## baldymcgee6 Group Title 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

1. campbell_st Group Title

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.

2. baldymcgee6 Group Title

|dw:1353436712645:dw|

3. campbell_st Group Title

thats correct.....

4. baldymcgee6 Group Title

|dw:1353436831708:dw|

5. baldymcgee6 Group Title

@campbell_st

6. baldymcgee6 Group Title

|dw:1353437026561:dw|

7. campbell_st Group Title

ok.... this is where indices and trig seem to clash $(\sin(x))^2 = \sin^2(x)$ and $(\frac{(\sin(x))}{1})^{-2} = \frac{1}{(\sin(x))^2} = \frac{1}{\sin^2(x)}$

8. campbell_st Group Title

hope that makes some sense... the inverse trig is only ever written as $\sin^{-1}(a)$

9. baldymcgee6 Group Title

Okay, thanks so much for the clarification.