- WELCOME TO AN INTRO TO THE INVERSE FUNCTIONS
OF SINE, COSINE AND TANGENT.
THE GOALS OF THE VIDEO ARE TO DEFINE THE INVERSE FUNCTIONS
FOR SINE, COSINE AND TANGENT,
GRAPH THE INVERSE FUNCTIONS OF SINE, COSINE AND TANGENT,
AND THEN USE THE INVERSE FUNCTIONS TO SOLVE PROBLEMS.
SO REMEMBER FROM THE VIDEO, INVERSE FUNCTIONS,
ONLY ONE-TO-ONE FUNCTIONS HAVE INVERSE FUNCTIONS.
IN SINE, COSINE AND TANGENT IN THEIR CURRENT FORM
ARE NOT ONE-TO-ONE FUNCTIONS.
EACH OF THESE FAILS THE HORIZONTAL LINE TEST
BECAUSE THIS HORIZONTAL LINE INTERSECTS THE GRAPH
IN MORE THAN ONE POINT.
SO WE CANNOT FIND THE INVERSE FUNCTIONS OF THESE
IN THEIR CURRENT FORM.
WHAT WE CAN DO IS RESTRICT THE DOMAIN OF A FUNCTION
TO MAKE IT ONE-TO-ONE WITHOUT CHANGING THE RANGE.
AND THIS ALLOWS US TO DETERMINE INVERSE FUNCTIONS
THAT WE OTHERWISE WOULD NOT BE ABLE TO.
SO, FOR EXAMPLE, IF WE CONSIDER THE FUNCTION F OF X = X SQUARED
ON THE INTERVAL FROM ZERO TO INFINITY
THIS IS A ONE-TO-ONE FUNCTION,
WHICH ALLOWS US TO FIND THE INVERSE FUNCTION
WHICH WE OTHERWISE COULD NOT
ON THE INTERVAL FROM NEGATIVE INFINITY TO INFINITY
AND THE INVERSE OPERATION OF SQUARING THE SQUARE ROOTING.
SO HERE IN RED WE SEE OUR INVERSE FUNCTION
F INVERSE OF X = SQUARE ROOT X.
AND IF WE GRAPH THEM ON THE COORDINATE PLANE WITH Y = X,
AGAIN, WE CAN SEE THE SYMMETRY ACROSS THE LINE Y = X.
NOW, FOR Y = SINE X WE RESTRICT THE DOMAIN TO THE CLOSE INTERVAL
FROM -PI OVER 2 TO PI OVER 2.
AND I'VE GRAPHED THAT HERE IN RED BELOW.
SO, AGAIN, FOR THIS GRAPH, OUR DOMAIN IS FROM -PI OVER 2
TO PI OVER 2 ON THE CLOSED INTERVAL
AND THE RANGE IS STILL FROM NEGATIVE ONE TO ONE
ON THE CLOSED INTERVAL.
NOW, THE INVERSE FUNCTION OF SINE IS GIVEN AS
Y = INVERSE SINE OF X OR Y = ARC SINE OF X.
THERE'S TWO WAYS TO WRITE THE INVERSE FUNCTION OF Y = SINE X.
IT'S GRAPHED HERE ON THE RIGHT BELOW.
AND THERE'S SOME IMPORTANT THINGS TO REMEMBER.
THE PROCESS FOR FINDING THIS INVERSE
IS TO INTERCHANGE THE X AND THE Y VARIABLES.
AND BY INTERCHANGING THE X AND THE Y VARIABLES
WE ARE INTERCHANGING THE DOMAIN AND THE RANGE.
NOTICE THE DOMAIN OF Y = SINE X
IS THE RANGE OF INVERSE SINE OF X.
AND THE RANGE OF SINE X IS THE DOMAIN OF INVERSE SINE X.
SO THE INPUT WILL BE THE SINE FUNCTION VALUE,
AND THE OUTPUT Y WILL BE THE ANGLE
THAT PRODUCES THAT SINE FUNCTION VALUE.
FOR Y = COSINE X
WE RESTRICT THE DOMAIN TO THE CLOSE INVERSE FROM 0 TO PI,
AND WE HAVE IT GRAPHED HERE IN BLUE BELOW.
AND NOTICE OUR RANGE IS STILL ON THE CLOSED INTERVAL
FROM -1 TO +1.
THE INVERSE FUNCTION OF Y = COSINE X
IS Y = INVERSE COSINE X, OR Y = ARC COSINE X.
AGAIN, THESE MEAN EXACTLY THE SAME THING.
THEY ARE THE INVERSE OF Y = COSINE X.
AND IT'S GRAPHED HERE ON THE RIGHT.
AGAIN, IT'S IMPORTANT THAT WE NOTICE THE DOMAIN
OF Y = COSINE X IS THE RANGE OF Y = INVERSE COSINE X.
