Solving Terminating Angles & Quandrants - f(pi)=-pi

[UrbanXrisis]

terminating angles??

If sin(x) = cos(x), in which quadrants can angle x terminate?
I have no clue what this question is asking.

also...

If {x+sin(x)}/cos(x) then f(pi) = ?

{pi+sin(pi)}/cos(pi)= pi+0/-1 = -pi

is that correct?

 

[Gokul43201]

1. Rephrasing : For what values of x does sin(x) = cos(x). Which quadrants are these values of x in ?

0 to 90 (pi/2 rad) is Q1
90 to 180 (pi rad) is Q2
180 to 270 (3pi/2 rad) is Q3
270 to 360 or 0 (2pi or 0 rad) is Q4

Have you tried drawing the curves of sin(x) and cos(x) ? What happens with these curves when sin(x) = cos(x) ?

2. Assuming you mean "If f(x) = {x+sin(x)}/cos(x) then f(pi) = ?", your answer is correct.

 

[recon]

I don't know if this will help, however, where sin(x) = cos(x), x is in the first and third quadrants.

You may want to take a look at figure 9 here:
http://mpec.sc.mahidol.ac.th/physmath/mat12/curve810.jpg

EDIT: Woops! Gokul beat me to helping you.

 

[UrbanXrisis]

I understand, so it's quadrants I and III.

For the second question, the question just says If {x+sin(x)}/cos(x), then f`(pi)= ?
the choices are... (a) 2 (B) 1 (C) -1 (D) -2 (E) 0

I got an answer of -pi which isn't any of the choices, is there a mistake in the question?

 

[Pyrrhus]

I understand, so it's quadrants I and III.

For the second question, the question just says If {x+sin(x)}/cos(x), then f`(pi)= ?
the choices are... (a) 2 (B) 1 (C) -1 (D) -2 (E) 0

I got an answer of -pi which isn't any of the choices, is there a mistake in the question?

Check again for the quadrants.

does F'(pi) stand for first derivative evaluating pi?

assuming it does.

$$F'(x) = \frac{1+\cos(x)+x\sin(x)}{\cos^2(x)}$$

$$F'(pi) = 0$$

-Cyclovenom

 

[UrbanXrisis]

according to recon's link, sinX=cosX in the 1st and 3rd quadrant. SinX and CosX intersect between 0 and 90 as well as between 180 and 270 which are the 1st and 3rd quadrants. What am I missing?

 

[ShawnD]

The answer might just be hiding in plane sight (sorry, bad math joke).

$$sin(x) = cos(x)$$

$$\frac{sin(x)}{cos(x)} = 1$$

$$tan(x) = 1$$