- WELCOME TO A VIDEO ON HALF ANGLE IDENTITIES.
THE GOAL OF THIS VIDEO IS TO USE HALF ANGLE IDENTITIES
TO DETERMINE FUNCTION VALUES.
SO HERE WE SEE ALL OF THE HALF ANGLE IDENTITIES.
NOTICE A LOT OF THESE SEEM VERY SIMILAR SO BE VERY CAREFUL
WHEN YOU COPY THEM DOWN.
NOTICE THAT THE FIRST THREE HAVE A PLUS OR MINUS SIGN ON THEM.
WHAT THIS TELLS US IS THAT WE HAVE TO DETERMINE
WHETHER THE FUNCTION VALUE WILL BE POSITIVE OR NEGATIVE
BASED UPON THE QUADRANT OF ANGLE A/2.
I DID INCLUDE THE DERIVATION OF THESE IDENTITIES.
WE DON'T HAVE TIME RIGHT NOW TO GO THROUGH THEM
BUT YOU MAY WANT TO PAUSE THE VIDEO
AND SEE HOW WE CAN OBTAIN THESE FORMULAS
AND THE NEXT SLIDE ALSO SHOWS
HOW TO FIND THE HALF ANGLE IDENTITY FOR TANGENT.
AGAIN, NOTICE THAT IT COMES FROM TAKING SINE OF 1/2 THETA
AND DIVIDING BY A COSINE 1/2 THETA.
BUT LET'S GO AHEAD AND TAKE A LOOK AT SOME PROBLEMS.
HERE WE WANT TO DETERMINE THE EXACT VALUE OF SINE PI/8.
SO WE KNOW WE'RE GOING TO USE THIS IDENTITY.
THE FIRST THING WE SHOULD DO IS FIND THE VALUE OF ANGLE "A."
THERE'S A COUPLE WAYS OF DOING THAT.
WE WANT A/2 TO EQUAL PI/8.
SO IF WE PERFORM CROSS PRODUCTS WE'LL HAVE 8A MUST EQUAL 2 x PI.
DIVIDING BY 8 WE CAN SEE THAT "A" MUST EQUAL 2 PI/8 OR PI/4.
SO WHEN WE SET THIS UP WE'RE GOING TO HAVE THE SINE OF PI/8
EQUALS THE SINE OF PI/4 DIVIDED BY 2.
SO WE'RE GOING TO USE PI/4 FOR ANGLE "A".
ANOTHER WAY TO THINK OF IT IS THAT
IF WE WANT PI/8 TO EQUAL A/2 WE'D HAVE TO DOUBLE THIS
AND THEN DIVIDE BY 2 FOR THEM TO BE EQUAL.
OKAY, SO SUBBING THIS INTO OUR FORMULA
WE'RE GOING TO HAVE + OR - THE SQUARE ROOT OF 1 - THE COSINE
OF ANGLE "A" WHICH IS PI/4 WHICH IS GOOD NEWS
BECAUSE THAT'S ONE OF THOSE REFERENCE ANGLES
AND THEN DIVIDED BY 2.
SO IF YOU DON'T HAVE YOUR UNIT CIRCLE HANDY
WE SHOULD SKETCH PI/4 IN STANDARD POSITION.
IT'S ACTUALLY A 45-45 RIGHT TRIANGLE.
SO THE COSINE OF PI/4 WOULD BE 1/SQUARE ROOT 2
OR IF WE RATIONALIZE THAT WE WOULD HAVE SQUARE ROOT 2/2.
SO WE'LL HAVE 1 - SQUARE ROOT 2/2 ALL OVER 2.
NOW LETS CLEAR THE FRACTIONS IN THE NUMERATOR.
SO WE'RE GOING TO MULTIPLY THE TOP AND BOTTOM BY 2
UNDERNEATH THE SQUARE ROOT.
WHEN WE DO THIS WE'LL HAVE 2 x 1
THAT WILL GIVE US 2 - SQUARE ROOT OF 2 DIVIDED BY 2 x 2
WILL GIVE US THE SQUARE ROOT OF 2 AND THEN 2 x 2 OF COURSE IS 4.
