Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=\frac1{2\sqrt2}$$
$$\sin\left(\alpha-\frac{\pi}{4}\right)=\frac1{2\sqrt2}$$
$$\alpha-\frac{\pi}{4}=\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)$$
I calculated the value of $\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)\approx 20.705^\circ$,
so I got $\alpha\approx 45^\circ+20.705^\circ=65.705^\circ$
I calculated
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\frac{1}{\sin^365.705^\circ}-\frac{1}{\cos^3 65.705^\circ}\approx -13.0373576$$
My question: Can I find the value of above trigonometric expression without using calculator? Please help me solve it by simpler method without solving for $\alpha$. Thanks

With $s=\sin \alpha$ and $c=\cos \alpha$ we have
$$\frac 1{s^3} -\frac 1{c^3}=\left(\frac 1 s - \frac 1c\right)\left(\frac 1{s^2} + \frac 1{sc} + \frac 1{c^2}\right)$$
$$ = \frac{c-s}{sc}\left(\frac 1{sc}+\frac 1{(sc)^2}\right)$$
Now, since $s-c=\frac 12$, you have $\frac 14 = 1-2sc \Leftrightarrow sc = \frac 38$. So, you get
$$\frac 1{s^3} -\frac 1{c^3} = -\frac 12\cdot \frac 83\left(\frac 83 + \left(\frac 83\right)^2\right) = -\frac{352}{27}$$