AND THE RANGE OF Y = COSINE X IS THE DOMAIN
OF Y = INVERSE COSINE X.
SO THE INPUT FOR Y = INVERSE COSINE OF X
IS THE COSINE FUNCTION VALUE
AND THE OUTPUT OR THE Y VALUE IS THE ANGLE
THAT PRODUCES THE COSINE FUNCTION VALUE.
AND LASTLY WE HAVE Y = TANGENT X.
AND FOR THIS FUNCTION WE RESTRICT THE DOMAIN
TO THE OPEN INTERVAL FROM -PI OVER 2 TO PI OVER 2,
AGAIN, GRAPHED HERE IN GREEN.
AND FOR THE RANGE IT STILL IS FROM NEGATIVE INFINITY
TO POSITIVE INFINITY.
AND THE INVERSE OF Y = TANGENT X
IS GIVEN AS Y = INVERSE TANGENT X
OR Y = ARC TANGENT X.
AND IT'S GRAPHED HERE ON THE RIGHT.
AND ONCE AGAIN NOTICE THAT THE DOMAIN OR THE INPUT
IS GOING TO BE THE TANGENT FUNCTION VALUE,
AND THE OUTPUT OR THE RANGE WILL BE THE ANGLE
THAT PRODUCES THAT TANGENT FUNCTION VALUE.
OKAY, LET'S TAKE A LOOK AT SOME PROBLEMS.
WE WANT TO FIND THE EXACT VALUE FOR EACH REAL NUMBER Y
AND THEN USE THE CALCULATOR TO CHECK OUR WORK.
NOW, WHAT I DID HERE IN GREEN WAS LIST THE OUTPUT OR THE RANGE
FOR EACH OF THESE INVERSE TRIG FUNCTIONS.
SO WE KNOW THAT THE VALUE OF Y MUST BE IN THIS GIVEN INTERVAL.
WHAT I'VE ALSO DONE IS INCLUDE THE REFERENCE TRIANGLES,
THE 30, 60, 90, AND THE 45, 45, 90 REFERENCE TRIANGLES.
AND THESE WILL BE VERY HELPFUL IN SOLVING THESE PROBLEMS.
SO WHEN WE LOOK AT NUMBER ONE,
Y = ARC SINE NEGATIVE SQUARE ROOT OF 3 OVER 2,
THE FIRST THING WE SHOULD DO IS TAKE A LOOK
AT WHAT WOULD THE REFERENCE BE FOR SQUARED OF 3 OVER 2.
AND HOPEFULLY YOU ARE PRETTY GOOD AT THIS NOW.
YOU CAN SEE THE REFERENCE ANGLE
WOULD BE A 60 DEGREE REFERENCE ANGLE
BECAUSE THAT GIVES YOU THE OPPOSITE OVER HYPOTENUSE
OF SQUARE ROOT 3 OVER 2.
EVEN THOUGH THIS IS NEGATIVE,
THIS WILL TELL US WHAT OUR REFERENCE ANGLE IS,
AND THEN WE CAN DETERMINE WHICH QUADRANT IT WOULD BE IN.
SO IF OUR ANGLE MUST BE IN THIS INTERVAL,
WE MUST BE IN EITHER THE FIRST OR THE FOURTH QUADRANT.
WE CAN'T BE IN THE FIRST
BECAUSE ALL VALUES ARE POSITIVE IN THAT QUADRANT.
SO WE'RE GOING TO BE IN QUADRANT TWO
WITH THE -60 DEGREE REFERENCE ANGLE.
AND, IN FACT, -60 DEGREES IS IN THIS INTERVAL.
SO THAT IS THAT THE VALUE OF ARC SINE NEGATIVE SQUARE ROOT
OF 3 OVER 2.
IF WE WANT THIS IN RADIANS IT WOULD BE -PI OVER 3.
OKAY. FOR NUMBER TWO WE HAVE Y = ARC COSINE SQUARE ROOT 2 OVER 2.
REMEMBER THE RANGE OR THE OUTPUT FOR ARC COSINE
IS ON THE CLOSED INTERVAL FROM ZERO TO PI.
SO NOW OUR ANGLE MUST BE IN EITHER THE FIRST
OR THE SECOND QUADRANT.
NEXT, THE COSINE FUNCTION VALUE IS SQUARE ROOT 2 OVER 2
WHICH SHOULD REMIND YOU OF A REFERENCE ANGLE OF 45 DEGREES.
ADJACENT OVER HYPOTENUSE WOULD BE ONE OVER SQUARE ROOT OF TWO.
RATIONALIZED WOULD BE SQUARE ROOT OF 2 OVER 2.
SO WE HAVE A 45 DEGREE REFERENCE ANGLE,
AND SINCE WE HAVE A POSITIVE SQUARE ROOT OF 2 OVER 2,
WE'RE IN THE FIRST QUADRANT.
