Inequalities with trigonometric functions

[tigerd12]

Homework Statement

Three functions are defined as follows:

f:x> cos x for the domain 0< (or equal to) x < (or equal to) 180
g:x> sin x for the domain 0< (or equal to) x < (or equal to) 90
h:x>tan x for the domain p< (or equal to) x < (or equal to) q


Find the range of f.

-1<(or equal to) x < (or equal to) 1 (correct?)

Given that the range of h is the same as the range of g, find a value of p and a value of q.

this is the one i don't quite understand, i got p = 0, q= 90, is that right?

if the domain is the same therefore the range is the same, yes?

 

[EnumaElish]

if the domain is the same therefore the range is the same, yes?

No, not necessarily. And, isn't the question "given identical ranges, find the domain of h"?

 

[tigerd12]

No, not necessarily. And, isn't the question "given identical ranges, find the domain of h"?

Yes, so I am right. ?

 

[EnumaElish]

Can you graph sin and tan between 0 and 360? Or between 0 and 90? What is Sin(0)? What is Sin(90)?

What is Tan(0)? Tan(90)?

 

[HallsofIvy]

Homework Statement

Three functions are defined as follows:

f:x> cos x for the domain 0< (or equal to) x < (or equal to) 180
g:x> sin x for the domain 0< (or equal to) x < (or equal to) 90
h:x>tan x for the domain p< (or equal to) x < (or equal to) q

A bit peculiar, actually! I assume they mean 180 degrees, and 90 degrees.
Normally, sine and cosine, as functions are interpreted as in radians. The way sine and cosine are defined, as functions, x is "dimensionless" but radians give the correct values. Anyway, I'll go with degrees.

Find the range of f.

-1<(or equal to) x < (or equal to) 1 (correct?)

Given that the range of h is the same as the range of g, find a value of p and a value of q.

this is the one i don't quite understand, i got p = 0, q= 90, is that right?

if the domain is the same therefore the range is the same, yes?

When p= 0, tan(p)= 0. That's not what you want is it?
What is tan(90)? That's also not what you want is it?

For what p is tan(p)= -1?
For what q is tan(q)= 1?

 

[tigerd12]

I think I got it

Wait.. I think I got this..

so sin(0) - tan(0

sin(90) = 1
tan(45) = 1

therefore the P and Q are 0 and 45?

 

[HallsofIvy]

Wait.. I think I got this..

so sin(0) - tan(0

sin(90) = 1
tan(45) = 1

therefore the P and Q are 0 and 45?

I'm sorry, how did we get to talking about sin(x)- tan(x)? I thought the question was about the range of tan(x).
"h:x>tan x for the domain p< (or equal to) x < (or equal to) q" and you were to find the domain given that the range was the same as the range of sin(x) (-1 to 1).
You said you thought the domain would be the same as long as the range was the same. That is certainly not true! Different functions can take different domains (x-value) to the same range (y-value).
Yes, it is true that tan(45)= 1 so the upper limit is 45. But since the lower limit on the range of sin(x) is -1, you need to determine where tan(x)= -1, not 0!