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
 A: You can easily evaluate it without calculator as follows
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\frac{\cos^3\alpha-\sin^3\alpha}{\sin^3\alpha\cos^3\alpha}$$
$$=\frac{(\cos\alpha-\sin\alpha)(\cos^2\alpha+\sin^2\alpha+\cos\alpha\sin\alpha)}{\sin^3\alpha\cos^3\alpha}$$
$$=\frac{-(\sin\alpha-\cos\alpha)(1+\cos\alpha\sin\alpha)}{\frac18(2\sin\alpha\cos\alpha)^3}$$
$$=\frac{-4(\sin\alpha-\cos\alpha)(3-(\sin\alpha-\cos\alpha)^2)}{(1-(\sin\alpha-\cos\alpha)^2)^3}$$
$$=\frac{-4(\frac12)(3-(\frac12)^2)}{(1-(\frac12)^2)^3}$$
$$=-\frac{352}{27}$$
A: 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
$$\frac 1{s^3} -\frac 1{c^3} = -\frac 12\cdot \frac 83\left(\frac 83 + \left(\frac 83\right)^2\right) = -\frac{352}{27}$$
A: $$\frac{1}{\sin^3\alpha} -\frac{1}{\cos^3 \alpha}=\frac{\cos^3\alpha-\sin^3\alpha}{\sin^3\alpha \cos^3 \alpha}= -\frac{(\sin\alpha-\cos\alpha)(1+\sin\alpha\cos \alpha)}{(\sin \alpha \cos\alpha)^3}$$ 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.
A: Hint:
$$\left(\dfrac12\right)^2=(\sin\alpha-\cos\alpha)^2=?$$
So, we know $\sin\alpha\cos\alpha=?$
$$\dfrac1{\sin\alpha}-\dfrac1{\cos\alpha}=\dfrac{?}{\sin\alpha\cos\alpha}=?$$
Finally use $$\left(\dfrac1{\sin\alpha}-\dfrac1{\cos\alpha}\right)^3=\dfrac1{\sin^3\alpha}-\dfrac1{\cos^3\alpha}-\dfrac3{\sin\alpha\cos\alpha}\left(\dfrac1{\sin\alpha}-\dfrac1{\cos\alpha}\right)$$
A: $$\sin\alpha-\cos\alpha=\frac{1}{2}$$
$$(\sin\alpha-\cos\alpha)^2=\frac{1}{4}$$,
$$\sin\alpha\cos\alpha=\frac38$$
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\left(\frac 1{\sin\alpha}-\frac 1{\cos\alpha}\right)\left(\frac 1{\sin^2\alpha}+\frac 1{\cos^2\alpha}+\frac 1{\sin\alpha \cos\alpha}\right)$$
$$=\left(\frac {-(\sin\alpha-\cos\alpha)}{\sin\alpha\cos\alpha}\right)\left(\frac {\cos^2\alpha+\sin^2\alpha+\sin\alpha \cos\alpha}{\sin^2\alpha\cos^2\alpha}\right)$$
$$=\frac {-(\sin\alpha-\cos\alpha)(1+\sin\alpha\cos\alpha)}{(\sin\alpha\cos\alpha)^3}$$
$$=\frac{-{1\over2}(1+ {3\over8})}{({3\over8})^3}$$
$$=\frac{-352}{27}$$
