View Full Version : terminating angles??
UrbanXrisis
Sep9-04, 10:03 PM
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
Sep9-04, 10:33 PM
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.
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
Sep9-04, 10:48 PM
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 isnt any of the choices, is there a mistake in the question?
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 isnt 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
Sep9-04, 11:43 PM
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?
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
:wink:
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