If $\sin\theta+\cos\theta=1$ prove that $\cos\theta-\sin\theta=\pm1$ So my work, 
Squaring both sides $$(\sin\theta+\cos\theta)^2=1$$
$$1+2\sin\theta\cos\theta=1\ \ \ \ \ \text{-------(i)}$$
$$\sin\theta\cos\theta=0 \ \ \ \ \ \text{------(ii)}$$
So reverting back to $(i)$, 
$$\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-4\sin\theta\cos\theta=1-4\sin\theta\cos\theta$$
$$(\cos\theta-\sin\theta)^2=1-4\sin\theta\cos\theta$$
$$\cos\theta-\sin\theta=\pm1$$

But my teacher says that there is a shorter solution than that, so please can someone help me find that?
 A: After ii), you can say that one of the $\sin \theta$ and $\cos \theta$ has to be $0$, and this implies the other one to be $\pm 1$.
So also the difference $\sin \theta - \cos \theta = \pm 1$.
A: Notice, we have $$\cos \theta+\sin\theta=1$$
$$(\cos \theta+\sin\theta)^2=1$$$$\cos^2\theta+\sin^2\theta+2\sin \theta\cos \theta=1$$ $$1+2\sin \theta\cos \theta=1$$
$$\iff \sin\theta\cos \theta=0\tag 1$$
Now, we have
$$(\cos \theta-\sin\theta)^2=\cos^2\theta+\sin^2\theta-2\sin \theta\cos \theta$$
$$=(\cos^2\theta+\sin^2\theta+2\sin \theta\cos \theta)-4\sin \theta\cos \theta$$
$$=(\cos \theta+\sin\theta)^2-4\sin \theta\cos \theta$$
Substituting the corresponding values
$$=(1)^2-4(0)=1$$ $$\cos\theta-\sin\theta=\pm\sqrt 1=\pm 1$$
A: Let $x = \cos\theta$ and $y = \sin\theta$. Then $(x,y)$ is a point on
the unit circle. But the equation
$$\sin \theta + \cos \theta = 1$$
says that $x + y = 1$. So $(x,y)$ must be on the line given by $x + y = 1$,
that is, the line that intersects the unit circle at $(1,0)$ and $(0,1)$.
In fact, since $(x,y)$ is on that circle and on that line,
it must be one of those two points.
Case 1: $(x,y) = (0,1)$.
\begin{align}
\cos \theta &= 0 \\ \sin \theta &= 1 \\ \cos\theta - \sin\theta &= -1
\end{align}
Case 2: $(x,y) = (1,0)$
\begin{align}
\cos \theta &= 1 \\ \sin \theta &= 0 \\ \cos\theta - \sin\theta &= 1
\end{align}
And those are the only two possible cases that can occur.
A: $\cos \theta + \sin \theta=1$ is easily solved for $\theta$ in a graphical way, since it describes the intersection between a line and the goniometric circle:
$$
\begin{cases}
X+Y=1\\
X^2+Y^2=1,
\end{cases}
$$
where $X=\cos\theta$, $Y=\sin\theta.$
So either $\cos\theta =0$, $\sin\theta=1$, or viceversa. Plugging this into $\cos\theta-\sin\theta$ you either get $+1$ or $-1$.
