Hey sakonpure6.

Is sin^nx = [sin(x)]^n? If so then the answer is yes based on exponents where a^b * a^c = a^(b+c)

Results 1 to 3 of 3

- Nov 21st 2013, 02:39 PM #1

- Nov 21st 2013, 02:56 PM #2

- Joined
- Sep 2012
- From
- Australia
- Posts
- 6,282
- Thanks
- 1636

- Nov 21st 2013, 05:26 PM #3

- Joined
- Apr 2005
- Posts
- 18,451
- Thanks
- 2532