Could any one help me factorising this term. I know the answer but I don't know how to get there. Thank you for your time.
$$-x+x\cos(x)+\cos(x)\sin(x)-\sin(x)=(\cos(x)-1)(x+\sin(x))$$
How do I get there? Is there a simple way to factorise terms like this?
On
First factor the terms with $x$ $$-x+x\cos(x)+\cos(x)\sin(x)-\sin(x)=x[\cos(x)-1]+\cos(x)\sin(x)-\sin(x)$$ Then factor terms with sine $$x[\cos(x)-1]+\cos(x)\sin(x)-\sin(x)=x[\cos(x)-1]+\sin(x)[\cos(x)-1]$$ Now factor $\cos(x)-1$ to obtain your result $$x[\cos(x)-1]+\sin(x)[\cos(x)-1]=(\cos(x)-1)(x+\sin(x))$$
On
Just begin with obvious partial factorisations, in the hope there will be a common factor: \begin{align} -x+x\cos x+\cos x\sin x-\sin x&=-x+\cos x(x+\sin x)-\sin x\\ &=\cos x(x+\sin x)-(x+\sin x)\\ &=(\cos x-1)(x+\sin x). \end{align}
On
There is not a general rule, you to practice with this kind of manipulations.
A good hint is try to group toghether step by step similar terms as follow:
$$-x+x\cos(x)+\cos(x)\sin x-\sin x$$
$$-x+\cos x(x+\sin x)-\sin x$$
$$\cos x(x+\sin x)-(x+\sin x)$$
$$(\cos x-1) (x+\sin x)$$
On
Taking it from where everyone else got:
Look at the left part($\cos(x)-1$). Set $2a=x$ to get $$\cos(2a)-1$$ now multiply by $-1$ twice to get $-(1-\cos(2a))$, this should ring a bell to you, remember the double angle formulas $$\cos(2\theta)=1-2\sin^2(\theta)$$hence we get $\cos(2a)-1=-(1-\cos(2a))=-(1-(1-2\sin^2(a)))=-2\sin^2(a)$ plug $x=2a$ to get $-2\sin^2(\frac x2)$ hence you have the reduced form of $$\boxed{-2\sin^2\left(\frac x2\right)(x+\sin(x))}$$
This is "factoring by grouping." Factor $x$ out of the first two terms and factor $\sin x$ out of the last two terms. The two blobs left both have a factor of $(\cos x -1)$, so factor that out.