OUR LAST STEP, SINCE PI/8 IS IN THE 1st QUADRANT
WE KNOW OUR FUNCTION VALUE WILL BE POSITIVE.
NEXT, THE SQUARE ROOT OF 4 WOULD EQUAL 2
AND THEN OUR NUMERATOR DOES NOT SIMPLIFY
AND WE HAVE NOW FOUND OUR FUNCTION VALUE
THAT WE WERE LOOKING FOR.
SINE OF PI/8 IS EQUAL TO THE SQUARE ROOT OF 2
- THE SQUARE ROOT OF 2 ALL DIVIDED BY 2.
LET'S GO AHEAD AND CHECK THIS ON THE GRAPHING CALCULATOR.
HERE YOU CAN SEE I'VE COMPARED THE SINE OF PI/8 IN RADIANS
TO THE EXACT VALUE THAT WE FOUND
AND THEY ARE THE SAME WHICH VERIFIES OUR WORK.
WE WANT TO FIND THE EXACT VALUE OF COSINE 105 DEGREES.
SO WE'LL USE THIS IDENTITY.
AGAIN, IF WE WANT A/2 TO EQUAL 105 DEGREES
WE'D HAVE TO DOUBLE THIS.
SO 210 DIVIDED BY 2 WILL GIVE US 105 DEGREES.
SO WE'RE GOING TO USE 210 DEGREES FOR ANGLE "A."
AGAIN IF WE DON'T HAVE OUR UNIT CIRCLE HANDY
COSINE OF 210 DEGREES THAT WOULD BE 30 DEGREES
PAST 180 MAKES THIS IS A 30-60-90 RIGHT TRIANGLE
SO WE HAVE 1 2 SQUARE ROOT 3 BUT BOTH X AND THE Y COORDINATES
WOULD BE NEGATIVE.
SO THE COSINE OF 210 DEGREES
WOULD BE THE ADJACENT OVER THE HYPOTHENUSE OR -SQUARE ROOT 3/2.
NEXT, LETS CLEAR THE FRACTIONS
AGAIN SO I'LL MULTIPLY THE TOP AND THE BOTTOM
INSIDE THE SQUARE ROOT BY 2.
WELL 2 x 1 WOULD BE 2,
2 x SQUARE ROOT 3 DIVIDED BY 2 WOULD BE -SQUARE ROOT 3
AND THEN 2 x 2 OF COURSE IS 4
AND THEN LASTLY, 105 DEGREES FALLS IN THE 2nd QUADRANT
WHERE THE X COORDINATE WOULD BE NEGATIVE
THEREFORE THE COSINE VALUE WOULD BE NEGATIVE
SO THIS WILL BE NEGATIVE.
THE SQUARE ROOT OF 4 WOULD GIVE US 2.
THE NUMERATOR DOES NOT SIMPLIFY.
WE HAVE THE SQUARE ROOT OF 2 - SQUARE ROOT OF 3
AND THERE'S OUR FUNCTION VALUE.
OKAY, WE HAVE ONE MORE QUESTION.
HERE WE'RE GIVEN COSINE "A" IS EQUAL TO -2/3
IN THE 2nd QUADRANT.
WE WANT TO DETERMINE COSINE A/2, SINE OF A/2 AND TANGENT A/2.
FIRST THING I WANT TO MAKE A NOTE OF IS ANGLE "A"
IS IN THE 2nd QUADRANT WHICH MEANS
IT'S BETWEEN 90 DEGREES AND 180 DEGREES.
SO A/2 WOULD BE BETWEEN 90 DEGREES AND 45 DEGREES
WHICH MEANS IT'S IN QUADRANT 1.
WHAT THAT TELLS US IS ALL OF THESE FUNCTION VALUES
WILL BE POSITIVE SO LETS JUST MAKE A NOTE OF THAT HERE.
