- WELCOME TO A LESSON ON USING INVERSE SINE, COSINE AND TANGENT
TO EVALUATE EXPRESSIONS AND SOLVE PROBLEMS.
LET'S DO A QUICK REVIEW ON INVERSE TRIG FUNCTIONS.
Y EQUALS ARC SINE X OR Y EQUALS INVERSE SINE X
MEAN THE SAME THING,
WHICH IS EQUIVALENT TO X EQUALS SINE Y,
WHERE Y IS AN ANGLE IN THE CLOSED INTERVAL
FROM -PI/2 TO PI/2.
Y EQUALS ARC COSINE OF X OR Y EQUALS INVERSE SINE OF X
MEANS X EQUALS COSINE Y
WHERE Y IS IN THE CLOSED INTERVAL FROM ZERO TO PI.
AND LASTLY, Y EQUALS ARC TANGENT X OR Y EQUALS INVERSE TANGENT X
MEANS X EQUALS TANGENT Y,
AND Y IS ON THE OPEN INTERVAL FROM -PI/2 TO PI/2.
SO THE MAIN THING WE HAVE TO KEEP IN MIND HERE IS THAT
FOR INVERSE TRIG FUNCTIONS THE INPUT, OR THESE X VALUES,
ARE ACTUALLY TRIG RATIOS OR TRIG FUNCTION VALUES,
AND THE OUTPUT Y IS ACTUALLY THE ANGLE
THAT HAS THE GIVEN TRIG FUNCTION VALUES.
SO, FOR EXAMPLE, HERE ON NUMBER ONE
WE HAVE THE SINE OF ARC COSINE 1/4.
ARC COSINE 1/4 IS GOING TO RETURN AN ANGLE
ON THE CLOSED INTERVAL FROM ZERO TO PI
THAT HAS A COSINE FUNCTION VALUE OF 1/4.
ONCE WE FIND THAT ANGLE, OR AT LEAST MODEL IT,
THEN WE'LL FIND THE SINE OF IT.
SO LET'S GO AHEAD AND SKETCH A MODEL OF THIS ANGLE THETA.
WE KNOW THE ANGLE MUST BE IN EITHER THE FIRST
OR SECOND QUADRANT,
BUT SINCE THE COSINE FUNCTION VALUE IS POSITIVE
WE'RE GOING TO BE IN THE FIRST QUADRANT.
SO WE'LL SKETCH A REFERENCE TRIANGLE IN QUADRANT ONE.
HERE IS OUR ANGLE THETA.
AND THE RATIO OF THE ADJACENT TWO HYPOTENUSE IS ONE TO FOUR.
SO WE HAVE ONE HERE AND FOUR HERE.
WHAT WE WANT TO DO IS FIND THE SINE RATIO FOR THETA
WHICH IS THE RATIO OF THE OPPOSITE SIDE TO THE HYPOTENUSE.
WE CAN FIND THE LENGTH OF IT BY USING THE PYTHAGOREAN THEOREM,
WHICH STATES THAT 4 SQUARED MUST EQUAL Y SQUARED + 1 SQUARED.
SOLVING THIS FOR Y SQUARED WE'LL HAVE Y SQUARED = 15.
SQUARE ROOT BOTH SIDES.
Y = SQUARE ROOT OF 15.
SO THE SINE OF ANGLE THETA WILL BE SQUARE ROOT 15/4.
AND THAT'S THE VALUE WE'RE LOOKING FOR.
LET'S GO AHEAD AND TAKE A LOOK AT NUMBER TWO NOW.
COSINE OF ARC TANGENT 4/3.
SO OUR ANGLE THETA HAS A TANGENT FUNCTION VALUE OF 4/3.
AND, AGAIN, IT'S POSITIVE
SO WE'RE GOING TO BE IN THE FIRST QUADRANT AGAIN.
LET'S SKETCH OUR REFERENCE TRIANGLE.
HERE'S OUR ANGLE THETA.
THE RATIO OF THE OPPOSITE TO ADJACENT SIDE IS 4, 3.
AND THIS IS ACTUALLY A 3, 4, 5 RIGHT TRIANGLE.
SO WE KNOW THE HYPOTENUSE WILL EQUAL FIVE UNITS.
NOW THAT WE'VE MODELED OUR ANGLE THETA
WE WANT TO FIND THE COSINE OF IT.
COSINE IS ADJACENT OVER HYPOTENUSE.
SO IN THIS CASE WE'LL HAVE 3/5.
OKAY, NUMBER THREE IS A LITTLE BIT DIFFERENT.
WE WANT TO FIND THE SINE OF THE QUANTITY
ARC SINE 1/2 + ARC TANGENT NEGATIVE SQUARE ROOT 3.
ARC SINE OF +1/2,
WHAT ANGLE HAS A SINE FUNCTION VALUE OF 1/2?
WE CAN SEE FROM THIS 30, 60, 90 RIGHT TRIANGLE
THAT WOULD BE 30 DEGREES.
SO WE'LL HAVE THE SINE OF 30 DEGREES PLUS.
NOW WE NEED TO FIND THE ARC TANGENT
OF NEGATIVE SQUARE ROOT 3.
LOOKING AT OUR REFERENCE TRIANGLE
WE CAN SEE THAT A 60 DEGREE ANGLE
HAS A TANGENT RATIO OF SQUARE ROOT 3/1 OR SQUARE ROOT 3.
