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- WE WANT TO FIND THE EXACT TRIG FUNCTION VALUES
USING REFERENCE TRIANGLES.
NOTICE HOW THE ANGLES WERE GIVEN IN RADIANS,
SO WE COULD CONVERT THESE TO DEGREES.
BUT I THINK IT'S IMPORTANT TO LEARN
HOW TO WORK WITH RADIANS AS WELL.
SO LET'S SKETCH THESE ANGLES IN CENTER POSITION,
FORM THE REFERENCE TRIANGLE,
AND THEN DETERMINE THE TRIG FUNCTION VALUES.
SO WE'LL FIRST SKETCH 7 PI/4 RADIANS IN STANDARD POSITION,
SO HERE'S THE INITIAL SIDE.
NOW, WE'LL START TO ROTATE COUNTER CLOCKWISE.
WELL, HALF A ROTATION COUNTER CLOCKWISE WOULD BE PI RADIANS.
BUT BECAUSE OUR ANGLE IS A MULTIPLE OF PI/4 RADIANS,
IT'S HELPFUL TO VIEW THIS AS 4 PI/4 RADIANS
WHICH MEANS FROM HERE,
WE NEED TO ROTATE ANOTHER 3 PI/4 RADIANS.
WELL, 1 COMPLETE ROTATION COUNTER CLOCKWISE
WOULD BE 2 PI RADIANS WHICH IS EQUAL TO 8 PI/4 RADIANS.
SO WE COULD ALSO THINK OF THIS
AS WE HAVE TO ROTATE 1 PI/4 RADIANS
LESS THAN 8 PI/4 RADIANS
WHICH MEANS WE WOULD ROTATE SOMEWHERE IN HERE
PI/4 RADIANS SHORT OF 1 COMPLETE ROTATION.
SO THIS WOULD BE THE TERMINAL SIDE.
AND WE SHOULD RECOGNIZE THAT THE ANGLE
BETWEEN THE TERMINAL SIDE AND THE X AXIS IS PI/4 RADIANS.
IF WE ROTATED ONE MORE PI/4 RADIANS,
WE WOULD BE AT 8 PI/4 RADIANS.
SO NOW, WE CAN FORM OUR REFERENCE TRIANGLE
BY SKETCHING IT PERPENDICULAR TO THE X AXIS HERE.
SO HERE'S OUR REFERENCE TRIANGLE IN BLACK.
AND BECAUSE WE KNOW THAT PI/4 RADIANS IS EQUAL TO 45 DEGREES,
WE KNOW WE LABEL THE 2 LEGS OF THIS RIGHT TRIANGLE AS 1
AND THE HYPOTENUSE AS SQUARE ROOT 2.
BUT WE ALSO HAVE TO MAKE SURE
THAT WE HAVE THE RIGHT SINES ON THE LEGS,
AND SINCE WE'RE IN THE 4th QUADRANT
WHERE THE X COORDINATE IS POSITIVE
AND THE Y COORDINATE IS NEGATIVE,
THIS 1 MUST BE NEGATIVE.
AND NOW, WE CAN USE THIS REFERENCE ANGLE
AND THIS REFERENCE TRIANGLE TO DETERMINE
THESE FIRST 2 TRIG FUNCTION VALUES.
SO THE SINE OF 7 PI/4 RADIANS IS EQUAL TO THE RATIO
OF THE OPPOSITE SIDE TO THE HYPOTENUSE
WHICH IS -1 DIVIDED BY SQUARE ROOT 2.
SO THIS IS THE EXACT VALUE FOR THIS TRIG FUNCTION.
BUT SOMETIMES WE ARE ASKED TO RATIONALIZE THE DENOMINATOR,
SO LET'S ALSO DO THAT.
TO RATIONALIZE THE DENOMINATOR,
WE MULTIPLY BY THE SQUARE ROOT OF 2/THE SQUARE ROOT OF 2,
SO THIS WOULD BE -SQUARE ROOT 2 DIVIDED BY 2.
