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msingh

  • 3 years ago

find sintheta

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  1. msingh
    • 3 years ago
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  2. shubhamsrg
    • 3 years ago
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    Can you write this expression as : sin@ + sqrt(sec^4 a) = sec^4 a ? (@ = theta a = alpha)

  3. ParthKohli
    • 3 years ago
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    Deja vu.

  4. Shadowys
    • 3 years ago
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    the key is in the infinite.

  5. sauravshakya
    • 3 years ago
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    |dw:1359202205624:dw|

  6. ParthKohli
    • 3 years ago
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    \[\sec^4 \alpha = \sin\theta + \sec^3 \alpha\]Nice

  7. sauravshakya
    • 3 years ago
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    So, sin(theta) =sec^4(alpha) -sec^3(alpha)

  8. ParthKohli
    • 3 years ago
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    Nice!

  9. msingh
    • 3 years ago
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    |dw:1359204540645:dw|

  10. msingh
    • 3 years ago
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    how this came

  11. Yahoo!
    • 3 years ago
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    put y = sec ^4 theta

  12. Yahoo!
    • 3 years ago
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    |dw:1359205220430:dw|

  13. msingh
    • 3 years ago
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    how, i didn;t understand

  14. Yahoo!
    • 3 years ago
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    \[\sin \theta = \sec^4\theta \pm \sec^2 \theta \]

  15. msingh
    • 3 years ago
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    its secθ or secalpha

  16. sauravshakya
    • 3 years ago
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    I guess I did a mistake

  17. Yahoo!
    • 3 years ago
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    Sorry Alpha

  18. msingh
    • 3 years ago
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    |dw:1359205684540:dw|

  19. sauravshakya
    • 3 years ago
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    |dw:1359205682988:dw|

  20. msingh
    • 3 years ago
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    |dw:1359205748804:dw|

  21. msingh
    • 3 years ago
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    x^2-x+x=sec^4alpha x^2=sec^4alpha x=sec^2alpha

  22. msingh
    • 3 years ago
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    if we put the value of sintheta in given equation, we get, |dw:1359205875734:dw|

  23. sauravshakya
    • 3 years ago
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    x^2 =sec^4alpha x=+- sec^2 alpha

  24. msingh
    • 3 years ago
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    yes, u r right after that putting the value of x in x^2-x=sintheta

  25. sauravshakya
    • 3 years ago
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    yep

  26. msingh
    • 3 years ago
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    answer is sec^2alpha tan^2alpha

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