ALL FUNCTION VALUES WILL BE POSITIVE.
NEXT, LETS GO AHEAD AND SKETCH THIS ANGLE
IN A STANDARD POSITION.
WE'RE IN THE 2nd QUADRANT
SO HERE'S OUR REFERENCE TRIANGLE AND THIS REFERENCE ANGLE
HAS A COSINE VALUE OF -2/3.
USING THE PYTHAGOREAN THEOREM WE CAN DETERMINE THIS THIRD SIDE
OF THIS REFERENCE TRIANGLE
AND IT WILL EQUAL THE SQUARE ROOT OF 5.
LET'S GO AHEAD AND TRY TO FIND THESE FUNCTION VALUES.
COSINE OF A/2 IS EQUAL TO THE SQUARE ROOT OF 1 + COSINE "A"
WHICH IS -2/3 ALL OVER 2.
REMEMBER WE ALREADY DETERMINED ALL OF THESE WILL BE POSITIVE.
LET'S GO AHEAD AND CLEAR THE FRACTIONS.
OUR DENOMINATOR WILL BE 6,
NUMERATOR WOULD BE 3 + -2
WHICH WOULD GIVE US THE SQUARE ROOT OF 1/6
WHICH IS EQUAL TO 1/SQUARE ROOT OF 6
AND IF WE RATIONALIZE THIS
WE WOULD HAVE THE SQUARE ROOT OF 6/6.
LET'S GO AHEAD AND TRY TO FIND THE SINE OF A/2.
WE HAVE SQUARE ROOT OF 1 - COSINE "A"
SO 1 - -2/3 ALL OVER 2.
AGAIN CLEARING THE FRACTION BY MULTIPLYING THE TOP
AND BOTTOM BY THE DENOMINATOR OF 3
WE'D HAVE 6 AGAIN IN OUR DENOMINATOR
AND HERE WE WOULD HAVE 3 + 2
WHICH EQUALS THE SQUARE ROOT OF 5/6
WHICH EQUALS THE SQUARE ROOT OF 5/THE SQUARE ROOT OF 6
AND WE RATIONALIZE THIS.
WE WILL HAVE THE SQUARED OF 30/6 AND WE HAVE ONE MORE.
WE WANT TO FIND THE TANGENT OF A/2.
NOTICE WE HAVE A VARIETY OF OPTIONS HERE.
I'M GOING TO GO AHEAD AND USE THE LAST ONE THIS TIME.
OF COURSE, IF WE HAVE SINE OF A/2 AND COSINE A/2
WE COULD JUST FIND THE QUOTIENT OF THE TWO.
BUT I'LL GO AHEAD AND USE THE IDENTITY.
1 - COSINE ANGLE "A" WHICH IS GIVEN AS -2/3
ALL OVER SINE OF ANGLE "A."
GOING BACK OVER TO OUR REFERENCE TRIANGLE THE SINE OF ANGLE "A"
WOULD BE OPPOSITE OVER HYPOTHENUSE
OR SQUARE ROOT OF 5/3.
1 - -2/3 THAT WOULD WORK OUT TO 5/3
AND THEN INSTEAD OF DIVIDING BY THE SQUARE ROOT OF 5/3
WE'LL MULTIPLY BY THE RECIPROCAL.
THIS SIMPLIFIES NICELY.
SO WE HAVE A FUNCTION VALUE OF 5/SQUARE ROOT OF 5
WHICH IS CORRECT BUT WE'LL RATIONALIZE IT ONE MORE TIME.
SO WE HAVE 5 SQUARE ROOT 5/5
WHICH SIMPLIFIES TO SQUARE ROOT OF 5 FOR THE TANGENT OF A/2.
OKAY, THAT WAS QUITE A BIT OF WORK
AND A LOT OF ALGEBRA INVOLVED
SO BE VERY CAREFUL WHEN YOU WORK THROUGH THESE PROBLEMS
AND I WISH YOU SUCCESS.
THANK YOU FOR WATCHING AND HAVE A GOOD DAY.