CAn someone explain how you get from:

sinx cosx^2 - sinx to sinx (cos^2x-1) to -sinx (1-cos^2x) to -sinx( sin^2x) to -sin^3x

and help me with factoring this equation:

sec^3x-sec^2x-secx+1

Thank you in advance(Bow)

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- Apr 28th 2008, 05:29 PM>_<SHY_GUY>_<Fundamental identities
CAn someone explain how you get from:

sinx cosx^2 - sinx to sinx (cos^2x-1) to -sinx (1-cos^2x) to -sinx( sin^2x) to -sin^3x

and help me with factoring this equation:

sec^3x-sec^2x-secx+1

Thank you in advance(Bow) - Apr 28th 2008, 05:44 PMo_O
$\displaystyle {\color{blue}\sin x} \cos^{2} x - {\color{blue}\sin x}$

$\displaystyle = {\color{blue}\sin x} \left(\cos^{2} x - 1\right)$

sin x was factored out of the expression just as you would if you had something like $\displaystyle ab + a = a(b + 1)$

$\displaystyle = -\sin x \left(1 - cos^{2} x\right)$ (-1 was factored out of the bracketed expression)

Recall that: $\displaystyle \cos^{2} x + \sin^{2} x = 1 \: \Rightarrow \: {\color{red}\sin^{2}x = 1 - \cos^{2} x}$

Substituting it in, we get:

$\displaystyle = -\sin x \cdot {\color{red} \sin^{2} x}$

$\displaystyle = -\sin^{3} x$ since $\displaystyle a^{m}a^{n} = a^{m+n}$

----------------------------

$\displaystyle {\color{red}\sec^{3} x - \sec^{2} x} \: {\color{blue}- \sec x + 1}$

Focusing on the red, let's factor out $\displaystyle \sec^{2} x$ and for the blue, let's factor out -1:

$\displaystyle = \sec^{2} x (\sec x - 1) -( \sec x - 1)$

Factor out (sec x - 1):

$\displaystyle = (\sec x - 1){\color{magenta}\left(sec^{2}x - 1\right)}$

And hopefully the expression in purple reminds you of a certain identity. - Apr 28th 2008, 05:48 PM>_<SHY_GUY>_<
so from that i use the pythagorean identities multiply, and thats it right?

- Apr 28th 2008, 05:49 PMo_O
I'm not sure what you mean by what you wrote. What I was referring to was this identity: $\displaystyle 1 + \tan^{2} x = \sec^{2} x$

- Apr 28th 2008, 05:55 PM>_<SHY_GUY>_<
which is derived from the pythagorean identity: sin^2+cos^2 = 1

and from there you multiply