Evaluate the two trigonometric functions using its period as an aid.

1. cos (9pi/2) = ?

Ans: 0

2. cos (-9pi/2) = ?

Ans: 0

once around is 4 pi/2

twice around is 8 pi/2

so this is twice around plus pi/2
which is pi/2
or
90 degrees

I suspect you can do the clockwise one now.

But isn't cos (pi/2) = 0, because cos(t) = x?

To evaluate trigonometric functions using their periods, you need to remember the properties of trigonometric functions and their periodicity.

1. For the function cos (9pi/2):

The period of the cosine function is 2pi, which means that the function repeats itself every 2pi units. To evaluate cos (9pi/2), we can use the concept of periodicity to find an equivalent angle whose cosine value we know.

Since the period is 2pi, we can subtract multiples of 2pi from the given angle until we obtain an angle within one period. In this case, we can subtract 4pi from 9pi/2 to get:

9pi/2 - 4pi = pi/2

The cosine of pi/2 is known to be 0:

cos (pi/2) = 0

Therefore, cos (9pi/2) = 0.

2. For the function cos (-9pi/2):

Similarly, we can use the concept of periodicity to find the equivalent angle within one period.

Since the period of cosine is 2pi, we can add multiples of 2pi to the given angle until we obtain an angle within one period. In this case, we can add 4pi to -9pi/2:

-9pi/2 + 4pi = -pi/2

The cosine of -pi/2 is also known to be 0:

cos (-pi/2) = 0

Therefore, cos (-9pi/2) = 0.