(New Version Available) Solving Trigonometric Equations IV

(New Version Available) Solving Trigonometric Equations IV


[ MUSIC ] – WELCOME TO THE FOURTH VIDEO
ON SOLVING TRIG EQUATIONS. THE GOAL OF THIS VIDEO
IS TO SOLVE TRIG EQUATIONS WITH HALF ANGLE
AND MULTIPLE ANGLES. WE WANT TO SOLVE THIS EQUATION
ON THE INTERVAL FROM ZERO TO 360 DEGREES. AND NOTICE WHAT’S DIFFERENT
ABOUT THIS ONE IS WE HAVE HALF ANGLES. BUT LET’S GO AHEAD AND
APPROACH THIS IN THE SAME WAY. LET’S ADD SINE X/2
TO BOTH SIDES OF THE EQUATION. THAT WOULD GIVE US 2 SINE OF X
DIVIDED BY 2=SQUARE ROOT 2. WE’LL DIVIDE BOTH SIDES BY 2. WE HAVE SINE OF X DIVIDED BY 2
=SQUARE ROOT 2/2. THIS LOOKS JUST LIKE
OTHER TYPES OF EQUATIONS THAT WE SOLVED, EXCEPT NOW WE HAVE
A HALF ANGLE. WHAT WE CAN DO HERE IS LET
THIS HALF ANGLE EQUAL U. LET’S REWRITE THIS
IN TERMS OF U NOW. WE HAVE SINE U
=SQUARE ROOT 2/2. WE’LL SOLVE THIS LIKE
WE NORMALLY WOULD TO FIND U, AND THEN WE’LL SOLVE FOR X
IN THIS EQUATION. A RATIO OF SQUARE ROOT 2/2
SHOULD REMIND YOU OF A 45, 45, 90
REFERENCE TRIANGLE. THE SINE OF 45 DEGREES
WOULD BE 1/SQUARE ROOT 2, WHICH WOULD RATIONALIZE
TO SQUARE ROOT 2/2. SO WE HAVE A REFERENCE ANGLE
OF 45 DEGREES. THE NEXT THING
WE HAVE TO REMEMBER IS THAT SINE THETA
WILL BE POSITIVE IN THE FIRST
AND THE SECOND QUADRANT. SO WE HAVE TO SKETCH A
REFERENCE ANGLE OF 45 DEGREES IN THE FIRST QUADRANT
AND ALSO THE SECOND QUADRANT. AND LET’S GO AHEAD AND JUST
LABEL ALL THESE SIDES. SO WE HAVE TWO POSSIBLE ANGLES
FOR U. WE CAN HAVE A 45 DEGREE ANGLE. WE COULD ALSO HAVE
A 135 DEGREE ANGLE. NOW, BE CAREFUL BECAUSE OUR SOLUTION NEEDS
TO BE IN TERMS OF X, NOT U. SO WE CAN GO AHEAD AND
REPLACE U INTO THIS EQUATION AND THEN SOLVE FOR X. WE CAN SEE HERE TO SOLVE FOR X WE MULTIPLY BOTH SIDES
OF THE EQUATION BY 2, SO X=90 DEGREES AND X=270
DEGREES FOR THIS PROBLEM. LET’S GO AHEAD
AND TRY ANOTHER ONE. HERE WE HAVE A TRIPLE ANGLE. LET’S FOLLOW
THE SAME PROCEDURE. LET’S SOLVE THIS
FOR COSINE 3X. WE’LL ADD SQUARE ROOT 3
TO BOTH SIDES, DIVIDE BY 2. SO, AGAIN, THIS LOOKS
VERY FAMILIAR, BUT NOW WE’RE GOING TO REPLACE
3X WITH U INSTEAD. SO WE’LL HAVE COSINE U
=SQUARE ROOT 3/2. WE’LL FIND U
AND THEN COME BACK HERE AND FIND THE VALUE OF X. AND THE RATIO
OF SQUARE ROOT 3/2 SHOULD REMIND YOU OF A 30,
60, 90 RIGHT TRIANGLE. THE COSINE OF 30 DEGREES
=SQUARE ROOT 2/2. SO WE HAVE A REFERENCE ANGLE
OF 30 DEGREES. COSINE IS POSITIVE IN THE
FIRST AND THE FOURTH QUADRANT. SO WE’LL SKETCH A REFERENCE
ANGLE OF 30 DEGREES IN THE FIRST
AND THE FOURTH QUADRANT. SO WE’LL SKETCH TWO DIFFERENT
