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

$$\frac{1}{\sin^3\alpha} -\frac{1}{\cos^3 \alpha}=\frac{\cos^3\alpha-\sin^3\alpha}{\sin^3\alpha \cos^3 \alpha}= -\frac{(\sin\alpha-\cos\alpha)(1+\sin\alpha\cos \alpha)}{(\sin \alpha \cos\alpha)^3}$$ Now,$$ (\sin \alpha -\cos \alpha )^2 =\frac 14 \implies 1-2\sin\alpha\cos\alpha =\frac 14 \\\implies \sin\alpha \cos\alpha =\frac 38$$ Just plug in the values of $\sin \alpha-\cos \alpha$ and $\sin \alpha\cos \alpha$ to finish.
