If $\sin x - \cos x = \frac{1}{2}$ then determine: $\sin^4 x + \cos^4 x$ If $$\sin x - \cos x = \frac{1}{2}$$ then determine: $$\sin^4 x + \cos^4 x$$
I tried making it $(\sin^2 x)^2+(\cos^2 x)^2$ but then I get nothing that can help. What is the trick to this?

I'm using the identity $2(u^2+v^2)=(u+v)^2+(u-v)^2$ repeatedly:
$$(\cos x+\sin x)^2=2(\cos^2 x+\sin^2 x)-(\cos x-\sin x)^2=2-\frac14=\frac74,$$ so
\begin{align}2(\cos^4 x+\sin^4 x)&=(\cos^2 x+\sin^2 x)^2+(\cos^2 x-\sin^2 x)^2\\&=(\cos^2 x+\sin^2 x)^2+(\cos x+\sin x)^2\,(\cos x-\sin x)^2\\&=1+\frac74\cdot\frac14=\frac{23}{16}\end{align} We get $$\cos^4 x+\sin^4 x=\frac{23}{32}.$$
