Problem with period in trig.?

Question

for the problem y = 4 cos (x/2 + pi/2) there are 2 ways to discover the period.

1. period = 2pi/b and in this case (x/2 + pi/2) = 1/2(x + pi)
thus 2pi/.5 = 4pi

2. however when I try and solve algebraically for the inequality:
0 less than equall to 1/2(x+pi) less than equall to 2pi
I always get an interval of -pi, pi wich has a period of 2pi.

What am I doing wrong in # 2?

Answer 1

Remember that trigonometric equations hit 0's three times per period: once at the start, once at the middle, and once at the end.

The middle is where the hump that goes up meets the hump that goes down on a graph.

It is important to take that into account. You should account for the humps when doing this. It is convenient to note that the length of the base of each hump is always equal to half a period, so all you need to do is double -pi and pi in part 2 in order to get a period of 4pi.

Answer 2

0 ≤ 1/2(x + π) ≤ 2π
0 ≤ x + π ≤ 4π
-π ≤ x ≤ 3π
and the interval between -π and 3π is 4π.

think about when cos x = 0. x = π/2, 3π/2, ....
so solve
x/2 + π/2 = π/2
x = 0
and then solve
x/2 + π/2 = 3π/2
x + π = 3π
x = 2π,
and then remember that's HALF a period.

Answer 3

properly, after one era, the function repeats itself. considering that csc is in straightforward terms a million/sin, whilst sin repeats itself, so will csc. In different words, csc(3x/2) has the comparable era as sin(3x/2). That it variety of seems such as you're pleased with, so i will bypass away you there. P.S. exceedingly much no person makes use of csc interior the real worldwide, all of us merely write a million/sin!