- WELCOME TO A VIDEO ON THE TRIGONOMETRIC FUNCTIONS
IN TERMS OF THE ANGLES OF ROTATION.
SO IF WE CONSIDER THE FOLLOWING ANGLE,
LET'S CALL THIS ANGLE FEDA FROM HERE TO HERE.
IN STANDARD POSITION,
THIS INTERSECTS A CIRCLE WITH RADIUS R AT THE POINT (X, Y).
YOU CAN THINK OF THE RADIUS
AS THE HYPOTENUSE OF THE TRIANGLE FORM
BY DROPPING THIS PERPENDICULAR TO THE X AXIS.
AND BECAUSE OF THE RELATIONSHIP
BETWEEN THE THREE SIDES OF A RIGHT TRIANGLE
BASED UPON THE PYTHAGOREAN THEOREM,
WE CAN STATE THAT R SQUARED IS EQUAL TO X SQUARED + Y SQUARED.
AGAIN, R BEING THE HYPOTENUSE,
SO R WOULD EQUAL THE PRINCIPLE SQUARE ROOT OF X SQUARED
+ Y SQUARED.
SO USING THIS RIGHT TRIANGLE FOR OUR ANGLE FEDA,
WHICH IS ACTUALLY CALLED A REFERENCE TRIANGLE,
WE CAN DEFINE THE TRIGONOMETRIC FUNCTIONS
IN TERMS OF X, Y, AND R.
AGAIN, JUST KEEP IN MIND THAT IF THIS IS OUR ANGLE FEDA
AND THIS OUR TRIANGLE,
Y WOULD ACTUALLY BE THE OPPOSITE SIDE,
R AGAIN IS THE HYPOTENUSE,
WHICH LEAVES X TO BE THE ADJACENT SIDE.
SO WE'RE GOING TO STATE THESE IN TERMS OF X, Y, AND R,
BUT ESSENTIALLY WE'RE USING THE SAME THING WE DID
IN THE LAST VIDEO
WHERE SIGN FEDA WOULD BE OPPOSITE OVER HYPOTENUSE,
BUT NOW THE OPPOSITE SIDE IS Y AND THE HYPOTENUSE IS R.
COSINE DATA, INSTEAD OF STATING IT AS ADJACENT OVER HYPOTENUSE,
IS NOW X OVER R AND TANGENT R,
INSTEAD OF STATING IT AS OPPOSITE OVER ADJACENT,
WOULD NOW BE Y/X.
AND THEN JUST REMEMBER THAT COSECANT DATA IS THE RECIPROCAL
OF SINE DATA,
SO INSTEAD OF Y/R WE HAVE R/Y.
AND FOR SECANT DATA, WHICH IS THE RECIPROCAL OF COSINE DATA,
WE HAVE R/X INSTEAD OF X/R.
AND AGAIN, COTANGENT DATA IS THE RECIPROCAL OF TANGENT DATA,
SO WE HAVE X/Y INSTEAD OF Y/X.
SO USING THIS ACRONYM SOHCAHTOA
WILL ALSO HELP YOU REMEMBER THESE.
YOU JUST HAVE TO REMEMBER THAT WHEN WE SKETCH OUR TRIANGLE
WE USE X, Y's, AND R's INSTEAD.
NOW IF THE TERMINAL SIDES OF THE ANGLES
DO NOT FALL IN THE FIRST QUADRANT,
WE NEED TO PAY CLOSE ATTENTION TO THE SINE OF X AND Y
BECAUSE THE SINE OF THE X AND Y
WILL AFFECT THE SINE OF THE TRIGONOMETRIC FUNCTIONS.
WE KNOW THAT BOTH X AND Y ARE POSITIVE IN THE FIRST QUADRANT
SO ALL TRIGONOMETRIC FUNCTIONS WILL BE POSITIVE,
BUT OVER IN THE SECOND QUADRANT THE X IS NEGATIVE
AND THE Y IS POSITIVE.
OF COURSE, THIS WILL AFFECT THE SINE OF COSINE AND TANGENT
AND THERE RECIPROCALS.
IN THE THIRD QUADRANT BOTH X AND Y ARE NEGATIVE
AND THIS WILL AFFECT SINE AND COSINE AND THERE RECIPROCALS.
AND THEN IN QUADRANT FOUR, THE X COORDINATE IS POSITIVE,
THE Y IS NEGATIVE,
SO THIS WILL AFFECT THE SINE OF THE SINE FUNCTION
AS WELL AS THE TANGENT FUNCTION AND THERE RECIPROCALS.
SO HERE'S AN EXAMPLE OF IT.
