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baldymcgee6
 4 years ago
Difference between negative exponent and inverse functions?
For example,
sine inverse: sin^1(x)
or
1/(sin(x)) = (sin(x))^1
baldymcgee6
 4 years ago
Difference between negative exponent and inverse functions? For example, sine inverse: sin^1(x) or 1/(sin(x)) = (sin(x))^1

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campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.1the 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353436712645:dw

baldymcgee6
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353436831708:dw

baldymcgee6
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353437026561:dw

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.1ok.... 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
 4 years ago
Best ResponseYou've already chosen the best response.1hope that makes some sense... the inverse trig is only ever written as \[\sin^{1}(a)\]

baldymcgee6
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, thanks so much for the clarification.
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