- WELCOME BACK FOR ANOTHER VIDEO THAT WILL SHOW
A PROOF OF ONE OF THE PROPERTIES OF HYPERBOLIC FUNCTIONS.
AGAIN, WE'VE ALREADY DEFINED
THE SIX HYPERBOLIC FUNCTIONS GIVEN HERE,
WE ALSO DISCUSSED HOW WHILE EVEN THOUGH
THESE ARE EXPONENTIAL FUNCTIONS THEY SHARE SIMILAR PROPERTIES
TO TRIG FUNCTIONS SOME OF WHICH WE CAN SEE HERE ON THIS SCREEN
BUT THERE'S ALSO SEVERAL OTHER PROPERTIES GIVEN HERE
ON THIS SCREEN.
IN THIS VIDEO WE'RE GOING TO BE PROVING THIS IDENTITY HERE
WHERE HYPERBOLIC TANGENT
SQUARED X PLUS HYPERBOLIC SECANT SQUARED X IS = TO +1.
SO LET'S GO AHEAD AND GET STARTED.
I WANT TO SHOW A PRETTY DETAILED PROOF
SO I'M GOING TO WRITE EACH OF THESE HYPERBOLIC FUNCTIONS
IN THE EXPONENTIAL FORM SQUARE THEM
AND THEN SHOW THE LEFT SIDE IS EQUAL TO +1.
SINCE HYPERBOLIC TANGENT X IS = TO HYPERBOLIC SINE X
DIVIDED BY HYPERBOLIC COSINE X
WE'RE GOING TO WRITE
HYPERBOLIC TANGENT SQUARED X AS E TO THE X - E
TO THE -X DIVIDED BY 2,
THIS IS = TO HYPERBOLIC SINE X DIVIDED BY HYPERBOLIC COSINE X,
WHICH IS = TO E TO THE X + E TO THE -X DIVIDED BY 2.
BUT SINCE WE HAVE HYPERBOLIC TANGENT SQUARED X
THIS IS ACTUALLY ALL BEING SQUARED
PLUS HYPERBOLIC SECANT SQUARED X WILL BE
HYPERBOLIC SECANT FUNCTION IS EQUAL TO THE RECIPROCAL
OF THE HYPERBOLIC COSINE FUNCTION
SO WE CAN WRITE THIS
AS 2 DIVIDED BY E TO THE X + E TO THE -X SQUARED.
AGAIN, THIS IS THE HYPERBOLIC SECANT FUNCTION
WHICH IS THE RECIPROCAL OF THE HYPERBOLIC COSINE FUNCTION.
WE WANT TO SHOW THAT THIS IS EQUAL TO +1.
LET'S START BY SIMPLIFYING INSIDE THE PARENTHESES HERE.
WE CAN ACTUALLY SIMPLIFY OUT THE DENOMINATORS OF 2
IF WE MULTIPLY THE NUMERATOR BY 2
AND THE DENOMINATOR BY 2,
WHICH IS JUST MULTIPLYING BY 2/2 WHICH IS = TO 1.
NOW, WHEN MULTIPLYING THE NUMERATOR BY 2,
THIS 2 AND THIS 2 IS SIMPLIFIED OF 1
AND THEN WHEN MULTIPLYING THE DENOMINATOR BY 2,
THIS 2 AND THIS 2 SIMPLIFY TO 1.
WHICH MEANS THIS FRACTION NOW HAS A NUMERATOR
THAT CONSISTS OF TWO FACTORS OF THE E TO THE X - E TO THE -X.
AND THE DENOMINATOR THAT CONTAINS
TWO FACTORS OF E TO THE X + E TO THE -X.
AND NOW FOR THE SECOND FRACTION WE CAN GO AHEAD AND SQUARE 2
THAT WOULD BE 4
AND THEN IF WE SQUARE THE DENOMINATOR
WE WOULD HAVE TWO FACTORS OF E TO THE X + E TO THE -X.
