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