Question

Question #72621

Answer 1

`sin(theta-pi/2)=-.54`
`tan(x-pi/2)=.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 `cosx=sin(x+pi/2)`. `cosx` therefore is simply a sine curve that has been shifted `pi/2` units to the left. You can see that visually below:

`f(x)=cosx`
graph{cosx [-2.98, 8.113, -1.75, 3.8]}

`f(x)=sin(x+pi/2)`
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 `pi/2` units to the left, but rather `pi/2` units to the right. That give us something that looks like this:

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

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

Therefore, `sin(theta-pi/2)=-.54` and `tan(x-pi/2)=.18`

Answer 2

Here's another way of doing it.

We know that `sin(A - B) = sinAcosB - sinBcosA`.

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 `tanx= sinx/cosx`, that `sin(A - B) = sinAcosB - sinBcosA` and `cos(A - B) = cosAcosB + sinAsinB`

`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 `cotx`, thus:

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

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

Hopefully this helps!