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baldymcgee6

Difference between negative exponent and inverse functions? For example, sine inverse: sin^-1(x) or 1/(sin(x)) = (sin(x))^-1

  • one year ago
  • one year ago

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  1. campbell_st
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    the negative indice means to find the reciprocal of sin(x) so given \[\sin(x) = \frac{\sin(x)}{1}\] then the reciprocal is \[(\frac{\sin(x)}{1})^{-1} = \frac{1}{\sin(x)}\] and for the inverse function if \[\sin(x) = \frac{a}{b}\] the sin of an angle is equal to a ratio then \[x = \sin^{-1}(\frac{a}{b})\] the angle is equal to the inverse sin of the ratio. its the 2 way connection between and angle and a ratio. hope it helps.

    • one year ago
  2. baldymcgee6
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    |dw:1353436712645:dw|

    • one year ago
  3. campbell_st
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    thats correct.....

    • one year ago
  4. baldymcgee6
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    |dw:1353436831708:dw|

    • one year ago
  5. baldymcgee6
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    @campbell_st

    • one year ago
  6. baldymcgee6
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    |dw:1353437026561:dw|

    • one year ago
  7. campbell_st
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    ok.... this is where indices and trig seem to clash \[(\sin(x))^2 = \sin^2(x)\] and \[(\frac{(\sin(x))}{1})^{-2} = \frac{1}{(\sin(x))^2} = \frac{1}{\sin^2(x)}\]

    • one year ago
  8. campbell_st
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    hope that makes some sense... the inverse trig is only ever written as \[\sin^{-1}(a)\]

    • one year ago
  9. baldymcgee6
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    Okay, thanks so much for the clarification.

    • one year ago
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