– WELCOME TO THE FIRST
OF SEVERAL VIDEOS WHERE WE’LL PROVE THE PROPERTIES
OF HYPERBOLIC FUNCTIONS. IN THE INTRODUCTORY VIDEO
WE DEFINED EACH OF THE SIX HYPERBOLIC
FUNCTIONS AS WE SEE HERE AND WE ALSO DISCUSSED
HOW THESE FUNCTIONS EVEN THOUGH
THEY’RE EXPONENTIAL FUNCTIONS HAVE SIMILAR PROPERTIES
TO TRIG FUNCTIONS. THE PROPERTIES
WE’RE TALKING ABOUT ARE THE PROPERTIES GIVEN HERE AND NOTICE IF YOU LOOK AT THESE
PROPERTIES OR IDENTITIES THEY LOOK VERY SIMILAR
TO SOME OF THE TRIG IDENTITIES. IN THIS VIDEO WE’RE GOING TO
FOCUS ON THIS FIRST PROPERTY HYPERBOLIC COSINE SQUARED X –
HYPERBOLIC SINE SQUARED X=1. NOTICE HOW THIS LOOKS VERY
SIMILAR TO THE TRIG IDENTITY COSINE SQUARED X + SINE
SQUARED X EQUALS 1 BUT NOTICE FOR THE HYPERBOLIC
FUNCTION IDENTITY WE HAVE A DIFFERENCE
RATHER THAN A SUM. SO TO PROVE THIS IDENTITY
WE’RE GOING TO WRITE HYPERBOLIC COSINE SQUARED X
IN EXPONENTIAL FORM, HYPERBOLIC SINE SQUARED X
IN EXPONENTIAL FORM THEN SQUARE THE FRACTIONS
AND THEN SUBTRACT THE FRACTIONS AND THEN HOPEFULLY SHOW THE LEFT
SIDE IS EQUAL TO +1. SO FOR HYPERBOLIC
COSINE SQUARED WE WOULD HAVE E TO THE X + E TO THE -X
DIVIDED BY 2 SQUARED AND THEN MINUS HYPERBOLIC SINE
SQUARED WHICH WOULD BE E TO THE X – E TO THE -X DIVIDED
BY 2 SQUARED. WE WANT TO SHOW THIS IS EQUAL
TO A +1. SO NOW WE’LL GO AHEAD AND SQUARE
EACH FRACTION. WELL, IF WE SQUARED THE
DENOMINATOR, 2 SQUARED IS=TO 4 SO FOR THIS FIRST FRACTION WE’D
HAVE A DENOMINATOR OF 4 AND THEN THE NUMERATOR
WOULD BE TWO FACTORS OF E TO THE X + E TO THE -X. AND THEN WE’LL HAVE MINUS
THIS FRACTION WOULD HAVE THE DENOMINATOR
OF 4 AS WELL AND THEN WE’D HAVE TWO FACTORS
OF E TO THE X – E TO THE -X IN THE NUMERATOR. NOW, WE’LL GO AHEAD AND MULTIPLY
THE NUMERATORS AND COMBINE LIKE-TERMS. AND JUST FOR REVIEW
WHEN MULTIPLYING AND THE BASES ARE THE SAME WE SHOULD REMEMBER THAT
WE WILL BE ADDING THE EXPONENTS. SO WHEN MULTIPLYING
THESE BINOMIALS TOGETHER WE’LL HAVE FOUR PRODUCTS,
ONE, TWO, THREE AND FOUR. SO E TO THE X x E TO THE X
WOULD BE E TO THE POWER OF X + X OR 2X THERE’S THAT PRODUCT AND THEN E TO THE X x E TO THE –
X WOULD BE E TO THE X + -X WHICH WOULD BE E TO THE 0,
WHICH IS EQUAL TO 1, WHICH WE’LL SIMPLIFY
IN THE NEXT STEP. AND THEN WE HAVE
E TO THE -X x E TO THE +X, AGAIN, THAT’S GOING TO BE
E TO THE 0. AND THEN WE HAVE E TO THE -X x E
TO THE -X WHICH WOULD BE +E TO THE -2X
SINCE -X + -X IS=TO -2X. LET’S GO AHEAD
AND PUT THIS IN PARENTHESIS AND NOW WE’RE GOING TO SUBTRACT
THE SECOND FRACTION AND AGAIN, SHOW THIS IS=TO +1 SO AGAIN,
WE’LL HAVE FOUR PRODUCTS ONE, TWO, THREE AND FOUR. SO AGAIN, E TO THE X x E
TO THE X IS E TO THE 2X. WE’RE SUBTRACTING THIS
ENTIRE PRODUCT HERE. AND THEN WE HAVE
E TO THE X x -E TO THE -X WHICH WOULD BE – E TO THE 0. AGAIN, THE 0 CAME FROM
ADDING X AND -X. AND THEN WE HAVE -E
TO THE -X x E TO THE X THAT WILL BE -E TO THE 0. AND THEN FINALLY, WE HAVE -E
TO THE -X x -E TO THE -X WHICH WOULD BE POSITIVE
OR +E TO THE -2X. AGAIN, THIS -2X CAME FROM
ADDING -X AND -X. NOW, FOR THIS NEXT STEP
WE WANT TO CLEAR THE PARENTHESIS AND THEN WE CAN GO AHEAD
AND COMBINE THE NUMERATORS BECAUSE WE HAVE
A COMMON DENOMINATOR. SO WE’LL HAVE A SINGLE FRACTION
WITH THE DENOMINATOR OF 4 AND TO CLEAR THE PARENTHESIS
WE CAN THINK OF DISTRIBUTING A 1 HERE BUT SINCE WE HAVE A MINUS HERE
WE CAN THINK OF DISTRIBUTING A -1 TO CLEAR THE SECOND SET
OF PARENTHESIS. SO WE’D HAVE E TO THE 2X,
REMEMBER E TO THE 0 IS=TO 1 SO THIS IS ACTUALLY 1+1
OR 2 AND 1 x 2 IS 2 AND THEN WE HAVE + E TO THE -2X. THEN WE’LL HAVE – E TO THE 2X AND AGAIN, THIS IS -1, -1
SO THIS IS ACTUALLY -2 OR NEGATIVE 2, -1 x -2
BECOMES +2 AND THEN WE HAVE – E TO THE -2X. SOME OF THESE TERMS
ARE OPPOSITES E TO THE POWER OF 2X – E
TO POWER OF 2X WOULD BE 0 AND SO WOULD E TO THE -2X – E
TO THE -2X SO WE HAVE 2 + 2, WHICH IS 4. SO WE HAVE 4/4=1 AND 4/4=1 THEREFORE WE HAVE OUR PROOF. WE’LL GO AHEAD AND STOP HERE
FOR THIS VIDEO. WE’LL GO AHEAD AND PROVE
ANOTHER IDENTITY INVOLVING HYPERBOLIC FUNCTIONS
IN THE NEXT VIDEO, I HOPE YOU FOUND THIS HELPFUL.