Proving a long-ish trig identity

Question

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.

Answer 1

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*}

Answer 2

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 $$\sin(x)^3-\cos(x)^3=(\sin(x)-\cos(x))(\sin(x)^2+\sin(x)\cos(x)+\cos(x)^2)$$

Answer 3

Note that \begin{align} \sin^3(x)-\cos^3(x) & = (\sin(x)-\cos(x))(\sin^2(x)+\cos^2(x)+\sin(x)\cos(x)) \\ & =(\sin(x)-\cos(x))(1+\sin(x)\cos(x)). \end{align} Therefore, \begin{align} \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} & = \frac{(\sin(x)-\cos(x))(1+\sin(x)\cos(x))}{\sin(x)+\cos(x)} \\ & = \frac{(\sin(x)-\cos(x))^2(1+\sin(x)\cos(x))}{\sin^2(x)-\cos^2(x)} \\ & = \frac{(1-2\sin(x)\cos(x))(1+\sin(x)\cos(x))}{\sin^2(x)-\cos^2(x)} \\ & = \frac{1-\sin(x)\cos(x)-2\sin^2(x)\cos^2(x)}{\sin^2(x)-\cos^2(x)} \\ & = \frac{\sin^2(x)}{\sin^2(x)}\frac{\csc^2(x)-\cot(x)-2\cos^2(x)}{1-\cot^2(x)}. \end{align}

Answer 4

I'd start with the right-hand side: \begin{align} \frac{\csc^2x-\cot x-2\cos^2x}{1-\cot^2x} &=\frac{\dfrac{1}{\sin^2x}-\dfrac{\cos x}{\sin x}-2\cos^2x}{1-\dfrac{\cos^2x}{\sin^2x}} \\[6px] &=\frac{1-\sin x\cos x-2\cos^2x\sin^2x}{\sin^2x-\cos^2x} \\[6px] &=\frac{(\sin^2x+\cos^2x)^2-(\sin^2x+\cos^2x)\sin x\cos x-2\sin^2x\cos^2x}{\sin^2x-\cos^2x} \\[6px] &=\frac{\sin^4x-\sin^3x\cos x-\sin x\cos^3x+\cos^4x}{\sin^2x-\cos^2x} \\[6px] &=\frac{\sin^3x(\sin x-\cos x)-\cos^3x(\sin x-\cos x)}{\sin^2x-\cos^2x} \\[6px] &=\frac{(\sin^3x-\cos^3x)(\sin x-\cos x)}{(\sin x+\cos x)(\sin x-\cos x)} \\[6px] &=\frac{\sin^3x-\cos^3x}{\sin x+\cos x} \end{align} The trick is to “make the numerator homogeneous”.