Trig. Identities?

**r_maths** · April 17th 2008, 01:56 PM

[not sure if this would come under trig identities, (part B of this question is differentiation, but i'm not needing help with that)]

**lllll** · April 17th 2008, 02:09 PM

$\frac{\sin^3(x)+\cos^2(x)}{\sin^2(x)}$

$\frac{\sin^3(x)+(1-\sin^2(x))}{\sin^2(x)}$

$\sin(x)+\sin^{-2}(x)-1$

**r_maths** · April 17th 2008, 02:25 PM

EDIT: How do you get -1?

**lllll** · April 17th 2008, 02:39 PM

you have the identity that:

$\sin^2(x)+\cos^2(x)=1$

so then $\cos^2(x)=1-\sin^2(x)$

so we then have:

$<br /> \frac{\sin^3(x)}{\sin^2(x)}+\frac{1}{\sin^2(x)}{\c olor{blue} -\frac{\sin^2(x)}{\sin^2(x)}}$

so:

${\color{blue} - \frac{\sin^2(x)}{\sin^2(x)}=-1}$

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