baldymcgee6
  • baldymcgee6
Difference between negative exponent and inverse functions? For example, sine inverse: sin^-1(x) or 1/(sin(x)) = (sin(x))^-1
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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campbell_st
  • campbell_st
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.
baldymcgee6
  • baldymcgee6
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campbell_st
  • campbell_st
thats correct.....

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baldymcgee6
  • baldymcgee6
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baldymcgee6
  • baldymcgee6
@campbell_st
baldymcgee6
  • baldymcgee6
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campbell_st
  • campbell_st
ok.... this is where indices and trig seem to clash \[(\sin(x))^2 = \sin^2(x)\] and \[(\frac{(\sin(x))}{1})^{-2} = \frac{1}{(\sin(x))^2} = \frac{1}{\sin^2(x)}\]
campbell_st
  • campbell_st
hope that makes some sense... the inverse trig is only ever written as \[\sin^{-1}(a)\]
baldymcgee6
  • baldymcgee6
Okay, thanks so much for the clarification.

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