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.
 A: $$(\sin t + \cos t )^2= a^2\rightarrow \sin t\cos t=\dfrac{a^2-1}{2}$$
then
$$\cos^4t + \sin^4t=(\cos^2t + \sin^2t)^2-2\cos^2t \sin^2t=1-2\left(\dfrac{a^2-1}{2}\right)^2$$
A: $$\sin t+\cos t=a\\(\sin t + \cos t)^2=a^2\\2\sin t \cos t=a^2-1
\\(\sin t + \cos t)^4=a^4\\ \sin^4 t + \cos^4 t+4\sin t \cos t (\sin^2 t+\cos^2 t)+6\sin^2 t \cos^2 t=a^4\\ \sin^4 t + \cos^4 t +2(a^2-1)+\frac 32(a^2-1)^2=a^4\\
\sin^4 t+\cos^4 t=a^4-2(a^2-1)-\frac 32(a^2-1)^2$$
A: If $\sin x+\cos x=a,a^2=1+2\sin x\cos x\iff\sin x\cos x=\dfrac{a^2-1}2 $
$$\sin^{n+2}x+\cos^{n+2}x=\sin^nx(1-\cos^2x)+\cos^nx(1-\sin^2x)$$
If $I_m=\sin^mx+\cos^mx,I_2=1,I_0=2$
$$I_{n+2}=I_n-(\sin x\cos x)^2I_{n-2}$$
Here $n=2$
A: 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:
$$\begin{align}\sin^4t+\cos^4t&=\frac{2}{16}\cdot \left(a^4+6a^2\left(2-a^2\right)+(2-a^2)^2\right)=\\
&=-\frac{a^4}{2}+a^2+\frac12. \end{align}$$
