# Proving a long-ish trig identity

Prove this identity without using cross multiplication by manipulating one side using trig identities:

$$\frac{\sin^3x-\cos^3x}{\sin x+\cos x} = \frac{\csc^2x-\cot x-2\cos^2x}{1-\cot^2x}$$

I first started off on the LHS and managed to get the denominator to become $1-\cot^2x$ by multiplying by $\sin x - \cos x$ and then dividing by $\sin^2 x$, but from there I had no idea how to continue.

• Start with the RHS, convert everything to sin and cos, it is easier to work that way. Then simplify. Jul 23 '18 at 20:31

Let $s= \sin x$ and $c=\cos x$ \begin{eqnarray*} \frac{s^3-c^3}{s+c} \frac{s-c}{s-c} = \frac{s^4+c^4-sc(s^2+c^2)}{s^2-c^2}. \end{eqnarray*} Now $s^4+c^4=(s^2+c^2)^2-2s^2 c^2$ and divide top & bottom by $s^2$ \begin{eqnarray*} \frac{1/s^2-c/s-2c^2}{1-c^2/s^2} . \end{eqnarray*}
Use that $$\sin^2(x)(1-\cot(x)^2)=\sin(x)^2-\cos(x)^2$$ then we get
$$\frac{\sin(x)^3-\cos(x)^3}{\sin(x)+\cos(x)}=\frac{2\cos(x)^4-2\cos(x)^2-\sin(x)\cos(x)+1}{(\sin(x)+\cos(x))(\sin(x)-\cos(x))}$$ and then use that $$\sin(x)^3-\cos(x)^3=(\sin(x)-\cos(x))(\sin(x)^2+\sin(x)\cos(x)+\cos(x)^2)$$