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Lukecrayonz

  • 2 years ago

@jim_thompson5910 help with trig identities real quick?

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  1. Lukecrayonz
    • 2 years ago
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    cot x sec^4 x=cot x + 2 tan x + tan^3 x

  2. surdawi
    • 2 years ago
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    sec^2=1+tan^2

  3. Lukecrayonz
    • 2 years ago
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    Steps?:O

  4. zordoloom
    • 2 years ago
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    Do you need help?

  5. Lukecrayonz
    • 2 years ago
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    Yes

  6. zordoloom
    • 2 years ago
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    That should help.

  7. Lukecrayonz
    • 2 years ago
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    Uhh what site is that.. :O

  8. Lukecrayonz
    • 2 years ago
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    @zordoloom

  9. jim_thompson5910
    • 2 years ago
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    cot x sec^4 x=cot x + 2 tan x + tan^3 x cot x sec^4 x=1/tan x + 2 tan x + tan^3 x cot x sec^4 x=1/tan x + 2 tan^2 x/tan x + tan^3 x cot x sec^4 x=(1 + 2 tan^2 x)/tan x + tan^3 x cot x sec^4 x=(1 + 2 tan^2 x)/tan x + tan^3 x/1 cot x sec^4 x=(1 + 2 tan^2 x)/tan x + tan^4 x/tan x cot x sec^4 x=(1 + 2 tan^2 x + tan^4 x)/tan x cot x sec^4 x=(1 + 2z + z^2)/tan x ... Let z = tan^2 x cot x sec^4 x=((z+1)^2)/tan x cot x sec^4 x=((tan^2 x+1)^2)/tan x cot x sec^4 x=((sec^2)^2)/tan x cot x sec^4 x=(sec^4 x)/tan x cot x sec^4 x=cot x sec^4 x

  10. Lukecrayonz
    • 2 years ago
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    ;__; you have different answers?

  11. jim_thompson5910
    • 2 years ago
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    there are different ways to do this, but they are both valid

  12. jim_thompson5910
    • 2 years ago
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    notice how I'm only manipulating the right side and I'm changing it into the left side

  13. Lukecrayonz
    • 2 years ago
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    1/tan x + 2 tan^2 x/tan x + tan^3 x (1 + 2 tan^2 x)/tan x + tan^3 x What allowed you to do that?

  14. jim_thompson5910
    • 2 years ago
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    1/tan x and 2 tan^2 x/tan x are fractions with the same denominator (tan x)

  15. jim_thompson5910
    • 2 years ago
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    I combined them to get (1 + 2 tan^2 x)/tan x

  16. Lukecrayonz
    • 2 years ago
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    Thanks a lot @jim_thompson5910 :D

  17. jim_thompson5910
    • 2 years ago
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    you're welcome

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