<s>[INST] 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.
 [/INST] $$\sin t+\cos t=a\\(\sin t + \cos t)^2=a^2\\2\sin t \cos t=a^2-1
\\(\sin t + \cos t)^4=a^4\\ \sin^4 t + \cos^4 t+4\sin t \cos t (\sin^2 t+\cos^2 t)+6\sin^2 t \cos^2 t=a^4\\ \sin^4 t + \cos^4 t +2(a^2-1)+\frac 32(a^2-1)^2=a^4\\
\sin^4 t+\cos^4 t=a^4-2(a^2-1)-\frac 32(a^2-1)^2$$
</s>