Question #72621

2 Answers
Feb 12, 2018

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

Feb 12, 2018

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!