- WE WANT TO SOLVE THE TRIG EQUATION SINE X = COSINE X
ON THE GIVEN INTERVAL.
THERE ARE SEVERAL WAYS TO SOLVE THIS EQUATION
SO WE WILL LOOK AT SEVERAL METHODS.
HOWEVER, IN GENERAL WHEN WE HAVE A TRIG EQUATION
IF WE WANT TO SOLVE IT ALGEBRAICALLY
IT'S HELPFUL TO HAVE THE TRIG EQUATION
IN TERMS OF ONE TRIG FUNCTION.
THERE'S NO OBVIOUS SUBSTITUTION HERE,
BUT IF WE DIVIDE BOTH SIDES OF THE EQUATION BY COSINE X,
WE CAN SUBSTITUTE TANGENT X FOR SINE X DIVIDED BY COSINE X.
SO NOW WE HAVE THE EQUATION TANGENT X = THIS WOULD BE +1.
SO NOW WE WANT ALL THE ANGLES IN THE GIVEN INTERVAL
THAT HAVE A TANGENT FUNCTION VALUE OF +1.
AND WE SHOULD RECOGNIZE THIS FROM OUR REFERENCE ANGLES
OR OUR REFERENCE TRIANGLES.
IF WE HAVE A 45, 45, 90 REFERENCE TRIANGLE
WE CAN LABEL THE TWO LEGS OF THE TRIANGLE ONE UNIT
AND THE LENGTH OF HYPOTENUSE SQUARE ROOT 2 UNITS.
FROM HERE WE SHOULD RECOGNIZE THAT THE TANGENT OF 45 DEGREES,
WHICH IS THE RATIO OF THE OPPOSITE SIDE
TO THE ADJACENT SIDE,
IS EQUAL TO +1.
SO ONE SOLUTION IS X = 45 DEGREES OR PI/4 RADIANS.
BUT WE WANT ALL THE ANGLES
THAT HAVE A TANGENT FUNCTION VALUE OF +1
ON THE GIVEN INTERVAL.
SO REMEMBER THAT TANGENT THETA IS EQUAL TO Y/X
ON THE COORDINATE PLANE
SO WE JUST FOUND THAT 45 DEGREES OR PI/4 RADIANS
HAS A TANGENT FUNCTION VALUE OF +1.
WE KNOW IT'S +1 BECAUSE THE X AND Y COORDINATES ARE POSITIVE
IN THE FIRST QUADRANT,
BUT IN THE THIRD QUADRANT BOTH X AND Y ARE NEGATIVE
AND, THEREFORE, TANGENT THETA WILL ALSO BE POSITIVE.
SO IF WE SKETCH A 45 DEGREE REFERENCE ANGLE
IN THE THIRD QUADRANT
THIS WILL GIVE US ANOTHER ANGLE
THAT HAS A TANGENT FUNCTION VALUE OF +1.
SO THIS WOULD BE 45 DEGREES,
WHICH MEANS 180 DEGREES + 45 DEGREES
WOULD ALSO BE ANOTHER ANGLE
THAT HAS A TANGENT FUNCTION VALUE OF +1.
180 DEGREES + 45 DEGREES WOULD BE 225 DEGREES,
WHICH IS EQUAL TO 5PI/4 RADIANS.
SO THESE WOULD BE OUR SOLUTIONS TO THE EQUATION
ON THE GIVEN INTERVAL.
LET'S TRY TO SOLVE THIS USING THE UNIT CIRCLE AS WELL.
OUR ORIGINAL EQUATION WAS SINE X = COSINE X,
BUT REMEMBER ON THE UNIT CIRCLE SINE THETA IS EQUAL TO Y
AND COSINE THETA IS EQUAL TO X.
SO ANOTHER WAY TO THINK OF THIS IS LOOKING AT THE UNIT CIRCLE
WE WANT TO FIND WHERE THE Y COORDINATE
IS EQUAL TO THE X COORDINATE.
AGAIN, BECAUSE Y = SINE THETA AND X = COSINE THETA.
SO IF WE TAKE A LOOK AT PI/4 RADIANS
NOTICE HOW THE X AND Y COORDINATES ARE EQUAL,
THEREFORE, SINE THETA AND COSINE THETA ARE EQUAL.
AND IF WE LOOK AT 5PI/4 RADIANS
NOTICE HOW THE X AND Y COORDINATES ARE ALSO EQUAL,
THEREFORE, SINE THETA AND COSINE THETA ARE EQUAL.
SO FROM THE UNIT CIRCLE,
WE CAN EASILY SEE THAT OUR SOLUTIONS WOULD BE
X = PI/4 RADIANS OR X = 5PI/4 RADIANS.
AND THE LAST WAY TO SOLVE THIS WOULD BE TO GRAPH Y = SINE X
AND Y = COSINE X
ON THE SAME COORDINATE PLANE AS SEEN HERE.
IN BLUE WE HAVE Y = SINE X
AND IN RED WE HAVE Y = COSINE X.
IF WE'RE TRYING TO DETERMINE
WHEN THESE ARE EQUAL TO EACH OTHER,
THAT WOULD BE WHERE THEY INTERSECT.
NOTICE AT THIS POINT OF INTERSECTION
THE X COORDINATE WOULD BE PI/4 RADIANS
AND AT THIS POINT OF INTERSECTION
THE X COORDINATE WOULD BE 5PI/4 RADIANS.
SO AS YOU CAN SEE THERE ARE SEVERAL WAYS
TO APPROACH THIS PROBLEM.
HOPEFULLY IT WAS HELPFUL TO SEE IT SOLVED THREE DIFFERENT WAYS.
THANK YOU FOR WATCHING.