Hi,
cos x - sin x = 1, How do we solve it?
I think x = 0, 3(pi)/2, but how do we solve it mathematically?
Thank you!
Start by squaring both sides:
$\displaystyle (cos(x) - sin(x))^2 = 1$
$\displaystyle cos^2(x) - 2sin(x)cos(x) + sin^2(x) = 1$
$\displaystyle 1 - sin(2x) = 1$
$\displaystyle sin(2x) = 0$
For what values of x does this happen?
$\displaystyle x = 0$ and $\displaystyle x = \frac{\pi}{2}$ and $\displaystyle x = \pi$ and $\displaystyle x = \frac{3 \pi}{2}$.
Since we squared the original equation we need to go back to the original and see if we have any extra solutions. When you go back to check you will see that only two solutions work: $\displaystyle x = 0, \frac{3 \pi}{2}$.
-Dan
cosX - sinX = 1
Express one function in terms of the other.
Say, cosX into sinX terms.
sqrt[1 -sin^2(X)] -sinX = 1
Isolate the radical,
sqrt[1 -sin^2(X)] = 1 +sinX
Square both sides,
1 -sin^2(X) = 1 +2sinX +sin^2(X)
-sin^2(X) -2sinX -sin^2(X) = 0
2sin^2(X) +2sinX = 0
(2sinX)(sinX +1) = 0
sinX = 0 or -1
When sinX = 0,
X = arcsin(0) = 0 or pi
if X = 0, cos(0) -sin(0) =? 1 -----------yes.
if X = pi, cos(pi) -sin(pi) =? 1 ---------no.
So, here, X = 0 only.
When sinX = -1,
X = arcsin(-1) = 3pi/2
cos(3pi/2) -sin(3pi/2) =? 1 ------yes.
Therefore, X = 0 or 3pi/2 --------------------------answer.