Unit Circle Trigonometry Text 3

Exercises Find the exact values of the following: 1. cos (150D ) 2. tan ( 330D ) 3. sin ( 225D ) 4. cot 60D 5. csc 225D 6. tan −120D 7. ( ) cot ( −450 ) D 8. ( ) cos ( 495 ) D 9. ( ) sin ( 3630 ) D Methods of Finding Trigonometric Functions of Special Angles We can see that the unit circle assists us greatly in finding trigonometric functions of special angles. But what if we don’t have a diagram of the unit circle, and we don’t wish to draw one? It is assumed that the quadrantal angles (multiples of 90o -- on the axes) are fairly easy to visualize without a diagram. One option is to memorize the coordinates of the special angles in the first quadrant of the unit circle (30o, 45o, and 60o), and use those values to find the trigonometric functions of angles in other quadrants. We will discuss a few ways of finding the basic trigonometric functions of 30o, 45o, and 60o (other than memorizing the coordinates of these special angles in the first quadrant of the unit circle). One method is to use a basic 45o-45o-90o triangle and a 30o-60o-90o triangle to derive the trigonometric functions of 30o, 45o, and 60o. We learned earlier in the unit that the trigonometric functions are constant for any given angle; for example, sin 30D is always ( ) 1 2 , regardless of the size of the radius of the circle – or the size of the triangle drawn. So we can choose any 45o-45o-90o triangle and 30o-60o-90o triangle to work from. For simplicity, let us choose both triangles to have a shorter leg of length 1, as shown below: 60o 45o 1 45o 30o 1 Recall that in a 45o-45o-90o triangle, the legs are congruent, and the length of the hypotenuse is 2 times the length of either leg. In a 30o-60o-90o triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the longer leg is 3 times the length of the shorter leg. Unit Circle Trigonometry Methods of Finding Trigonometric Functions of Special Angles Filling in the missing sides from the diagram above, we obtain the following triangles: 60o 2 45o 2 1 1 45o 30o 1 3 Now recall the trigonometric ratios that we learned for right triangles (shown below in abbreviated form): SOH-CAH-TOA sin(θ ) = cos(θ ) = Opposite Hypotenuse Adjacent Hypotenuse tan(θ ) = Opposite Adjacent Using these ratios on the triangles above, we obtain the following: sin(45D ) = Opposite Hypotenuse = cos(45D ) = Adjacent Hypotenuse tan(45D ) = Opposite Adjacent sin(30D ) = Opposite Hypotenuse cos(30D ) = Adjacent Hypotenuse tan(30D ) = Opposite Adjacent sin(60D ) = Opposite Hypotenuse = 3 2 (Note: This is the y-coordinate of 60o on the unit circle.) cos(60D ) = Adjacent Hypotenuse = 1 2 (Note: This is the x-coordinate of 60o on the unit circle.) tan(60D ) = Opposite Adjacent 1 2 = = 1 2 = 1 2 1 2 ⋅ ⋅ 2 2 2 2 = = 2 2 (Note: This is the y-coordinate of 45o on the unit circle.) 2 2 (Note: This is the x-coordinate of 45o on the unit circle.) = 11 = 1 = = = 1 2 = (Note: This is the y-coordinate of 30o on the unit circle.) 3 2 = 1 3 (Note: This is the x-coordinate of 30o on the unit circle.) 1 3 ⋅ 3 3 = 3 3 3 1 The reciprocal trigonometric ratios cosecant, secant, and cotangent can be obtained by simply taking the reciprocals of the sine, cosine, and tangent ratios above. Unit Circle Trigonometry Methods of Finding Trigonometric Functions of Special Angles Another method of remembering the trigonometric functions of 30o, 45o, and 60o, along with 0o and 90o, is shown below. (This is more of a ‘trick’ for remembering the ratios, so no mathematical justification is given.) Step 1: Label the columns at the top of the chart for all of the special angles between 0o and 90o in ascending order, as shown. Also label the rows with the words “sine” and “cosine.” 0o 30o 45o 60o 90o sine cosine Step 2: This chart is not yet complete! Write the numbers 0, 1, 2, 3, 4 in the “sine” row, and the numbers 4, 3, 2, 1, 0 in the “cosine” row, as shown. 