AT THIS POINT, NOTICE HOW WE DO HAVE A COMMON DENOMINATOR
WHICH MEANS YOU CAN COMBINE THESE TWO FRACTIONS,
THE FRACTION WOULD HAVE A COMMON DENOMINATOR OF E TO THE X + E
TO THE -X SQUARED
AND THE NUMERATOR WILL CONSIST OF TWO FACTORS
OF E TO THE X -E TO THE -X + 4.
AND NOW FOR THE NEXT STEP WE'LL GO AHEAD
AND MULTIPLY THESE TWO BINOMIALS
AND THEN WE'LL COMBINE LIKE-TERMS.
REMEMBER WHEN MULTIPLYING AND THE BASES ARE THE SAME
WE ADD THE EXPONENTS.
SO E TO THE X x E TO THE X WOULD BE E TO THE 2X
SINCE X + X IS = TO 2X.
NEXT WE HAVE E TO THE X x -E TO THE -X
WHICH WILL BE -E TO THE X + -X WHICH WOULD BE E TO THE 0.
AND THEN WE HAVE -E TO THE -X x E TO THE X
WHICH AGAIN WOULD BE -E TO THE 0.
REMEMBER, THESE ARE BOTH = TO 1
SINCE ANYTHING TO THE 0 POWER IS EQUAL TO 1.
AND THEN FINALLY, WE HAVE -E TO THE -X x -E TO -X
WHICH WILL BE POSITIVE OR +E TO THE -2X
AND WE STILL HAVE +4 = 1.
LET'S GO AHEAD AND CONTINUE THIS ON THE NEXT SLIDE.
WELL, SINCE E TO THE 0 IS = TO 1 THIS IS -1 -1
OR -2, -2 + 4 IS = TO +2 SO I CAN WRITE THIS
AS E TO THE 2X + 2 + E TO THE - 2X ALL OVER OUR DENOMINATOR.
NOW AT THIS POINT IT GETS A LITTLE TRICKY
BUT IF WE KEEP IN MIND WE WANT TO SHOW THIS IS = TO +1
THE ONLY WAY THIS WILL = 1
IS IF THIS NUMERATOR IS = TO THE DENOMINATOR.
WHICH MEANS IF THIS FACTORS IT'LL FACTOR
INTO TWO EQUAL FACTORS.
SO LET'S GO AHEAD AND TRY IT.
THE TWO EQUAL FACTORS OF E TO THE 2X WOULD BE
E TO THE X AND E TO THE X,
THE TWO EQUAL FACTORS OF E TO - 2X WOULD BE
E TO THE -X AND E TO THE -X.
AND BECAUSE THE INNER TERM AND BECAUSE THE MIDDLE TERM
IS POSITIVE THESE WILL BOTH NEED TO BE PLUS.
NOW, TO CHECK TO MAKE SURE THIS WORKS WE KNOW
THAT THE INNER PRODUCT AND OUTER PRODUCT MUST HAVE A SUM OF +2.
SO LET'S GO AHEAD AND CHECK THAT.
THE OUTER PRODUCT WOULD BE E TO THE POWER OF X + -X
WHICH WOULD BE E TO THE 0
AND THE INNER PRODUCT WOULD BE E TO THE 0 AS WELL
AND SINCE E TO THE 0 IS = TO 1 THIS IS JUST 1 + 1
WHICH IS = TO 2, WHICH IS OUR MIDDLE TERM.
SO THIS DOES FACTOR INTO TWO EQUAL FACTORS
AND NOTICE HOW THESE FACTORS ARE THE SAME FACTORS
IN THE DENOMINATOR
SO THIS WILL = +1.
LET GO AHEAD AND WRITE THIS AS E TO THE X + E TO THE -X SQUARED
ALL OVER E TO THE X + E TO THE - X SQUARED = 1.
AND NOW WE CAN EASILY SEE THE LEFT SIDE IS = TO 1
AND OF COURSE, THE RIGHT SIDE IS = TO 1
AND THEREFORE WE HAVE OUR PROOF.
OKAY, I HOPE YOU FOUND THIS EXPLANATION HELPFUL.
WE'LL TAKE A LOOK AT ANOTHER PROOF IN THE NEXT VIDEO.