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- WELCOME TO A QUICK LESSON ON THE GRAPHING
OF THE TANGENT FUNCTION.
THE GOALS OF THIS VIDEO ARE TO GRAPH THE TANGENT FUNCTION
ON THE COORDINATE PLANE USING THE UNIT CIRCLE,
AND ALSO TO DETERMINE THE DOMAIN AND RANGE
OF THE TANGENT FUNCTION.
THE NAME OF THE TANGENT FUNCTION
ACTUALLY COMES FROM THE TANGENT LINE
THAT IS PERPENDICULAR TO THE RADIUS
AND INTERSECTS THE CIRCLE AT A SINGLE POINT.
TO CREATE A TANGENT SEGMENT,
WHAT WE DO IS WE DRAW A LINE PERPENDICULAR TO THE X-AXIS
AS WE SEE HERE,
AND THEN EXTEND THETA
SO THAT IT INTERSECTS THE TANGENT LINE.
THE TANGENT FUNCTION IS DEFINED AS THE LENGTH
OF THIS RED SEGMENT HERE
OF THE LENGTH OF THIS TANGENT SEGMENT.
NOW, YOU PROBABLY ALSO REMEMBER THAT TANGENT THETA
IS DEFINED BY Y/X
WHERE THE POINT WITH COORDINATES (X,Y) WOULD BE
WHERE THE TERMINAL SIDE OF THIS ANGLE
INTERSECTS THE UNIT CIRCLE.
SO YOU MIGHT BE WONDERING
WHY IS THERE A DIFFERENT DEFINITION
FOR THE TANGENT FUNCTION NOW AND IN FACT THERE ISN'T.
THERE'S ACTUALLY TWO WAYS TO DEFINE THE VALUE
OF TANGENT THETA,
USING A TANGENT SEGMENT OR USING THE RATIO Y/X.
LET'S TAKE A LOOK AT WHY THESE ARE EQUAL TO EACH OTHER.
AND, IN FACT, THIS YELLOW TRIANGLE
AND THIS BLUE TRIANGLE ARE SIMILAR TRIANGLES.
AND NOTICE THAT THIS BLUE TRIANGLE
HAS THIS LENGTH LEG ONE
BECAUSE THIS IS A UNIT CIRCLE.
SO WHAT WE HAVE ARE TWO SIMILAR TRIANGLES HERE
WHERE TANGENT THETA IS EQUAL TO Y/X IN THE YELLOW TRIANGLE,
BUT IT'S ALSO EQUAL TO T/1 IN THE BLUE TRIANGLE.
AND SINCE THESE TWO TRIANGLES ARE SIMILAR
THIS FORMS A PROPORTION,
WHICH MEANS THEY'RE CROSS PRODUCTS MUST BE EQUAL.
SO WE'D HAVE Y x 1 WHICH IS Y = X x T.
AND, OF COURSE, IF WE SOLVE THIS EQUATION FOR T
THIS SHOWS THAT THE LENGTH OF THE SEGMENT T
IS EQUAL TO THE RATIO OF Y/X.
NOW LET'S GO AHEAD AND TAKE A LOOK AT THE UNIT CIRCLE.
REMEMBER THAT IF WE HAVE THE UNIT CIRCLE,
OUR TANGENT THETA WOULD BE EQUAL TO THE RATIO OF Y/X.
SO TO FIND FUNCTION VALUES,
WE'D HAVE TO TAKE EACH OF THESE ORDERED PAIRS
AND TAKE THE Y COORDINATE AND DIVIDE BY THE X COORDINATE.
AND THAT WOULD BE ANOTHER WAY
THAT WE COULD FIND TANGENT FUNCTION VALUES.
LET'S GO AND TAKE A LOOK AT A WOLFRAM DEMONSTRATION
TO GET A FEEL FOR THE GRAPH OF TANGENT THETA.
NOW, THE GRAPH OF THIS FUNCTION
IS A LITTLE MORE COMPLICATED
THAN THE GRAPH OF SINE THETA AND COSINE THETA
BECAUSE WE'RE GRAPHING THE FUNCTION VALUES
AS THE RATIO OF Y/X.
SO IF WE LOOK AT THE ANGLE 0 RADIANS,
THIS POINT WOULD BE (1,0).
THEREFORE, Y/X WOULD BE 0/1 OR 0.
NOW, AS WE APPROACH PI/2 RADIANS OR 90 DEGREES,
NOTICE THAT THE X COORDINATE IS APPROACHING ZERO
WHILE THE Y COORDINATE IS APPROACHING ONE.
AND, IN FACT, THE VALUES GET LARGER AND LARGER.
AND THEN AT PI/2 ITS ACTUALLY UNDEFINED
BECAUSE WE WOULD HAVE THE POINT (0,1) OR 1 DIVIDED BY 0,
WHICH WE KNOW IS UNDEFINED.
AND THAT PRODUCES--AND THAT PRODUCES A VERTICAL ASYMPTOTE
BECAUSE TANGENT THETA IS UNDEFINED WHEN X = 0.
AND AS WE WORK AROUND THE UNIT CIRCLE,
NOTICE, AGAIN, AT 3PI/2 TANGENT THETA IS UNDEFINED
PRODUCING A VERTICAL ASYMPTOTE AND SO-ON.
LET'S GO AND TAKE A LOOK AT SOME NEGATIVE ANGLES NOW.
AGAIN, IT'S A LITTLE MORE INVOLVED THAN SINE AND COSINE
BECAUSE IT'S RATIO OF COORDINATES
RATHER THAN JUST THE X OR Y COORDINATE AND SO-ON.
LET'S TAKE A LOOK AT THE ENTIRE GRAPH.