AND THIS IS THE ANGLE WE'RE LOOKING FOR.
IT'S EQUAL TO 45 DEGREES WHICH IS PI OVER FOUR RADIANS.
WE HAVE Y = ARC TANGENT -1.
REMEMBER THE OUTPUT FOR THIS
WILL BE ON THE OPEN INTERVAL FROM -PI OVER 2 TO PI OVER 2.
SO NOW WE'RE IN EITHER THE FIRST OR THE FOURTH QUADRANT.
AND THIS -1 IS THE TANGENT FUNCTION VALUE.
WE'RE LOOKING FOR THE ANGLE.
HAVING A TANGENT FUNCTION VALUE OF -1
SHOULD REMIND YOU OF ANOTHER REFERENCE ANGLE OF 45 DEGREES.
THE TANGENT OF 45 DEGREES IS 1 OVER 1, WHICH GIVES US ONE.
BUT BECAUSE OUR TANGENT FUNCTION VALUE IS NEGATIVE
WE CAN'T BE IN THE FIRST QUADRANT.
WE'LL BE IN THE FOURTH QUADRANT
WITH A REFERENCE ANGLE OF 45 DEGREES.
BUT IN THIS INTERVAL
WE'RE ACTUALLY LOOKING AT AN ANGLE THAT IS -45.
IS EQUAL TO -PI OVER FOUR RADIANS.
SO WE CAN USE THE REFERENCE TRIANGLES
TO HELP US SOLVE THESE PROBLEMS.
WE CAN ALSO USE THE UNIT --
SO YOU MIGHT WANT TO TRY THAT AS WELL.
FOR NUMBER ONE I'LL PRESS SECOND SINE
AND THEN TYPE IN NEGATIVE SQUARE ROOT 3 OVER 2.
AND THAT CHECKS THE FIRST ONE.
NOW WE'LL PRESS SECOND COSINE FOR NUMBER TWO
AND THEN OUR COSINE FUNCTION VALUE
IS THE SQUARE ROOT 2 OVER 2.
SECOND ONE CHECKS OUT,
AND THE LAST ONE WE'LL PRESS SECOND TANGENT NEGATIVE ONE.
SO THEY ALL THREE CHECK OUT.
OKAY. LET'S TAKE A LOOK AT ONE MORE PROBLEM.
WE WANT TO SOLVE THIS RIGHT TRIANGLE.
SO WE'RE GO AHEAD AND START OUT BY FINDING THE HYPOTENUSE
BY USING THE PYTHAGOREAN THEOREM.
LET'S CALL THIS X.
WE HAVE X SQUARED = 5 SQUARED + 15 SQUARED.
SO X SQUARED = 25 + 225. THAT'S 250.
WE'LL TAKE THE SQUARE ROOT OF BOTH SIDES,
LET'S GO AHEAD AND ROUND TO TWO DECIMAL PLACES.
THIS COMES OUT TO APPROXIMATELY 15.81.
OKAY, SO THAT'S REVIEW,
BUT REMEMBER TO SOLVE A RIGHT TRIANGLE
WE HAVE TO FIND THE LENGTH OF ALL THREE SIDES
AND THE MEASURE OF ALL THREE ANGLES.
WE KNOW THIS IS 90 DEGREES.
LET'S CALL THIS ANGLE A,
AND THE TANGENT OF ANGLE A MUST EQUAL OPPOSITE OVER HYPOTENUSE,
OR 5 OVER 15,
WHICH DOES SIMPLIFY TO 1 OVER 3.
NOW, TO SOLVE THIS EQUATION FOR A
WE CAN TAKE THE INVERSE TANGENT OF BOTH SIDES.
AND ON THE LEFT SIDE
IF WE TAKE THE INVERSE TANGENT OF TANGENT A,
THESE TWO UNDO EACH OTHER.
WE HAVE ANGLE A AS AN APPROXIMATE EQUAL TO
INVERSE TANGENT OF 5 OVER 15.
SO ANGLE A IS APPROXIMATELY 18.43 DEGREES.
AND TO FIND ANGLE B,
ANGLE A AND B MUST BE COMPLIMENTARY
OR ADD TO 90 DEGREES.
SO THAT LEADS US WITH ANGLE B EQUAL TO 71.57 DEGREES.
AND ONE THING YOU SHOULD BE CAREFUL ABOUT
IS THIS IS A ROUNDED VALUE.
SO IT IS BEST TO USE ARC TANGENT OR INVERSE TANGENT
TO FIND THESE ANGLES
BECAUSE THESE TWO ARE PRECISE, AND THIS ONE IS ROUNDED.
SO IT MAY RESULT IN AN ERROR IF WE USE A HYPOTENUSE
TO FIND THE MEASURE OF ANGLE A AND ANGLE B.
OKAY, I HOPE YOU FOUND THIS VIDEO HELPFUL.
THANK YOU FOR WATCHING.