SO AGAIN DEPENDING ON YOUR DIRECTIONS,
YOU MAY BE ABLE TO EXPRESS
YOUR ANSWER IN THIS FORM OR THIS FORM.
AND NOW FOR THE COSINE OF 7 PI/4 RADIANS,
WE NEED THE RATIO OF THE ADJACENT SIDE TO THE HYPOTENUSE
WHICH WOULD BE 1 DIVIDED BY SQUARE ROOT 2.
SO IF WE RATIONALIZE THE DENOMINATOR LIKE WE DID UP HERE,
THE DIFFERENCE IS WE WOULD HAVE SQUARE ROOT 2/2.
SO AGAIN THESE ALWAYS ARE EQUAL,
ONE HAS AN IRRATIONAL DENOMINATOR
AND ONE HAS A RATIONAL DENOMINATOR.
NOW FOR THE NEXT 2,
WE'LL SKETCH -3 PI/4 RADIANS AND FORM A REFERENCE TRIANGLE.
SO BECAUSE THEIR ANGLE IS NEGATIVE,
WE'LL NOW ROTATE CLOCKWISE 3 PI/4 RADIANS.
WELL, ONE-QUARTER ROTATION CLOCKWISE
WOULD BE -PI/2 RADIANS.
BUT BECAUSE OUR ANGLE IS A MULTIPLE OF PI/4 RADIANS,
IT'S HELPFUL TO VIEW THIS AS -2 PI/4 RADIANS
WHICH MEANS WE NEED TO ROTATE ANOTHER PI/4 RADIANS CLOCKWISE
WHICH WOULD BRING US TO HERE,
SO HERE'S OUR TERMINAL SIDE.
SO IF THIS IS -3 PI/4 RADIANS AND HALF A ROTATION
WOULD BE -PI OR -4 PI/4 RADIANS,
WE KNOW THAT OUR REFERENCE ANGLE HERE MUST BE PI/4 RADIANS.
SO NOW, WE'LL FORM THE REFERENCE TRIANGLE.
AGAIN BECAUSE WE HAVE A PI/4,
PI/4 RIGHT TRIANGLE OR 45-45 RIGHT TRIANGLE,
WE CAN LABEL THE 2 LEGS 1, THE HYPOTENUSE SQUARE ROOT 2.
BUT NOW, WE'RE IN THE 3rd QUADRANT
WHERE BOTH THE X COORDINATE AND THE Y COORDINATE ARE NEGATIVE,
SO THIS WILL BE -1 AND SO WILL THIS.
SO THE SINE OF -3 PI/4 RADIANS
IS EQUAL TO THE RATIO OF THE OPPOSITE SIDE TO THE HYPOTENUSE,
SO WE'D HAVE -1 DIVIDED BY SQUARE ROOT 2,
OR IF WE WANT TO RATIONALIZE THE DENOMINATOR,
WE WOULD HAVE -SQUARE ROOT 2/2.
AND NOTICE FOR THE COSINE FUNCTION VALUE,
WE'D HAVE THE RATIO OF THE ADJACENT SIDE
OF THE HYPOTENUSE WHICH IS STILL -1 DIVIDED BY SQUARE ROOT 2,
OR -SQUARE ROOT 2/2.
THAT'S GOING TO DO IT FOR THESE TWO EXAMPLES.
BUT I WILL LEAVE YOU WITH SOME NOTES ON THE 45-45-90
REFERENCE TRIANGLE OR THE PI/4, PI/4 REFERENCE TRIANGLE.
THIS IS THE TRIANGLE THAT WE USED,
BUT OF COURSE WE COULD MULTIPLY EACH LENGTH BY A CONSTANT.
SO IF THE CONSTANT WAS X,
WE COULD ALSO EXPRESS THIS REFERENCE TRIANGLE
IN THIS FORM HERE.
OKAY, I HOPE YOU FOUND THIS HELPFUL.