0o sine cosine 0 30o 45o 60o 90o 1 2 3 4 1 0 This chart is not yet complete! 4 3 2 This chart is not yet complete! Step 3: This chart is correct, but not yet simplified… For each of the numbers in the “sine” and “cosine” rows, take the square root of the number and then divide by 2. sine cosine 0o 30o 45o 60o 90o 0 2 4 2 1 2 3 2 2 2 2 2 3 2 1 2 4 2 0 2 We then simplify each of the numbers in the chart above. Unit Circle Trigonometry Methods of Finding Trigonometric Functions of Special Angles Step 4: This is the final version of the chart! Simplify each of the values from the table above, and we obtain our final chart: 0o 30o 45o 60o sine 0 1 2 2 2 cosine 1 3 2 2 2 3 2 1 2 90o 1 y-values on the unit circle 0 x-values on the unit circle Examples Use the methods learned in this section (special right triangles or the chart above) to find the exact values of the following trigonometric functions. Note that these examples were also included in the previous section, but will now be solved using a different method. 1. sin ( 240D ) 2. cos ( 315D ) ( 4. sec ( 30D ) Solutions: 1. sin 240D ( 5. tan 135D 3. cot ( 210D ) ) ) The terminal side of a 240o angle measures 60o to the x-axis, so we use a 60o reference angle. Using either a 30o-60o-90o triangle or our chart from above, we find that sin 60D = 23 . Since a 240o angle is in the third quadrant, and y- ( ) ( ) values in the third quadrant are negative, sin 240D must be negative. Since ( ) sin 60D = ( 2. cos 315D 3 2 ( ) , we conclude that sin 240D = − 3 2 . ) The terminal side of a 315o angle measures 45o to the x-axis, so we use a 45o reference angle. Using either a 45o-45o-90o triangle or our chart from above, we find that cos 45D = 22 . Since a 315o angle is in the fourth quadrant, and ( ) ( ) x-values in the fourth quadrant are positive, cos 315D must be positive. ( ) Since cos 45 = D Unit Circle Trigonometry 2 2 ( ) , we conclude that cos 315 = D 2 2 . Methods of Finding Trigonometric Functions of Special Angles 3. cot ( 210D ) The terminal side of a 210o angle measures 30o to the x-axis, so we use a 30o cos θ reference angle. Remember that cot (θ ) = tan1(θ ) = sin ((θ )) . If we use the 30o-60o- ( ) 90o triangle, we find that tan 30D = 1 3 , so cot ( 30D ) = 1 tan (θ ) instead use our chart from above, we find that cos ( 30D ) = sin ( 30D ) = 12 , so cot 30D = ( ) ( ) = sin ( 30 ) cos 30D D 3 2 1 2 = 3 2 = 3 . If we 3 2 and ⋅ 12 = 3 . Since a 210o angle is in the third quadrant, and both the x and y-values in the third quadrant are negative, this means that both cos ( 210D ) and sin ( 210D ) are negative, therefore their quotient cot ( 210D ) = we conclude that cot ( 210D ) = 3 . ( ( cos 210D sin 210D ) is positive. Since cot ( 30D ) = 3 , ) ( ) 4. sec 30D The terminal side of a 30o angle measures 30o to the x-axis, so we use a 30o reference angle. (A reference angle is not really necessary in this case, since any angle which is located in the first quadrant can be used as-is.) Remember that sec (θ ) = cos1(θ ) . Using either a 30o-60o-90o triangle or our chart from ( ) above, we find that cos 30D = 3 2 , so sec ( 30D ) = Rationalizing the denominator, sec ( 30D ) = 2 3 ⋅ 1 cos 30D ( ) 3 3 = 2 3 3 = 1 3 2 = 1⋅ 2 3 = 2 3 . . 5. tan (135D ) The terminal side of a 135o angle measures 45o to the x-axis, so we use a 45o sin θ reference angle. Remember that tan (θ ) = cos((θ )) . If we use the 45o-45o-90o ( ) triangle, we find that tan 45D = 11 = 1 . If we instead use our chart from above, sin ( 45D ) = 2 2 and sin ( 45D ) = 2 2 , so tan ( 45D ) = ( ) = cos ( 45 ) sin 45D D 2 2 2 2 = 2 2 ⋅ 2 2 = 1 . Since a 135o angle is in the second quadrant, the x-value is negative and the y-value is positive, which means that cos 135D is negative and sin 135D is positive; ( ) ( therefore their quotient tan (135 ) = ( conclude that tan (135 ) = −1 . D ( D sin 135 cos 135D ) ) ) is negative. Since