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Lindsey11x16

  • 2 years ago

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

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

  2. jishan
    • 2 years ago
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    yeahhhhh right hartnn

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

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

  5. hartnn
    • 2 years ago
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    and -1 won't come.

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

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

  8. Lindsey11x16
    • 2 years ago
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    should i use factorization in this

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

  10. Lindsey11x16
    • 2 years ago
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    should i turn secant into cos

  11. hartnn
    • 2 years ago
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    yes., do that...what u get ?

  12. Lindsey11x16
    • 2 years ago
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    |dw:1355024922967:dw|

  13. hartnn
    • 2 years ago
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    1 is correct.

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

  15. Lindsey11x16
    • 2 years ago
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    equals sinx^2

  16. Lindsey11x16
    • 2 years ago
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    ?

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

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

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

  20. Lindsey11x16
    • 2 years ago
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    yep

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

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

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

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

  25. hartnn
    • 2 years ago
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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

  26. Lindsey11x16
    • 2 years ago
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    |dw:1355025560186:dw|

  27. hartnn
    • 2 years ago
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    yup.

  28. Lindsey11x16
    • 2 years ago
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    thank you ! <3

  29. hartnn
    • 2 years ago
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    welcome ^_^

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