WHAT WE'LL NOTICE IS THERE ARE VERTICAL ASYMPTOTES
WHEN TANGENT THETA IS UNDEFINED,
AND THAT OCCURS WHERE X IS EQUAL TO ZERO.
AND THAT AS WE APPROACH PI/2 RADIANS
TANGENT THETA APPROACHES INFINITY.
AND AS THETA APPROACHES NEGATIVE PI OVER 2
THE FUNCTION VALUE APPROACHES NEGATIVE INFINITY.
OKAY, LET'S GO AHEAD AND TAKE A LOOK
AT HOW WE'RE GOING TO GRAPH THIS BY HAND.
LET'S GO AND GRAPH THIS BY HAND NOW.
I'VE ALREADY TAKEN THE TIME TO COMPLETE THIS TABLE
FOR VARIOUS ANGLES AND THE VALUE OF TANGENT DATA,
BUT LET ME SHOW YOU HOW I CREATED THIS
ON THE GRAPHING CALCULATOR.
FIRST STEP, I TYPED IN TANGENT X INTO Y1.
NEXT, I SET MY WINDOWS SO THAT IT STARTED AT ZERO
AND THE INDEPENDENT VARIABLE WAS ON "ASK".
AND LASTLY, I MADE SURE
THAT MY CALCULATOR WAS IN DEGREE MODE,
AND THEN I WENT TO THE TABLE.
AND YOU CAN SEE I'VE ALREADY TYPED IN SEVERAL ANGLES
IN DEGREES.
AND YOU WON'T SEE ALL OF THESE IN THE TABLE,
BUT I CAN CHOOSE ANY ANGLE I WANT NOW
AND FIND THE FUNCTION VALUE.
NOTICE AT 90 DEGREES THE FUNCTION IS UNDEFINED
BECAUSE THAT'S WHERE THE X COORDINATE
WOULD BE EQUAL TO ZERO.
LET'S GO AHEAD AND PLOT SOME FO THESE FUNCTIONS VALUES.
SO AT 0 DEGREES TANGENT THETA IS 0.
AT 30 DEGREE IT'S APPROXIMATELY 0.58.
AT 60 DEGREES IT'S 1.7.
AND THEN NOTICE AT 90 DEGREES IT'S UNDEFINED.
AGAIN, THAT'LL PRODUCE THE VERTICAL ASYMPTOTE.
SO AS THIS FUNCTION APPROACHES PI/2
IT INCREASES WITHOUT BOUND.
NOW, NOTICE I WANTED TO SEE
WHAT WAS HAPPENING TO THE LEFT OF ZERO,
SO I CHOSE -30 DEGREES
AND PRODUCED A FUNCTION VALUE OF APPROXIMATELY -0.58.
DIVIDE THIS INTO THIRDS. -60 IS -1.7.
AND, AGAIN, -90 IT'S UNDEFINED,
SO WE HAVE VERTICAL ASYMPTOTE.
AND SO OUR FUNCTION WOULD LOOK LIKE THIS TO THE LEFT OF ZERO.
THIS IS ONE COMPLETE CYCLE OF TANGENT THETA,
SO OUR PERIOD IS GOING TO BE PI RADIANS.
BUT TAKE A LOOK AT TANGENT THETA AT PI RADIANS
OR 180 DEGREES AT ZERO.
AND TO THE LEFT OF THIS
IT'S GOING TO RESEMBLE THIS PIECE HERE.
AND TO THE RIGHT OF THIS
IT'S GOING TO RESEMBLE THIS PIECE HERE.
SO WE'LL HAVE ANOTHER VERTICAL ASYMPTOTE AT 3PI/2
AND THE FUNCTION WILL LOOK LIKE THIS TO THE RIGHT.
AND NOW WE COULD JUST COPY AND PASTE THIS
INTO EACH INTERVAL OF LENGTH PI.
SO WHAT WE'LL NOTICE ON THIS FUNCTION
IS AT THE RANGE THE Y VALUES GO UP FOREVER
AND DOWN FOREVER,
THEREFORE THE RANGE IS ALL REALS.
BUT THE DOMAIN WOULD BE ALL X's
SUCH THAT X DOES NOT EQUAL SOME ODD MULTIPLE OF PI/2.
NOTICE WE HAVE 1PI/2, 3PI/2, NEXT WOULD BE 5PI/2.
AND WE CAN STATE THAT BY SAYING 2N + 1 x PI/2.
SO AS LONG AS X DOESN'T EQUAL AN ODD MULTIPLE OF PI/2
IT'S IN THE DOMAIN.
TANGENT THETA DOES NOT HAVE AN AMPLITUDE
BECAUSE THERE'S NO MAX OR MIN,
AND THE PERIOD WOULD BE PI RADIANS.
AGAIN, LET'S SUMMARIZE WHAT WE HAVE DISCOVERED.
TANGENT THETA IS GRAPHED HERE IN BLUE.
OUR DOMAIN, ANY X VALUE
THAT IS NOT AN ODD MULTIPLE OF PI/2 RADIANS,
BECAUSE THAT'S WHERE WE HAVE OUR VERTICAL ASYMPTOTES,
THE GRAPH IS DISCONTINUOUS AND HAS VERTICAL ASYMPTOTES.
THE X INTERCEPTS OCCUR AT MULTIPLES OF PI/2.
HERE WE (0PI,2PI) WHERE THE X INTERCEPTS.
THE PERIOD IS PI, THERE IS NO AMPLITUDE,
AND THE GRAPH IS SYMMETRICAL WITH RESPECTS TO THE ORIGIN,
AND THEREFORE IS AN ODD FUNCTION.
I HOPE YOU FOUND THIS VIDEO HELPFUL.
THANK YOU FOR WATCHING.
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