REFERENCE TRIANGLES, ONE HERE IN THE FIRST AND ONE
HERE IN THE SECOND AS WELL. AND, AGAIN, WE’LL LABEL
ALL THE SIDES OF THE TRIANGLE, AND THEN WE DETERMINE
WHICH ANGLES THESE ARE. WELL, OBVIOUSLY THE FIRST ONE
IS 3 DEGREES, AND THE SECOND ANGLE WOULD BE
360 – 30 OR 330 DEGREES. AGAIN, THAT’S WHAT U
IS EQUAL TO. WE WANT TO FIND THE VALUE
OF X. SO USING THIS EQUATION NOW
WE HAVE 30=3X. DIVIDING BOTH SIDES BY 3,
WE HAVE X=10 DEGREES. AND FOR U=330,
WE HAVE 330=3X, DIVIDING BOTH SIDES BY 3,
X=110 DEGREES. SO IT’S IMPORTANT TO REMEMBER
WHEN WE DO THIS SUBSTITUTION WE HAVE TO GO BACK
AND FIND THE VALUE OF X. NOW, THIS PROBLEM
IS A LITTLE BIT DIFFERENT. WE HAVE COSINE 2 THETA
AND A COSINE THETA. SINCE THESE ANGLES
ARE NOT THE SAME, WE’RE GOING TO HAVE TO PERFORM
A SUBSTITUTION. REMEMBER THAT COSINE 2 THETA
=2 COSINE SQUARE THETA – 1. SO WE’RE GOING TO PERFORM
THE SUBSTITUTION FOR COSINE 2 THETA. THIS IS IN QUADRATIC FORM, SO LET’S GO AHEAD
AND PUT THESE TERMS IN DESCENDING ORDER. THIS IS FACTORABLE. 2 COSINE THETA, COSINE THETA. NOW WE HAVE TO PLAY AROUND
WITH THE FACTORS OF -1. IF WE PUT A – 1 HERE
AND + 1 HERE, THE OUTER PRODUCT IS -2
COSINE THETA + THE INNER PRODUCT OF 1
COSINE THETA=THE MIDDLE TERM. NOW WE’LL SET EACH OF THESE
FACTORS EQUAL TO ZERO. IF WE SET THIS EQUAL TO ZERO
AND SOLVE FOR A COSINE THETA, WE’D HAVE COSINE THETA EQUALS
SUBTRACT 1, DIVIDE BY 2. AND OVER HERE WE’D HAVE COSINE
THETA, ADD 1 TO BOTH SIDES, YOU HAVE COSINE THETA=1. BACK TO OUR REFERENCE
TRIANGLES FOR A COSINE THETA=-1/2. BACK TO THE 30, 60, 90, THE
COSINE OF 60 DEGREES IS 1/2. BUT WE WANT -1/2. COSINE IS NEGATIVE IN
THE SECOND AND THIRD QUADRANT. SO WE’LL SKETCH A REFERENCE
ANGLE OF 60 DEGREES IN THOSE TWO QUADRANTS. THIS WOULD BE 120 DEGREES. THE SECOND ANGLE WOULD BE
180 + 60 OR 240 DEGREES. NEXT WE WANT TO KNOW
WHERE COSINE THETA=1. REMEMBER ON THE UNIT CIRCLE
OF COSINE THETA=THE X COORDINATE. X=1 RIGHT HERE AT THIS POINT
ON THE UNIT CIRCLE. AND THIS ANGLE WOULD BE–
ON THE GIVEN INTERVAL WE’RE GOING TO USE
ZERO DEGREES OR ZERO RADIANS. AND I ALSO NOTICED
THAT THEY DID WANT OUR ANSWERS IN RADIANS. LET’S GO AHEAD
AND CONVERT THESE TO RADIANS. 2 PI/3 RADIANS AND THIS
WOULD BE 4PI/3 RADIANS. AND I THINK WE’RE GOOD
ON THIS PROBLEM. I THINK WE HAVE TIME
FOR ONE MORE. AND, AGAIN, NOTICE WE HAVE
SINE THETA – SINE 2 THETA. SINCE THIS IS A SINGLE ANGLE
AND THIS IS A DOUBLE ANGLE, WE CAN’T JUST ISOLATE
ONE OF THE TRIG FUNCTIONS. WE’RE GOING TO HAVE TO DO
A SUBSTITUTION USING SINE 2 THETA=2