CONSIDER THE TERMINAL SIDE OF THE ANGLE IN STANDARD POSITION
THAT PASSES THROUGH THE POINT (8, -6),
FIND THE VALUES OF THE SIX TRIGONOMETRIC FUNCTIONS.
SO FIRST WE WOULD PLOT OUR ANGLE IN STANDARD POSITION,
WHICH WOULD THIS,
BUT REMEMBER TO FORM THE REFERENCE TRIANGLE
THAT WE'RE GOING TO USE TO FIND THESE VALUES,
WE DROP A PERPENDICULAR TO THE CLOSEST X AXIS, WHICH IS HERE.
SO WE'RE ACTUALLY GOING TO USE THIS ANGLE
TO DETERMINE THE VALUES OF THE TRIGONOMETRIC FUNCTIONS
FOR THIS ANGLE.
SO WE CAN SEE THAT THE X COORDINATE IS EIGHT,
SO THIS DISTANCE HERE WOULD BE EIGHT.
AND THE Y COORDINATE IS -6,
SO EVEN THOUGH WE THINK OF DISTANCE AS POSITIVE
WE ARE GOING TO USE A SIGNED INTEGER TO REPRESENT THIS AS -6.
NOW THE NEXT THING WE SHOULD RECOGNIZE
IS THE HYPOTENUSE IS MISSING, WHICH WE WILL CALL R.
SO REMEMBER FROM THE PREVIOUS SCREEN,
R IS EQUAL TO THE SQUARE ROOT OF X SQUARED + Y SQUARED.
LET'S FIND THAT.
SO WE HAVE 64 + 36 WHICH EQUALS THE SQUARE ROOT OF 100.
SO R IS EQUAL TO 10 AND R IS ALWAYS POSITIVE.
THAT'S ALL THE INFORMATION WE NEED TO FIND
THE SIX TRIGONOMETRIC FUNCTIONS FOR THIS ANGLE.
SINE DATA WOULD BE Y/R OR -6/10, WHICH SIMPLIFIES TO -3/5.
COSINE DATA WOULD BE X/R, WHICH WOULD BE 8/10 OR 4/5.
AND TANGENT DATA IS Y/X, -6/8 WHICH SIMPLIFIES TO -3/4.
ONCE WE HAVE THESE
WE CAN JUST IDENTIFY THE RECIPROCAL FUNCTIONS,
SO COSECANT DATA WOULD BE -5/3, SECANT DATA WOULD BE 5/4,
AND COTANGENT DATA WOULD BE -4/3.
SO JUST PAY CLOSE ATTENTION TO THE SINE OF X AND Y
AND IT'S PRETTY MUCH THE SAME AS THE LAST VIDEO.
NOW THE CIRCLE, THE TERMINAL SIDE OF ANGLE INTERSECTS,
CAN BE ANY CIRCLE.
HOWEVER, OFTEN IT IS USEFUL TO USE A CIRCLE WITH RADIUS ONE
AND THIS IS CALLED THE UNIT CIRCLE.
AND USING THE UNIT CIRCLE TO FIND THE VALUES
OF THE TRIGONOMETRIC FUNCTIONS DOES SIMPLIFY OUR WORK SLIGHTLY
BECAUSE NOW IF WE KNOW R IS ONE, WE CAN CONCLUDE THE FOLLOWING.
INSTEAD OF STATING SINE THETA AS Y/R,
WELL THAT WOULD JUST BE Y/1, WHICH IS EQUAL TO Y.
SO WE CAN FIND THE VALUE OF SINE THETA
BY LOOKING AT JUST THE Y COORDINATE
AS LONG AS WE KNOW IT'S THE UNIT CIRCLE.
AND THE SAME THING FOR COSINE THETA,
INSTEAD OF STATING IT AS X/R,
WELL IF R IS ONE, COSINE THETA EQUALS X ON THE UNIT CIRCLE,
AND THERE'S NO CHANGE FOR TANGENT DATA,
IT WILL STILL BE Y/X.
AND THEN AGAIN HERE
HAVE THE RECIPROCAL TRIGONOMETRIC FUNCTIONS.
INSTEAD OF Y WE HAVE 1/Y FOR COSECANT THETA.
INSTEAD OF X WE HAVE 1/X FOR SECANT THETA.
AND FOR COTANGENT THETA WE HAVE X/Y.
HERE'S A NICE TOOL TO VISUALIZE THE UNIT CIRCLE.
HERE WE HAVE ANY ANGLE THAT INTERSECTS THE UNIT CIRCLE
AND AS YOU SEE THE X VALUE WOULD BE COSINE THETA
AND THE Y VALUE WOULD BE SINE THETA
FOR ANY ANGLE THAT INTERSECTS THIS UNIT CIRCLE.
