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Section 6.3 Notes Page 1 6.3 Properties of Trigonometric Functions The following Even – Odd Properties will allow you to not deal with negative angles. Even – Odd Properties cos(t ) cos t sec(t ) sec t sin(t ) sin t csc(t ) csc t tan(t ) tan t cot(t ) cot t Periodic Properties If we start at an angle and go around one revolution ( 360 or 2 radians) we will end up at the same angle we started with. The k value is any integer, and represents how many revolutions are going around. If you want to use degrees, replace the 2k in the equations below with 360k. sin(t 2k ) sin t csc(t 2k ) csc t cos(t 2k ) cos t sec(t 2k ) sec t tan(t 2k ) tan t cot(t 2k ) cot t EXAMPLE: Find the EXACT value of sin 1485 . We need to divide 1485 by 360 to see how many revolutions we have. If you divide 1485 by 360 you will get 4 with a remainder of 45. So we can rewrite our problem as: sin(45 360(4)) . From our definitions above we know that sin(45 360(4)) sin 45 . From our previous values we know that sin 45 that sin 1485 2 . So we know 2 2 . 2 17 EXAMPLE: Find the EXACT value of tan . 4 First we will use our even-odd property to change the negative angle into a positive angle: 17 180 17 17 765 . Dividing this by tan tan . It would be easier to change this into degrees: 4 4 4 360 we will get 2 with a remainder of 45. So our problem becomes tan(45 360(2)) . From our periodic properties, tan(45 360(2)) tan 45 . From our previous values we know that tan 45 1 . So we know that 17 tan 1 . 4 Section 6.3 Notes Page 2 EXAMPLE: Find the exact value of tan 14 using reference angles. Draw the angle in standard position and 3 indicate the reference angle. 14 180 840 . We can subtract two revolutions from this: 3 840 360 360 120 . From our rules, we can rewrite this problem as: tan(120 360(2)) tan120 2 In radian form it would look like this: tan . 3 We can change this into degrees: 1.) First we will draw this angle in standard position. The reference angle is indicated by the double curved lines. 2 We found the reference angle by taking . (180 – 120) 3 3 2.) We need to apply the trig function to our reference angle. We have tan 3 3. 3.) We need to apply the appropriate sign. This angle is in the second quadrant, so tangent needs to be negative here. So now we 14 can write our answer: tan 3. 3 EXAMPLE: Find the reference angle of tan(225 ) and find its EXACT value. Using the even and odd properties, tan(225 ) tan 225 . We have changed the problem, so now we will look at positive 225 degrees. This is in the third quadrant, so our reference angle is 225 180 45 . We will now calculate tan 45 . The value for tan 45 is 1. So now we have tan(45 ) 1 In the third quadrant, tangent is positive, so we don’t need to change the sign of our answer, so tan(225 ) 1 . EXAMPLE: Find the reference angle of sec(210 ) and find its EXACT value. Using the even and odd properties, sec(210 ) sec 210 . We have changed the problem, so now we will look at positive 210 degrees. This is in the third quadrant, so our reference angle is 210 180 30 . We will do 1 . Now we can put in the value off our table, and we will get sec 30 . This is the same as cos 30 2 3 1 2 2 3 . Secant is negative in the 3rd quadrant, so we need to change the sign: sec( 210 ) . 3 3 3 3 2 Section 6.3 Notes Page 3 11 EXAMPLE: Find the reference angle of sin and find its EXACT value. 3 11 180 660 . So now our problem becomes: sin(660 ) . 3 Using the even and odd properties, sin(660 ) sin 660 . We have changed the problem, so now we will look at positive 660 degrees. We can rewrite the problem as sin(300 360 ) which becomes sin(300 ) . We want to change our angle into degrees: This is in the fourth quadrant, so our reference angle is 360 300 60 , or . We will now look at 3 3 . In the fourth quadrant, secant is 2 3 11 negative, so we need to change the sign of our answer, so sin . 3 2 sin 60 . We can put in the value off our table, and we will get 19 EXAMPLE: Find the reference angle of csc and find its EXACT value. 4 19 180 855 . Using the even and odd properties, 4 csc(855 ) csc 855 . This can be rewritten as csc(135 360 (2)) . This reduces to csc(135 ) . This is in the second quadrant, so the reference angle is 180 135 45 . So now we have csc 45 . This is the 1 2 2 2 1 same as . From our table we have: 2 . Since we are in the second 2 sin 45 2 2 Changing this into degrees we get: 2 19 quadrant, sine is positive, so we don’t need to change our sign. Our answer is csc 4 2. Fundamental Identities sin t 1 csc t csc t 1 sin t sin 2 t cos 2 t 1 cos t 1 sec t sec t 1 cos t tan t tan t 1 cot t cot t 1 tan t sin t cos t 1 tan 2 t sec 2 t cot t cos t sin t 1 cot 2 t csc 2 t Section 6.3 Notes Page 4 EXAMPLE: Given sin t 5 2 and cos t , find the other 4 trig values using identities. 3 3 We just use the identities and plug in our values for sine and cosine where necessary: 1 1 3 3 5 cos t 5 5 3 5 csc t 1 1 3 sin t 2 / 3 2 cot t 1 1 5 5 tan t 2 5 5 2 5 2 EXAMPLE: Given sin t sec t tan t 23 sin t 2 3 2 2 5 , cos t 5 5 3 3 5 5 7 and 0 t , use the Pythagorean Identity sin 2 t cos 2 t 1 to find cos t . 8 2 2 7 7 We first start with the identity sin t cos t 1 and plug in for sin t . You will get cos 2 t 1 . 8 8 49 49 Simplifying will give you from both sides. You will cos 2 t 1 . Now isolate the cosine by subtracting 64 64 15 15 get cos 2 t . Square root both sides to get cos t . Since the angle for t needs to be 0 t , we 64 2 8 know that this angle must be in the first quadrant. What we know from this is that the x-coordinate of the unit 15 circle must be positive in the first quadrant, so cos t . 8 2 2