SINE THETA COSINE THETA. SO LET’S START OFF WITH THAT. NOW, IT MAY BE DIFFICULT TO DETERMINE WHICH IDENTITY
TO USE TO PERFORM THE SUBSTITUTION. I’VE ALREADY PROVIDED IT, SO THAT DOES MAKE IT
A LITTLE BIT EASIER. SO WHEN YOU’RE DOING
THESE PROBLEMS YOU SHOULD HAVE
A LIST OF IDENTITIES HANDY SO YOU CAN REFERENCE THEM. NOW, EVEN THOUGH THIS IS
IN TERMS OF TWO DIFFERENT TRIGONOMETRIC FUNCTIONS, THERE’S A COMMON FACTOR
OF SINE THETA THAT WE CAN FACTOR OUT. SO WE’D BE LEFT
WITH 1 – 2 COSINE THETA. SETTING EACH OF THESE FACTORS
EQUAL TO ZERO, WE’D HAVE SINE THETA=0
OR 1 – 2 COSINE THETA=0. LET’S SOLVE THIS
FOR A COSINE THETA. SO WE’LL SUBTRACT 1
AND THEN DIVIDE BY -2, THAT WOULD BE A POSITIVE +1/2. LET’S FIND OUR ANGLES NOW. SINE THETA=0. REMEMBER ON THE UNIT CIRCLE
SINE THETA=Y. THE Y COORDINATE WOULD BE ZERO
ON THE X AXIS. SO SINE THETA WOULD BE ZERO
HERE AND HERE. AND, AGAIN, ON THIS INTERVAL
WE’D USE ZERO RADIANS. AND OVER HERE WE’D HAVE
PI RADIANS. AND, AGAIN,
THIS IS THE POINT (1,0). AND THIS IS THE POINT (-1,0). THE Y COORDINATE
IS EQUAL TO ZERO. NEXT FOR 1/2 WE’LL GO BACK TO
OUR 30, 60, 90 RIGHT TRIANGLE. WE’RE LOOKING FOR AN ANGLE
THAT HAS A COSINE OF 1/2 AND THAT WOULD BE 60 DEGREES. COSINE IS POSITIVE IN THE FIRST
AND THE FOURTH QUADRANT BECAUSE IT’S EQUAL TO X/R. X IS POSITIVE IN THE FIRST
AND FOURTH QUADRANT. LET’S SKETCH A 60 DEGREE
REFERENCE ANGLE IN BOTH THE FIRST
AND THE FOURTH QUADRANT, WHICH WOULD TELL US THAT ONE
ANGLE WOULD BE 60 DEGREES AND THE OTHER ANGLE
WOULD BE 300 DEGREES. THEY DID REQUEST OUR ANSWERS
TO BE IN RADIANS, SO LET’S CONVERT THESE
TO RADIANS. THIS WOULD BE PI/3. 300 DEGREES WOULD BE 5PI/3. OKAY. I THINK THAT’LL DO IT
FOR THIS VIDEO. I HOPE YOU FOUND IT HELPFUL. HAVE A GOOD DAY. [ MUSIC ]  

5 thoughts on “(New Version Available) Solving Trigonometric Equations IV

  • March 14, 2012 at 5:28 pm
    Permalink

    Thank you so much! =)

    Reply
  • April 13, 2012 at 1:35 am
    Permalink

    3:03 you said the cos of 30 deg is √2/2. Great vids though.

    Reply
  • October 5, 2012 at 7:56 am
    Permalink

    it is root 3 /2

    Reply
  • February 8, 2015 at 1:11 am
    Permalink

    there are 6 solutions of cos(3x) = √3/2 on the interval [0,2π].  use your graphing calculator to check for yourself

    Reply
  • January 27, 2019 at 4:33 pm
    Permalink

    Is there a video somewhere that gives details about when we really do need to have all the parenthesis in these trig problems?

    Reply

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