# Question #72621

Feb 12, 2018

$\sin \left(\theta - \frac{\pi}{2}\right) = - .54$
$\tan \left(x - \frac{\pi}{2}\right) = .18$

#### Explanation:

Bear with me here. I'm going to try to explain this the only way I know how, although I'm sure there is a formal proof.

One of the trig identities you learn is $\cos x = \sin \left(x + \frac{\pi}{2}\right)$. $\cos x$ therefore is simply a sine curve that has been shifted $\frac{\pi}{2}$ units to the left. You can see that visually below:

$f \left(x\right) = \cos x$
graph{cosx [-2.98, 8.113, -1.75, 3.8]}

$f \left(x\right) = \sin \left(x + \frac{\pi}{2}\right)$
graph{sin(x+pi/2) [-2.98, 8.113, -1.75, 3.8]}

Now, the interesting thing about your function is that the sine curve hasn't been shifted $\frac{\pi}{2}$ units to the left, but rather $\frac{\pi}{2}$ units to the right. That give us something that looks like this:

$f \left(x\right) = \sin \left(x - \frac{\pi}{2}\right)$
graph{sin(x-pi/2) [-2.98, 8.113, -1.75, 3.8]}

Basically, every point on the graph of $f \left(x\right) = \sin \left(x - \frac{\pi}{2}\right)$ is a negative reflection of the corresponding point on the graph of $f \left(x\right) = \sin \left(x + \frac{\pi}{2}\right)$. The same is true of the graphs of $\cot \left(x\right)$ and $\tan \left(x - \frac{\pi}{2}\right)$. Pick some other values for x and see that this relationship is true.

Therefore, $\sin \left(\theta - \frac{\pi}{2}\right) = - .54$ and $\tan \left(x - \frac{\pi}{2}\right) = .18$

Feb 12, 2018

Here's another way of doing it.

We know that $\sin \left(A - B\right) = \sin A \cos B - \sin B \cos A$.

Therefore:

$\sin \left(\theta - \frac{\pi}{2}\right) = \sin \theta \cos \left(\frac{\pi}{2}\right) - \sin \left(\frac{\pi}{2}\right) \cos \theta$

$\sin \left(\theta - \frac{\pi}{2}\right) = - \cos \theta$

$\sin \left(\theta - \frac{\pi}{2}\right) = - 0.54$

Now to 2. Recall that $\tan x = \sin \frac{x}{\cos} x$, that $\sin \left(A - B\right) = \sin A \cos B - \sin B \cos A$ and $\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$

$\tan \left(x - \frac{\pi}{2}\right) = \frac{\sin \left(x - \frac{\pi}{2}\right)}{\cos \left(x - \frac{\pi}{2}\right)}$
$\tan \left(x - \frac{\pi}{2}\right) = \frac{\sin x \cos \left(\frac{\pi}{2}\right) - \cos x \sin \left(\frac{\pi}{2}\right)}{\cos x \cos \left(\frac{\pi}{2}\right) + \sin x \sin \left(\frac{\pi}{2}\right)}$

$\tan \left(x - \frac{\pi}{2}\right) = \frac{- \cos x}{\sin} x$

$\tan \left(x - \frac{\pi}{2}\right) = - \cot x$

We're given the value of $\cot x$, thus:

$\tan \left(x - \frac{\pi}{2}\right) = - \left(- 0.18\right)$

$\tan \left(x - \frac{\pi}{2}\right) = 0.18$

Hopefully this helps!