# 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.

• first that $2\cos^2+2\sin^2=2$ Commented Sep 5, 2018 at 4:46
• By the fundamental theorem of symmetric polynomials, once you've written $\sin t \cos t$ in terms of $a$ you can write any polynomial expression which is symmetric in $\sin t$ and $\cos t$ in terms of $a$. Commented Sep 5, 2018 at 4:55
• $$a^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 2)=a^4 - 2\dfrac{a^2-1}{2}(3\dfrac{a^2-1}{2} + 2)=\color{blue}{\dfrac{-a^4+2a^2+1}{2}}$$ Commented Sep 5, 2018 at 4:57

$$(\sin t + \cos t )^2= a^2\rightarrow \sin t\cos t=\dfrac{a^2-1}{2}$$ then $$\cos^4t + \sin^4t=(\cos^2t + \sin^2t)^2-2\cos^2t \sin^2t=1-2\left(\dfrac{a^2-1}{2}\right)^2$$

• Very nice, don't know how I missed that first line, but that's always math in hindsight. Very smart limiting next expansion to squares. Commented Sep 5, 2018 at 4:42

$$\sin t+\cos t=a\\(\sin t + \cos t)^2=a^2\\2\sin t \cos t=a^2-1 \\(\sin t + \cos t)^4=a^4\\ \sin^4 t + \cos^4 t+4\sin t \cos t (\sin^2 t+\cos^2 t)+6\sin^2 t \cos^2 t=a^4\\ \sin^4 t + \cos^4 t +2(a^2-1)+\frac 32(a^2-1)^2=a^4\\ \sin^4 t+\cos^4 t=a^4-2(a^2-1)-\frac 32(a^2-1)^2$$

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 $n=2$

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}