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VeroZarate

  • one year ago

Verify that (sin x)(tan x cos x – cot x cos x) = 1 – 2 cos2x

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  1. binarymimic
    • one year ago
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    did you write this down correctly

  2. VeroZarate
    • one year ago
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    yes

  3. VeroZarate
    • one year ago
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    its very confusing

  4. binarymimic
    • one year ago
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    are you sure the last one shouldn't be \[1 - 2 \cos^{2}(x) \] instead of 1 - 2 cos 2x

  5. VeroZarate
    • one year ago
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    yes that is what i meant sorry it didnt square it

  6. binarymimic
    • one year ago
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    Ok try starting by rewriting tan x as sin x / cos x and cot x as cos x / sin x

  7. binarymimic
    • one year ago
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    \[(\sin x)(\frac{ \sin x }{ \cos x } * \cos x - \frac{ \cos x }{ \sin x } * \cos x) = 1 - 2 \cos^{2} x\]

  8. VeroZarate
    • one year ago
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    ok so cos^2x/sinx ?

  9. VeroZarate
    • one year ago
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    at the end ?

  10. binarymimic
    • one year ago
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    it would become \[(\sin x)(\sin x - \frac{ \cos^{2} x }{ \sin x }) = 1 - 2 \cos^{2} x\] then \[\sin^{2} x - \cos^{2} x = 1 - 2 \cos^{2} x\]

  11. binarymimic
    • one year ago
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    last step would just be add 2 cos^2(x) to both sides

  12. VeroZarate
    • one year ago
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    i dont understand where the 2cos^2x came from

  13. binarymimic
    • one year ago
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    there is already a -2 cos^2 x in the problem. the last step is to add 2 cos^2 x to both sides this way the right side becomes: 1 and the left side becomes: sin^2 x + cos^2 x

  14. binarymimic
    • one year ago
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    and we have the identity \[\sin^{2} x + \cos^{2} x = 1\]

  15. VeroZarate
    • one year ago
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    oh ok i understand! thanks! do you think you can help me out with another one?

  16. binarymimic
    • one year ago
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    sure but they recommend you close this question and start another one

  17. binarymimic
    • one year ago
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    brb though

  18. VeroZarate
    • one year ago
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    ok so ill open another

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