Our teacher gave us this problem to work on this week, but I think either he or someone else copied it down wrong. I have tried everything I can think of; see if any of you can get it:
1-sin(t) / cos(t) = cos(t) / 1-sin(t)
Our teacher gave us this problem to work on this week, but I think either he or someone else copied it down wrong. I have tried everything I can think of; see if any of you can get it:
1-sin(t) / cos(t) = cos(t) / 1-sin(t)
Let's see: (And please use parenthesis!)
$\displaystyle \frac{1 - sin(t)}{cos(t)}$
$\displaystyle = \frac{1 - sin(t)}{cos(t)} \cdot \frac{1 + sin(t)}{1 + sin(t)}$
$\displaystyle = \frac{1 - sin^2(t)}{cos(t)(1 + sin(t))}$
$\displaystyle = \frac{cos^2(t)}{cos(t)(1 + sin(t))}$
$\displaystyle = \frac{cos(t)}{1 + sin(t)}$
You are right. It was copied wrong.
-Dan
Hello, Gambit!
The classic identity is: .$\displaystyle \frac{1-\sin t}{\cos t} \;=\;\frac{\cos t}{1+ \sin t}$ . . . or one its variations.
. . The two binomials must have opposite signs.
I've created dozens of identities for my student's exams.
Start with $\displaystyle \sin^2\!x + \cos^2\!x \:=\:1$ . . .
. . $\displaystyle \cos^2\!x\:=\:1-\sin^2\!x \quad \Rightarrow\quad (\cos x)(\cos x) \:=\1-\sin x)$$\displaystyle (1 + \sin x) \quad \Rightarrow \quad \frac{\cos x}{1-\sin x} \:=\:\frac{1+\sin x}{\cos x}$
. . $\displaystyle \sin^2\!x\:=\:1-\cos^2\!x \quad\Rightarrow\quad (\sin x)(\sin x) \:=\1-\cos x)$$\displaystyle (1 + \cos x) \quad\Rightarrow\quad \frac{\sin x}{1+\cos x} \:=\:\frac{1-\cos x}{\sin x}$
. . . See?