# Question #72621

##### 2 Answers

#### 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

graph{cosx [-2.98, 8.113, -1.75, 3.8]}

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 *right*. That give us something that looks like this:

graph{sin(x-pi/2) [-2.98, 8.113, -1.75, 3.8]}

Basically, every point on the graph of

Therefore,

Here's another way of doing it.

We know that

Therefore:

#sin(theta-pi/2) = sinthetacos(pi/2) - sin(pi/2)costheta#

#sin(theta- pi/2) = -costheta#

#sin(theta- pi/2) = -0.54#

Now to 2. Recall that

#tan(x -pi/2) = (sin(x - pi/2))/(cos(x- pi/2))#

#tan(x - pi/2) = (sinxcos(pi/2) - cosxsin(pi/2))/(cosxcos(pi/2) + sinxsin(pi/2))#

#tan(x - pi/2) = (-cosx)/sinx#

#tan(x- pi/2) = -cotx#

We're given the value of

#tan(x- pi/2) = -(-0.18)#

#tan(x - pi/2) = 0.18#

Hopefully this helps!