- WE'RE GIVEN F OF X = 2 
COSINE X - 3 COSECANT X.
WE WANT TO FIND F PRIME OF X 
OR TH DERIVATIVE FUNCTION,
AND THEN F PRIME OF 2PI 
DIVIDED BY 3.
SO WE'LL BEGIN BY FINDING 
THE DERIVATIVE FUNCTION.
F PRIME OF X WILL BE EQUAL TO 2
x THE DERIVATIVE OF COSINE X - 
3 x THE DERIVATIVE COSECANT X.
AND SINCE THE DERIVATIVE 
OF COSINE X = -SINE X
WE'D HAVE 2 x -SINE X 
OR -2 SINE X
AND THEN - 3 x THE DERIVATIVE 
OF COSECANT X,
WHICH IS -COSECANT X 
x COTANGENT X.
HERE WE HAVE A NEGATIVE 
TIMES A NEGATIVE,
SO THIS WILL SIMPLIFY TO 
+ 3 COSECANT X COTANGENT X.
AND NOW THAT WE HAVE 
OUR DERIVATIVE FUNCTION,
WE CAN DETERMINE F PRIME 
OF 2PI DIVIDED BY 3.
F PRIME OF 2PI DIVIDED BY 3 
WOULD BE EQUAL TO -2 x SINE 2PI
DIVIDED BY 3 + 3 x COSECANT 2PI
DIVIDED BY 3 x COTANGENT 2PI 
DIVIDED BY 3.
SO TO FIND THESE 
TRIG FUNCTION VALUES,
WE CAN EITHER USE A UNIT CIRCLE 
OR SKETCH A REFERENCE TRIANGLE.
NOTICE THAT 2PI DIVIDED BY 3 
WOULD BE EQUAL TO 120 DEGREES.
LET'S BEGIN BY SKETCHING 
A REFERENCE TRIANGLE FOR 2PI
DIVIDED BY 3 RADIANS.
WE'D FIRST SKETCH THE ANGLE 
IN STANDARD POSITION,
SO IT WOULD START HERE 
ALONG THE +X-AXIS.
AND THEN WE'D ROTATE 2PI DIVIDED 
BY 3 RADIANS COUNTERCLOCKWISE
OR 120 DEGREES COUNTERCLOCKWISE 
TO SOMEWHERE IN HERE.
SO THIS WOULD BE 
THE TERMINAL SIDE OF 2PI
DIVIDED BY 3 RADIANS 
OR 120 DEGREES.
SO IF WE SKETCH 
A REFERENCE TRIANGLE
THAT WOULD BE THIS TRIANGLE 
HERE.
NOTICE HOW THE REFERENCE ANGLE 
WOULD BE 60 DEGREES
OR PI/3 RADIANS,
AND, THEREFORE, 
WE SHOULD RECOGNIZE
THIS IS A 30-60-90 
RIGHT TRIANGLE.
SO WE CAN LABEL THE SIDES 
1, 2 SQUARE ROOT 3,
BUT BECAUSE WE'RE 
IN THE SECOND QUADRANT
WHERE X IS NEGATIVE, 
THIS ONE WOULD BE -1.
SO NOW WE CAN USE 
THIS REFERENCE ANGLE
TO DETERMINE OUR TRIG FUNCTION 
VALUES,
WHICH WERE SINE 2PI 
DIVIDED BY 3,
COSECANT 2PI DIVIDED BY 3, 
AND COTANGENT 2PI DIVIDED BY 3.
WHERE SINE WOULD BE THE RATIO 
OF THE OPPOSITE SIDE
OF THE HYPOTENUSE,
WHICH WOULD BE SQUARE ROOT 3 
DIVIDED BY 2.
COSECANT 2PI DIVIDED BY 3 
WOULD BE THE RECIPROCAL
OF THE SINE FUNCTION VALUE
OR THE RATIO OF THE HYPOTENUSE 
TO THE OPPOSITE SIDE,
WHICH WOULD BE 2 
DIVIDED BY SQUARE ROOT 3.
AND THEN THE COTANGENT 2PI 
DIVIDED BY 3
WOULD BE THE RATIO 
OF THE ADJACENT SIDE
TO THE OPPOSITE SIDE,
WHICH WOULD BE -1 
DIVIDED BY SQUARE ROOT 3.
