Thread: Find Exact Value

Let t = theta for short

If f(t) = csc (t) and f(a) = 2, find the exact value of:

(A) f(-a)

(B) f(a) + f(a + 2pi) + f(a + 4pi)

Hello,

Originally Posted by magentarita

Let t = theta for short

If f(t) = csc (t) and f(a) = 2, find the exact value of:

(A) f(-a)

(B) f(a) + f(a + 2pi) + f(a + 4pi)

csc(t)=1/sin(t)

We know that sin(-t)=-sin(t) (you can check in on an unit circle)
And the sine function is periodic, that is sin(t+2k*pi)=sin(t), where k is any integer.

Originally Posted by Moo

Hello,

csc(t)=1/sin(t)

We know that sin(-t)=-sin(t) (you can check in on an unit circle)
And the sine function is periodic, that is sin(t+2k*pi)=sin(t), where k is any integer.

This is not a matter of knowing the periodic functions.

This is just plugging and chugging but I have no idea how to break it down.

Originally Posted by magentarita

This is not a matter of knowing the periodic functions.

This is just plugging and chugging but I have no idea how to break it down.

Why don't you try ?
I'm telling you that the sine function is periodic. That is to say, for example, .
Thus . This helps for question B.

Originally Posted by magentarita

Let t = theta for short

If f(t) = csc (t) and f(a) = 2, find the exact value of:

(A) f(-a)

(B) f(a) + f(a + 2pi) + f(a + 4pi)

f(t)=csc(t)
f(a)=csc(a)=2
so f(-a)=csc(-a)=-csc(a)=-2 (as csc(a)=2 )
2) f(a) + f(a + 2pi) + f(a + 4pi)
=csc(a)+csc(a+2pi)+csc(a+4pi)
=csc(a)+csc(a)+csc(a)
(csc(t+2n pi) where n is a integer =csc(t))
=2+2+2=6 hence
f(a) + f(a + 2pi) + f(a + 4pi)=6
hope this helps

Originally Posted by magentarita

This is not a matter of knowing the periodic functions.

This is just plugging and chugging but I have no idea how to break it down.

Originally Posted by nikhil

f(t)=csc(t)
f(a)=csc(a)=2
so f(-a)=csc(-a)=-csc(a)=-2 (as csc(a)=2 )
2) f(a) + f(a + 2pi) + f(a + 4pi)
=csc(a)+csc(a+2pi)+csc(a+4pi)
=csc(a)+csc(a)+csc(a)
(csc(t+2n pi) where n is a integer =csc(t))
=2+2+2=6 hence
f(a) + f(a + 2pi) + f(a + 4pi)=6
hope this helps

Excellent reply. Thanks!