YOU CAN ALSO SEE THE NICE REFERENCE TRIANGLES HERE
THAT ARE FORMED WITH THE X AXIS.
AGAIN, BECAUSE WE HAVE THE UNIT CIRCLE,
WE WOULD ONLY NEED THE X AND Y COORDINATE
SINCE WE KNOW R IS EQUAL TO ONE.
LET'S USE THE UNIT CIRCLE TO DETERMINE THE VALUES
OF EACH TRIGONOMETRIC FUNCTION LISTED HERE.
SINE 90 DEGREES, IF WE SKETCH THAT IN STANDARD POSITION,
WOULD LOOK SOMETHING LIKE THIS
WHERE WE CAN'T ACTUALLY FORM A RIGHT TRIANGLE.
BUT THAT'S OKAY BECAUSE NOW WE KNOW THAT SINE THETA
IS JUST EQUAL TO Y ON THE UNIT CIRCLE.
AND WE CAN SEE THIS IS A UNIT CIRCLE
SO SINE 90 DEGREES WOULD JUST EQUAL 1,
COSINE 90 DEGREES WOULD EQUAL X, WHICH IS 0
AND TANGENT 90 DEGREES WOULD BE Y/X, WHICH WOULD BE UNDEFINED.
FOR THE COSECANT OF 180 DEGREES, THE TERMINAL SIDE WOULD BE HERE.
REMEMBER THE COSECANT FUNCTION IS THE RECIPROCAL
OF THE SINE FUNCTION
AND SINCE SINE DATA IS EQUAL TO Y THIS WOULD EQUAL 1/Y.
AGAIN, IF WE HAVE 180 DEGREES IT WOULD LOOK SOMETHING LIKE THIS.
SO WE'RE LOOKING AT THIS POINT AND THE Y COORDINATE IS 0, 1/0
AS WE SAID BEFORE IS UNDEFINED.
NOW IF WE LOOK AT 270 DEGREES, WE HAVE OUR INITIAL SIDE HERE,
OUR TERMINAL SIDE WOULD BE HERE.
SO WE'RE ONLY LOOKING AT THIS ORDERED PAIR.
REMEMBER THAT COTANGENT IS A RECIPROCAL OF TANGENT,
TANGENT IS Y/X SO THIS WOULD BE X/Y.
WELL X/Y WOULD BE 0/-1, WHICH IS EQUAL TO 0.
AND OUR LAST ANGLE IS 0 DEGREES.
SO WE'RE LOOKING AT THIS ORDERED PAIR.
REMEMBER COSINE THETA IS EQUAL TO X,
SO WE KNOW THAT COSINE ZERO DEGREES IS EQUAL TO ONE.
NOW, ANY ANGLE THAT DOES HAVE ITS TERMINAL SIDE
ON THE X OR Y AXIS IS CALLED A QUADRANTAL ANGLE.
HERE'S A LIST OF ALL THE TRIGONOMETRIC FUNCTIONS
THAT ARE QUADRANTAL ANGLES FOR YOUR REFERENCE.
YOU MAY WANT TO PAUSE THE VIDEO AND TAKE A LOOK AT THEM.
LET'S TAKE A LOOK AT ANOTHER PROBLEM NOW.
A 60 DEGREE ANGLE INTERSECTS A UNIT CIRCLE
AT THE POINT (1/2, SQUARE ROOT 3/2).
LET'S GO AHEAD AND SKETCH THAT.
AND THE QUESTION ASKS TO DETERMINE THE VALUES
OF THE SIX TRIGONOMETRIC FUNCTIONS.
SO IF YOU KNOW IT'S A UNIT CIRCLE,
THIS GOES RATHER QUICKLY.
SINE 60 DEGREES WOULD JUST BE THE Y COORDINATE.
SO SQUARE ROOT 3/2, COSECANT 60 DEGREES WOULD BE THE RECIPROCAL
OR 2/SQUARE ROOT 3.
WE COULD RATIONALIZE THIS, BUT I WONT DO THAT KNOW DUE TO TIME.
COSINE 60 DEGREES WOULD JUST BE THE X COORDINATE.
THE RECIPROCAL WOULD BE TWO.
AND FOR TANGENT THETA WE WOULD HAVE THE SQUARE ROOT OF 3/2
AND THEN 1/2 IN THE DENOMINATOR,
WHICH DOES SIMPLIFY NICELY TO THE SQUARE ROOT OF 3.
AND THEN FOR COTANGENT WE'D HAVE THE RECIPROCAL OF THIS,
OR 1/SQUARE ROOT OF 3
AND YOU MAY BE REQUIRED TO RATIONALIZE THIS.
I HOPE YOU FOUND THIS VIDEO HELPFUL.
THANK YOU FOR WATCHING.