    • one year ago
  14. sabika13 Group Title
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    okay, then why isnt :(cosx+sinx)^2 --> (cosx-sinx)(cosx+sinx)

    • one year ago
  15. matricked Group Title
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    bcoz a^2 =a*a

    • one year ago
  16. matricked Group Title
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    so (a+b)^2 =(a+b)(a+b)

    • one year ago
  17. matricked Group Title
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    but if u multiply (a+b)(a-b) u get (a^2 -b^2)

    • one year ago
  18. sabika13 Group Title
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    so (a^2-b^2) does not equal (a-b)^2 ?

    • one year ago
  19. matricked Group Title
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    sahi hai..

    • one year ago
  20. sabika13 Group Title
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    oh i get it..

    • one year ago
  21. sabika13 Group Title
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    =(1-tanx)/(1+tanx) = (tan(pi/4) -tanx)/(1+tan(pi/4)tanx)=tan(pie/4 -x) can u explain that? .. where did pi/4 come from ?

    • one year ago
  22. Dido525 Group Title
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    Tan (pi/4) = 1.

    • one year ago
  23. sabika13 Group Title
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    is that a trig identity?

    • one year ago
  24. Dido525 Group Title
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    No. If you do that on a calculator you get that.

    • one year ago
  25. Dido525 Group Title
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    Tan(pi/4) is always 1. Look on a unit circle.

    • one year ago
  26. sabika13 Group Title
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    Okay, thanks!!

    • one year ago
  27. Dido525 Group Title
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    Welcome :) .

    • one year ago
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