UNLESS WE'RE DEALING WITH SINE, 
COSINE, AND SOMETIMES TANGENT,
I THINK SKETCHING A REFERENCE 
TRIANGLE IS OFTEN EASIER
TO FIND SPECIFIC 
TRIG FUNCTION VALUES.
NOTICE IF WE TAKE A LOOK 
AT THE UNIT CIRCLE FOR A MOMENT,
HERE'S THE TERMINAL SIDE 
OF 2PI DIVIDED BY 3 RADIANS
WHERE X IS EQUAL TO COSINE THETA 
AND Y IS EQUAL TO SINE THETA.
WE CAN EASILY SEE 
THAT THE COSINE FUNCTION VALUE
WILL BE -1/2
AND THE SINE FUNCTION VALUE 
WOULD BE SQUARE ROOT 3
DIVIDED BY 2.
BUT WE STILL HAVE TO PUT 
SOME THOUGHT
INTO DETERMINING 
THE COSECANT FUNCTION VALUE
AND THE COTANGENT 
FUNCTION VALUE.
SO NOW THAT WE HAVE 
THESE THREE FUNCTION VALUES,
WE'LL GO BACK 
AND PERFORM SUBSTITUTION
TO DETERMINE F PRIME OF 2PI 
DIVIDED BY 3.
SO F PRIME OF 2PI DIVIDED BY 3 
WOULD BE EQUAL TO -2
OR -2/1 x SINE 2PI DIVIDED BY 3,
WHICH IS SQUARE ROOT 3 
DIVIDED BY 2.
AND THEN + 3 OR 3/1 
x COSECANT 2PI DIVIDED BY 3,
WHICH IS 2 DIVIDED 
BY SQUARE ROOT 3
x COTANGENT 2PI DIVIDED BY 3,
WHICH IS -1 
DIVIDED BY SQUARE ROOT 3.
SO NOW WE'LL FIND THESE PRODUCTS 
AND THEN FIND THE SUM.
NOTICE HERE THAT 2 SIMPLIFY OUT, 
SO WE HAVE -SQUARE ROOT 3.
AND THEN HERE WE'D HAVE + -6 
DIVIDED BY--
WELL, SQUARE ROOT 3 X 
SQUARE ROOT 3
WOULD JUST BE THE SQUARE ROOT 
OF 9 OR 3,
SO THIS SIMPLIFIES TO -SQUARE 
ROOT + -2 OR JUST - 2.
SO F PRIME OF 2PI DIVIDED BY 3 
IS EQUAL TO -SQUARE ROOT 3 - 2,
WHICH WILL THE SLOPE 
OF THE TANGENT LINE
AT X = 2PI DIVIDED BY 3 ON 
THE ORIGINAL FUNCTION F OF X.
LET'S ALSO GET A DECIMAL 
APPROXIMATION FOR THIS.
BUT KEEP IN MIND 
WE SHOULD NOT ROUND
UNLESS OUR DIRECTIONS 
TELL US TO.
WE HAVE -SQUARE ROOT 3 - 2.
IF WE ROUND 
TO THREE DECIMAL PLACES
THIS WILL BE APPROXIMATELY 
- 3.732,
WHICH, AGAIN, WOULD BE 
THE APPROXIMATE SLOPE
WITH A TANGENT LINE AT X = 2PI 
DIVIDED BY 3.
LET'S LOOK AT THE GRAPH 
OF OUR FUNCTION
AND LOCATE THE POINT 
WHEN X = 2PI DIVIDED BY 3
AND LOOK AT THE SLOPE 
WITH A TANGENT LINE.
HERE WE SEE THE GRAPH OF F OF X 
IN BLUE.
NOTICE WHEN X = 2PI 
DIVIDED BY 3 RADIANS
THIS WOULD BE THE POINT 
ON THE FUNCTION.
AND IF WE SKETCH THE TANGENT 
LINE AT THAT POINT,
THE SLOPE WITH THIS TANGENT LINE 
WOULD BE -SQUARE ROOT 3 - 2
OR APPROXIMATELY -3.732.
I HOPE YOU FOUND THIS HELPFUL.
