• June 13th 2009, 08:14 PM
darkryder341
There's a question that I just can't figure out, and I'm sure it's really simple but it's just not working for me. Some help would be greatly appreciated. I've tried looking for a similar question, but to no avail. Anyhow, here it is:

sinx - cosx = 1

• June 13th 2009, 08:22 PM
gosualite
Quote:

Originally Posted by darkryder341
There's a question that I just can't figure out, and I'm sure it's really simple but it's just not working for me. Some help would be greatly appreciated. I've tried looking for a similar question, but to no avail. Anyhow, here it is:

sinx - cosx = 1

Just look at the graphs of the two functions and see where they are exactly one unit apart. This occurs in many places. One such is when $x = \frac{\pi}{2}$. Do you see how to get all the rest?
• June 13th 2009, 08:44 PM
Soroban
Hello, darkryder341!

There are number of approaches to this problem (none are obvious).

Quote:

Solve for $x\!:\quad \sin x - \cos x \:=\: 1$
I will assume that the answers are on the interval $[0,\:2\pi)$

Square both sides: . $(\sin x - \cos x)^2 \:=\:1^2$

We have: . $\sin^2x - 2\sin x\cos x + \cos^2x \;=\;1$

. . . . . . . $\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}} - \underbrace{2\sin x\cos x}_{\text{This is }\sin2x} \;=\;1$

And we have: . $1 - \sin2x \:=\:1 \quad\Rightarrow\quad \sin2x \:=\:0$

Then: . $2x \:=\:0,\:\pi,\:2\pi,\:3\pi$

Hence: . $x \:=\:0,\:\tfrac{\pi}{2},\:\pi,\:\tfrac{3\pi}{2}$

But we find that $x \:=\:0,\:\tfrac{3\pi}{2}$ are extraneous roots.

Therefore, the answers are: . $x \;=\;\frac{\pi}{2},\:\pi$

• June 13th 2009, 09:06 PM
darkryder341
Hmm...makes sense now! Thanks very much for both of your help :D I really do appreciate it.
• June 14th 2009, 12:50 AM
Trig equation
Hello darkryder341
Quote:

Originally Posted by darkryder341
There's a question that I just can't figure out, and I'm sure it's really simple but it's just not working for me. Some help would be greatly appreciated. I've tried looking for a similar question, but to no avail. Anyhow, here it is:

sinx - cosx = 1

An alternative approach.

Expressions like $\sin x - \cos x$ can always be expressed in terms of a single trig function, like this:

Suppose $\sin x - \cos x = r\sin(x-\alpha)$

Then $\sin x - \cos x = r\sin x \cos \alpha - r\cos x \sin\alpha$

So, comparing coefficients:

$r\cos\alpha = 1$ and $r\sin\alpha = 1$

If we square and add, $r^2(\cos^2\alpha + \sin^2\alpha) = 2$

$\Rightarrow r = \sqrt 2$ (taking the positive value)

and, dividing:

$\tan\alpha = 1$

$\alpha = \tfrac{\pi}{4}$

So the original equation becomes:

$\sqrt2\sin(x- \tfrac{\pi}{4}) = 1$

$\Rightarrow \sin(x- \tfrac{\pi}{4}) = \frac{1}{\sqrt2}$

$\Rightarrow x -\frac{\pi}{4} = \frac{\pi}{4}, \frac{3\pi}{4}, ...$

$\Rightarrow x = \frac{\pi}{2}, \pi, ...$