HOWEVER, SINCE WE HAVE NEGATIVE SQUARE ROOT 3,
AND THE INVERSE TANGENT FUNCTION RETURNS AN ANGLE
BETWEEN -PI/2 AND PI/2,
WE'RE GOING TO SKETCH A 60 DEGREE REFERENCE ANGLE
IN THE FOURTH QUADRANT,
WHICH MEANS OUR ANGLE THETA IS GOING TO BE -60 DEGREES.
SO GOING BACK TO OUR PROBLEM
WE HAVE SINE OF 30 DEGREES + -60 DEGREES,
WHICH IS EQUAL TO THE SINE OF -30 DEGREES.
LET'S GO AHEAD AND MODEL THIS.
A -30 DEGREE ANGLE WOULD LOOK SOMETHING LIKE THIS.
SO OUR REFERENCE TRIANGLE WOULD BE 1, 2, SQUARE ROOT OF 3,
WHERE THE Y COORDINATE WOULD BE -1.
SO THE SINE OF THIS ANGLE
IS GOING TO BE OPPOSITE OVER HYPOTENUSE OR -1/2.
LET'S GO AHEAD AND TAKE A LOOK AT ANOTHER PROBLEM.
A 10-FOOT LADDER IS LEANING UP AGAINST THE WALL.
THE FOOT OF THE LADDER IS 3 FEET FROM THE BASE OF THE WALL.
HOW FALL WILL THE LADDER REACH
AND WHAT IS THE ANGLE THE LADDER MAKES WITH THE GROUND?
SO IF THE LADDER IS LEANING AGAINST THE WALL
THIS IS GOING TO BE 10 FEET.
IT'S 3 FEET FROM THE BASE OF THE WALL,
SO THIS WILL BE 3 FEET.
SO THIS WOULD BE ACTUALLY THE HEIGHT THE LADDER WILL REACH,
AND THIS WOULD BE THE ANGLE THAT IT MAKES WITH THE GROUND.
WE WANT TO FIND H AND WE ALSO WANT TO FIND THETA.
WELL, WE CAN USE THE PYTHAGOREAN THEOREM TO FIND H
WHERE 10 SQUARED MUST EQUAL H SQUARED + 3 SQUARED.
SO WE'LL SUBTRACT 9 ON BOTH SIDES,
AND THEN TAKE THE SQUARE ROOT OF BOTH SIDES.
THE SQUARE ROOT OF 91 IS APPROXIMATELY 9.54 FEET.
NOW, TO FIND ANGLE THETA
I'M GOING TO USE THE SIDES OF THE RIGHT TRIANGLE
LABELED 3 FEET AND 10 FEET.
SO WE COULD STATE THE COSINE OF ANGLE THETA
IS 3 FEET OVER 10 FEET OR 3/10.
NOW, TO SOLVE THIS EQUATION FOR THETA
WE CAN TAKE THE INVERSE COSINE OF BOTH SIDES.
ON THE LEFT THESE TWO UNDO EACH OTHER
AND WE'RE LEFT WITH THETA
AND ON THE RIGHT WE CAN EVALUATE THIS WITH OUR CALCULATORS.
INVERSE COSINE 3/10.
SO SECOND COSINE 3/10,
AND OUR ANGLE IS APPROXIMATELY 72.54 DEGREES.
LET'S GO AHEAD AND TAKE A LOOK AT ONE MORE.
THE OPTIMAL ANGLE OF ELEVATION A SHOT PUTTER SHOULD AIM FOR
TO THROW THE GREATEST DISTANCE
DEPENDS ON THE VELOCITY IN FEET PER SECOND
AND THE INITIAL HEIGHT, H, IN FEET OF THE SHOT
IS GIVEN BY THIS EQUATION.
SO OUR QUESTION IS, IF A SHOT PUTTER CAN THROW THE SHOT PUT
WITH HEIGHT OF 6.6 FEET
AND A VELOCITY OF 42 FEET PER SECOND,
AT WHAT ANGLE SHOULD THE ATHLETE THROW TO MAXIMIZE THE DISTANCE?
SO ESSENTIALLY WE'RE GOING TO JUST DO SOME SUBSTITUTION HERE
AND THEN EVALUATE THE ARC SIGN.
SO, AGAIN, OR VELOCITY IS 42 FEET PER SECOND.
SO WE HAVE 42 SQUARED HERE.
AND WE HAVE 2 x 42 SQUARED + 64
x THE HEIGHT WHICH IS GIVEN AS 6.6.
REMEMBER, ALL OF THIS IS UNDERNEATH THE SQUARE ROOT.
LET'S SEE IF WE CAN DO THIS ON OUR CALCULATOR.
IF WE'RE CAREFUL WE CAN DO THIS ALL IN ONE STEP.
SO 2nd, SINE.
WE'VE HAVE THE SQUARE ROOT.
WE HAVE A NUMERATOR OF 42 SQUARED
DIVIDED BY OUR DENOMINATOR, ANOTHER SET OF PARENTHESIS.
WE CLOSE THIS FOR OUR DENOMINATOR
AND THEN WE CLOSE THIS FOR THE INVERSE SINE,
AND WE HAVE AN ANGLE THAT'S APPROXIMATELY 41.93 DEGREES.
I HOPE YOU FOUND THIS VIDEO HELPFUL.
THANK YOU FOR WATCHING.