If $\sin\theta+\cos\theta=1$ prove that $\cos\theta-\sin\theta=\pm1$

Question

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?

$(\cos \theta - \sin \theta)^2 = 1 - 4 \sin\theta \cos\theta$? What? — Najib Idrissi, Sep 12 '15 at 13:37
@Rememberme, I know that, but my teacher says, there is a solution shorter than it. — Aditya Agarwal, Sep 12 '15 at 13:44
As far as I know, $(a-b)^2 = a^2 + b^2 - 2ab$. Where did your $4$ come from? — Najib Idrissi, Sep 12 '15 at 13:46
You can avoid the $2\sin \theta \cos\theta$ by computing $$(\sin \theta+\cos \theta)^2+(\cos \theta-\sin \theta)^2.$$ The first term is 1 by assumption, and both together are 2 owing to the Pythagorean trig identity. (The spirit of the proof is, of course, the same as yours.) — Semiclassical, Sep 12 '15 at 14:40
@labbhattacharjee, I think that my question is sufficiently different from the one you have shown. — Aditya Agarwal, Sep 13 '15 at 09:30
@AdityaAgarwal, Why? Have you noticed the comment from Semiclassical? — lab bhattacharjee, Sep 13 '15 at 11:45

score 4 · Accepted Answer · answered Sep 12 '15 at 13:41

4

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 $\sin \theta - \cos \theta = \pm 1$.

answered Sep 12 '15 at 13:41

Ant

21,098

Harish Chandra Rajpoot · Answer 2 · 2015-09-12T14:06:34.720

0

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 $$=(1)^2-4(0)=1$$ $$\cos\theta-\sin\theta=\pm\sqrt 1=\pm 1$$

edited Sep 12 '15 at 14:06

answered Sep 12 '15 at 14:00

Harish Chandra Rajpoot

37,450

1

I cannot see a difference between mine and your solution, except the fact that you have written every step. – Aditya Agarwal Sep 12 '15 at 14:01
Yes, you are right. – Harish Chandra Rajpoot Sep 12 '15 at 14:07

score 0 · Answer 3 · answered Sep 12 '15 at 15:36

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.

score 0 · Answer 4 · answered Sep 12 '15 at 15:52

$\cos \theta + \sin \theta=1$ 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 $-1$.

If $\sin\theta+\cos\theta=1$ prove that $\cos\theta-\sin\theta=\pm1$

4 Answers4