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Lindsey11x16

cos^x(sec^2x-1) what pythagoream identiy to use?

  • one year ago
  • one year ago

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  1. hartnn
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    first multiply them out. like a(b-c) = ab-ac so whats cos^x(sec^2x-1) = ... - ... ?

    • one year ago
  2. jishan
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    yeahhhhh right hartnn

    • one year ago
  3. Lindsey11x16
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    cos^2x(sec^2x) - cos^2x-1

    • one year ago
  4. hartnn
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    yup, and you know what cos x * sec x = .. ?

    • one year ago
  5. hartnn
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    and -1 won't come.

    • one year ago
  6. hartnn
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    \(\large \cos^2x(\sec^2x-1)=\cos^2x\sec^2x-\cos^2x = ...?\)

    • one year ago
  7. hartnn
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    can u simplify that ^ ? use the fact that cos x = 1/ sec x.

    • one year ago
  8. Lindsey11x16
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    should i use factorization in this

    • one year ago
  9. hartnn
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    nopes. since, cos x = 1/ sec x what is cos x * sec x = ... ?

    • one year ago
  10. Lindsey11x16
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    should i turn secant into cos

    • one year ago
  11. hartnn
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    yes., do that...what u get ?

    • one year ago
  12. Lindsey11x16
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    |dw:1355024922967:dw|

    • one year ago
  13. hartnn
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    1 is correct.

    • one year ago
  14. hartnn
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    \(\large \cos^2x\sec^2x-\cos^2x = 1- \cos^2 x=...?\)

    • one year ago
  15. Lindsey11x16
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    equals sinx^2

    • one year ago
  16. Lindsey11x16
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    ?

    • one year ago
  17. hartnn
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    use th identity \(\huge \color{red}{\sin^2x+\cos^2x=1}\) yes, sin^2 x is correct :)

    • one year ago
  18. Lindsey11x16
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    thanks! this is hard cuz cuz i never know which identity to use :(

    • one year ago
  19. hartnn
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    do u wanna know alternative way to solve this ? using other identity...

    • one year ago
  20. Lindsey11x16
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    yep

    • one year ago
  21. hartnn
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    using \(\huge \color{red}{\sec^2x=1+\tan^2x}\) what can you write (sec^2x-1)=...?

    • one year ago
  22. Lindsey11x16
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    sec^2x-1 = tan

    • one year ago
  23. hartnn
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    its \(\tan^2x\)

    • one year ago
  24. hartnn
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    and you can write tan^2 x as sin^2 x/ cos^2x

    • one year ago
  25. hartnn
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    so, put (sec^2 x-1) as sin^2 x/ cos^2x cos^2 x*( sin^2 x/ cos^2x) will again give you sin^2 x

    • one year ago
  26. Lindsey11x16
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    |dw:1355025560186:dw|

    • one year ago
  27. hartnn
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    yup.

    • one year ago
  28. Lindsey11x16
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    thank you ! <3

    • one year ago
  29. hartnn
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    welcome ^_^

    • one year ago
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