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
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!