tan ( 45 ) = 1 , we D D Unit Circle Trigonometry Methods of Finding Trigonometric Functions of Special Angles Evaluating Trigonometric Functions of Other Angles Now that we have explored the trigonometric functions of special angles, we will briefly look at how to find trigonometric functions of other angles. First, we will derive an important trigonometric identity, known as a Pythagorean Identity. It can be related to the Pythagorean Theorem; consider the following right triangle (drawn in the first quadrant for simplicity). y x2 + y 2 = r 2 r y θ x x We can see that the equation x 2 + y 2 = r 2 (the equation of a circle with radius r) is also the Pythagorean Theorem as it relates to the right triangle above. Let us now direct our attention to the unit circle. Since the radius is 1, any point on the circle itself satisfies the equation x 2 + y 2 = 1 (the equation of a circle with radius 1). On the unit circle, we know that x = cos (θ ) and y = sin (θ ) . Substituting these into the equation x 2 + y 2 = 1 , we obtain the equation ( cos (θ ) ) + ( sin (θ ) ) 2 2 = 1 . It is standard practice in trigonometry to write these individual terms in shortened form. We write ( cos (θ ) ) as cos 2 (θ ) , and we write 2 ( sin (θ ) ) 2 as sin 2 (θ ) . We then obtain the following trigonometric identity: cos 2 (θ ) + sin 2 (θ ) = 1 The identity cos 2 (θ ) + sin 2 (θ ) = 1 applies to an angle drawn in any circle of radius r, not just the unit circle. A short justification is shown below. Unit Circle Trigonometry Evaluating Trigonometric Functions of Other Angles We begin with the general equation of a circle of radius r: x2 + y 2 = r 2 Dividing both sides by r 2 , we obtain the equation x2 y2 r 2 + = . r2 r2 r2 Simplifying the equation, x2 y2 + =1. r2 r2 The equation can then be rewritten as 2 2 ⎛x⎞ ⎛ y⎞ ⎜ ⎟ + ⎜ ⎟ = 1. ⎝r⎠ ⎝r⎠ For a circle of radius r, we know from our general trigonometric definitions that x y cos (θ ) = and sin (θ ) = . We substitute these into the above equation and r r conclude that cos 2 (θ ) + sin 2 (θ ) = 1 . We will now use this Pythagorean identity to help us to find trigonometric functions of angles. Example If sin (θ ) = 3 5 and 90D < θ < 180D , find the exact values of cos (θ ) and tan (θ ) . Solution: Since 90D < θ < 180D , the terminal side of θ is in the second quadrant. Looking at our unit circle, we can easily see that there is no special angle with a y-value of 53 . There are two methods by which we can solve this problem, both of which are shown below. Method 1: Our first method is to use the Pythagorean identity cos 2 (θ ) + sin 2 (θ ) = 1 . Since sin (θ ) = 53 , we plug it into the equation cos 2 (θ ) + sin 2 (θ ) = 1 . cos 2 (θ ) + ( 53 ) = 1 2 Unit Circle Trigonometry Evaluating Trigonometric Functions of Other Angles We then simplify the equation and solve for cos (θ ) : cos 2 (θ ) + 259 = 1 cos 2 (θ ) = 1 − 259 cos 2 (θ ) = 25 25 cos 2 (θ ) = 16 25 cos (θ ) = 4 5 − 259 OR cos (θ ) = − 54 cos (θ ) = − 54 (We need to choose; see below...) The terminal side of θ is in the second quadrant. We know that x values in the second quadrant are negative, therefore cos (θ ) is negative. We now want to find tan (θ ) . Using the definition tan (θ ) = tan (θ ) = sin (θ ) cos (θ ) = tan (θ ) = − 34 3 5 − 54 sin (θ ) cos (θ ) , = 53 ⋅ − 54 = − 34 . We conclude that cos (θ ) = − 54 and tan (θ ) = − 34 . Method 2: Our second method is to draw a right triangle in a circle of radius r. Since 90D < θ < 180D , we draw the right triangle in the second quadrant, as shown below. (When drawing the right triangle, we must make sure that the right triangle has one leg on the x-axis and that one acute angle of the triangle -- the reference angle of θ -has its vertex at the origin.) y x2 + y 2 = r 2 ( x, y ) P r y θ x Unit Circle Trigonometry x Evaluating Trigonometric Functions of Other Angles Since sin (θ ) = 53 , and we