Given $\sin(t) + \cos(t) = a$, derive an expression in '$a$' for $(\cos(t))^4 + (\sin(t))^4$ I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semesters. I wonder if there is a typo in the statement or if I just haven't been rigorous enough?  Here is the question:
$$\text{Given } \sin{t} + \cos{t} = a, \text{ find an equivalent expression for } \sin^4{t} + \cos^4{t} \text{ in terms of } a.$$
Has anybody seen this one before? I tried (among other things) this (which is an approximation of an earlier attempt):
$(\sin{t} + \cos{t})^4$ and using whatever identities I could remember;
$(\sin{t} + \cos{t})^4 = \sin^4{t} + \cos^4(t) + 4\sin{t}\cos^3{t} + 6\sin^2{t}\cos^2{t} + 4\sin^3{t}\cos{t}$
so then 
$\begin{align}
\sin^4{t} + \cos^4{t} &= (\sin{t} + \cos{t})^4 - 4\sin{t}\cos^3{t} - 6\sin^2{t}\cos^2{t} - 4\sin^3{t}\cos{t}\\
&= (\sin{t} + \cos{t})^4 - 2\sin{t}\cos{t}(2\cos^2{t} + 3\sin{t}\cos{t} + 2\sin^2{t})\\
&= (\sin{t} + \cos{t})^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 4)\\
&= a^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 4)
\end{align}$
Couldn't get further than this, felt like I was overthinking it.

Alternatively, note that:
$$\begin{align}\sin t+\cos t=a \Rightarrow \sin\left(t+\frac{\pi}{4}\right)=\frac a{\sqrt{2}} \Rightarrow t&=\arcsin \frac a{\sqrt{2}}-\frac{\pi}{4}\\
\sin t=\sin \left(\arcsin \frac a{\sqrt{2}}-\frac{\pi}{4}\right)&=\frac a{\sqrt{2}}\cdot \frac 1{\sqrt{2}}-\sqrt{1-\frac{a^2}{2}}\cdot \frac1{\sqrt{2}};\\
&=\frac12\left(a-\sqrt{2-a^2}\right);\\
\cos t=\cos \left(\arcsin \frac a{\sqrt{2}}-\frac{\pi}{4}\right)&=\sqrt{1-\frac{a^2}{2}}\cdot \frac1{\sqrt{2}}+\frac a{\sqrt{2}}\cdot \frac 1{\sqrt{2}}=\\
&=\frac12\left(\sqrt{2-a^2}+a\right).\end{align}$$
Hence:
$$\begin{align}\sin^4t+\cos^4t&=\frac{2}{16}\cdot \left(a^4+6a^2\left(2-a^2\right)+(2-a^2)^2\right)=\\
&=-\frac{a^4}{2}+a^2+\frac12. \end{align}$$
