How do I prove that if: $\cos^3(x) + \sin^3(x) = 1$ then: $\cos(x) = 0 ; \sin(x)=1$ or $\cos(x)=1 ; \sin(x)=0$ How do I prove that if:
$$\cos^3(x) + \sin^3(x) = 1$$
then:
$$\cos(x) = 0 ; \sin(x)=1 \text{ or } \cos(x)=1 ; \sin(x)=0?$$
Starting from the first expression, I couldn't figure out how to reach the conclusion. I replaced 1 by $\cos^2(x) + \sin^2(x) $ hoping to factor it but to no avail.
 A: Aha OK you were pretty close. As you say $cos^{2}(x)+sin^{2}(x)=1$. Now $cos^{3}(x)+sin^{3}(x)$ is less than or equal to $cos^{2}(x)+sin^{2}(x)=1$ as the absolute value of $sin(x)$ and $cos(x)$ are both less than $1$. The only time that they are equal is when $cos(x)=1$ (and that means $sin(x)=0$) or $sin(x)=1$ (in which case $cos(x)=0$).
A: This is what I tried after reading your answers, please correct me if there are mistakes!$$cos^3+sin^3=1$$ $$cos^3+sin^3=cos² +sin²$$ $$cos^3-cos²+sin^3-sin²=0$$ $$cos²(cos-1) + sin²(sin-1)=0$$ $$(1-sin²)(cos-1)+(1-cos²)(sin-1)=0$$ $$ (1+sin)(1-sin)(cos-1)+(1+cos)(1-cos)(sin-1)=0 $$ $$(1+sin)(1-sin)(cos-1)-(cos-1)(1-cos)(sin-1)=0$$ $$(cos-1)[ (1+sin)(1-sin)-(sin-1)(1-cos)]=0$$ $$(cos-1)[ (1+sin)(1-sin)+(1-sin)(1-cos)]=0$$ $$(cos-1)[ (1-sin)(1+sin)(1-cos)]=0$$ $$(cos-1)(1-sin²)(1-cos)=0$$ $$(1-sin²)(1-cos)=0$$ $$1-sin²=0$$ $$sin²=1$$ $$sin=1$$
or: $$1-cos=0$$ $$cos=1$$
If: $sin=1$ ⇒  $sin^3=1$
Then:$$cos^3+sin^3=1$$ $$cos^3+1=1$$ $$cos^3=0$$ $$cos=0$$
If: $cos=1$ ⇒  $cos^3=1$
Then:$$cos^3+sin^3=1$$ $$sin^3+1=1$$ $$sin^3=0$$ $$sin=0$$
A: Hint: $\cos^3x\le\cos^2x$, with equality if and only if $\cos x=0$ or $1$. And the same for $\sin$.
A: Rather than carry the notation we write $s=\sin(x)$ and $c=\cos(x)$. Subtracting the equation from $s^2+c^2=1$ yields $s^3-s^2 + c^3-c^2=0$ or $s(s^2-1)+c(c^2-1) = s(-c^2) + c(-s^2) = 0$ and so either $s=0$ or $c=0$.
A: Write $u = \cos x$, $v=\sin x$. From $u^2 + v^2=1$ we get  $u$, $v$ $\le 1$  so $u^3\le u^2$, $v^3 \le v^2$, and adding up we get $u^3 + v^3\le 1$. We have equality if and only if we have equality in the previous equalities, therefore $(u,v)=(1,0)$ or $(0,1)$.
It is interesting to sketch the curves $u^2+v^2=1$ and $u^3+v^3=1$. From the above, the function $u^3+v^3$ has maximum value $1$ on $u^2+v^2=1$, so the two curves are tangent at $(1,0)$ and $(0,1)$. In other words, the multiplicities of the intersection at $(1,0)$, $(0,1)$ are $2$. Bézout's theorem tells us there should be two more intersection points. They are complex,  $(-1+\frac{i}{\sqrt{2}},-1-\frac{i}{\sqrt{2}})$ and its conjugate $(-1-\frac{i}{\sqrt{2}},-1+\frac{i}{\sqrt{2}})$