know that sin (θ ) = y r , we can label the diagram with the values y = 3 and r = 5 , as shown below. (The opposite formula sin (θ ) = ry can be remembered by using the ratio hypotenuse , using the reference angle of θ which is inside the triangle.) Note that when choosing how to label the sides of the triangle, the radius of the circle is always chosen to be positive. y x2 + y 2 = r 2 5 3 θ x x To find the value of x, we then use the equation for the circle (or equivalently, the Pythagorean Theorem) x 2 + y 2 = r 2 : x 2 + 32 = 52 x 2 + 9 = 25 x 2 = 16 x = 4 OR x = − 4 (We need to choose; see below.) x = −4 The terminal side of θ is in the second quadrant. We know that x values in the second quadrant are negative, therefore we choose the negative value for x. To find cos (θ ) , we now use the formula cos (θ ) = rx . (The formula cos (θ ) = x r can be remembered by using the ratio adjacent hypotenuse , using the reference angle of θ which is inside the triangle.) Note that we can NOT use the formula cos (θ ) = x , since that trigonometric definition only applies to the unit circle, and the circle that we created has radius 5. cos (θ ) = rx = Unit Circle Trigonometry −4 5 = − 54 Evaluating Trigonometric Functions of Other Angles To find tan (θ ) , we now use the formula tan (θ ) = xy . (Remember that the formula tan (θ ) = opposite adjacent y x can be remembered by using the ratio , using the reference angle of θ which is inside the triangle.) tan (θ ) = xy = = − 34 3 −4 Note that we could have also used the formula tan (θ ) = sin (θ ) cos (θ ) = 3 5 −4 5 = 53 ⋅ −54 = − 43 We conclude that cos (θ ) = − 54 and tan (θ ) = − 34 . We now wish to derive two other Pythagorean identities. Both identities can be easily derived from the identity cos 2 (θ ) + sin 2 (θ ) = 1 . First, begin with the identity cos 2 (θ ) + sin 2 (θ ) = 1 . Now divide each term by cos 2 (θ ) : cos 2 (θ ) cos (θ ) 2 Since sin (θ ) cos (θ ) + sin 2 (θ ) cos (θ ) 2 = tan (θ ) and = 1 cos 2 (θ ) 1 cos (θ ) = sec (θ ) , we can simplify the above equation and obtain the following Pythagorean identity: 1 + tan 2 (θ ) = sec 2 (θ ) In a similar fashion, we again begin with the identity cos 2 (θ ) + sin 2 (θ ) = 1 . Now divide each term by sin 2 (θ ) : cos 2 (θ ) sin (θ ) 2 Since cos (θ ) sin (θ ) + sin 2 (θ ) sin (θ ) 2 = cot (θ ) and = 1 sin (θ ) 1 sin (θ ) 2 = csc (θ ) , we can simplify the above equation and obtain the following Pythagorean identity: cot 2 (θ ) + 1 = csc 2 (θ ) Unit Circle Trigonometry Evaluating Trigonometric Functions of Other Angles Pythagorean Identities cos 2 (θ ) + sin 2 (θ ) = 1 Example If cot (θ ) = 11 5 1 + tan 2 (θ ) = sec 2 (θ ) cot 2 (θ ) + 1 = csc 2 (θ ) and 90D < θ < 270D , find the exact values of sin (θ ) and cos (θ ) . Solution: Since 90D < θ < 270D , the terminal side of θ is either in the second or the third quadrant. Recall the definition for cotangent cot (θ ) = xy ; we are given that cot (θ ) = 11 5 , which is a positive number. In the second quadrant, x is negative and y is positive, so the quotient cot (θ ) = x y is negative; therefore the terminal side of θ can not lie in the second quadrant. In the third quadrant, on the other hand, both x and y are negative, so the quotient cot (θ ) = xy is positive, which is consistent with the given information. We then conclude that the terminal side of θ is in the third quadrant. There are two methods by which we can solve this problem, both of which are shown below. Method 1: Our first method is to use the Pythagorean identity cot 2 (θ ) + 1 = csc 2 (θ ) . Since cot (θ ) = 11 5 , we plug it into the equation cot (θ ) + 1 = csc (θ ) . 2 ( ) 11 5 2 2 + 1 = csc 2 (θ ) We then simplify the equation and solve for csc (θ ) : ( ) 11 5 2 + 1 = csc 2 (θ ) 11 25 + 1 = csc 2 (θ ) 11 25 2 + 25 25 = csc (θ ) 36 25 = csc 2 (θ ) Unit Circle Trigonometry Evaluating Trigonometric Functions of Other Angles