This was the first problem on my test

1. Jun 23, 2009

1. The problem statement, all variables and given/known data

Find the average rate of change of function g(t)=3+tan(t) under interval [-pi/4;pi/4]

2. Relevant equations

3. The attempt at a solution

I am stuck at (-tan(-pi/4))/pi

2. Jun 23, 2009

CompuChip

What are sin(pi/4) and cos(pi/4)?

3. Jun 23, 2009

Same. So tan(pi/4) is 1?

4. Jun 23, 2009

Staff: Mentor

Yes.

5. Jun 23, 2009

HallsofIvy

Staff Emeritus
Of course, you say your difficulty is with
$$-tan(-\pi/4)/\pi= -\frac{sin(-\pi/4)}{cos(-\pi/4)}$$
What is that?

Last edited: Jun 24, 2009
6. Jun 23, 2009

CompuChip

So my next question is: what happens to the sine and the cosine when you put a minus in the argument. In other words: how does sin(-x) relate to sin(x) and how does cos(-x) relate to cos(x)?

7. Jun 23, 2009

It is the inverse

8. Jun 23, 2009

rock.freak667

-x => you measure clockwise from the horizontal axis, draw for quadrants and draw an acute angle clockwise from the positive horizontal axis. What quadrant does it lie in? Is sin(x) positive or negative in this quadrant?

9. Jun 23, 2009

it lies on IV quadrant. sin is only positive in quadrant II, so it is negative.

10. Jun 23, 2009

rock.freak667

thus sin(-x)=-sinx, do the same for cos(x)

11. Jun 24, 2009

HallsofIvy

Staff Emeritus
??? Sine is positive in the I and III quadrants. It is negative in the II and IV quadrants.

12. Jun 24, 2009

HallsofIvy

Staff Emeritus
??? Sine is positive in the I and III quadrants. It is negative in the II and IV quadrants.

But, as Rockfreak667 pointed out, it is much simpler to use the fact that sine is an odd function, cosine an even function, that to worry about "quadrants".

13. Jun 24, 2009

Staff: Mentor

That's half right. Sine is positive in the quadrants I and II, and negative in quadrants III and IV. Cosine is positive in quadrants I and IV, and negative in quadrants II and III. As a result, tangent is positive in quadrants I and III, and negative in quadrants II and IV.

14. Jun 24, 2009

$$\frac{f(b) - f(a)}{b - a}$$