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Two special triangles Two families of angles often show up in computations on the unit circle and we use them to practice our understanding of trigonometry. In this presentation we examine the 30-60-90 triangle and the 45-90-45 triangle (and all triangles on the unit circle related to these.) Elementary Functions The 30-60-90 triangle The 30-60-90 triangle has a right angle (90◦ ) and two acute angles of 30◦ and 60◦ . We assume our triangle has hypotenuse of length 1 and draw it on the unit circle: Part 4, Trigonometry Lecture 4.2a, Two Special Triangles Dr. Ken W. Smith Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 1 / 70 Smith (SHSU) Elementary Functions 2013 2 / 70 The 30 − 60 − 90 triangle The 30 − 60 − 90 triangle Anytime we consider a 30-60-90 triangle, we imagine that triangle as half of an equilateral triangle. For example, we reflect the 30-60-90 triangle about the x-axis and so we imagine an equilateral triangle with one vertex at the origin and the other two vertices to the right of the origin, on the unit circle symmetrically spaced about the x-axis. Smith (SHSU) Elementary Functions 2013 3 / 70 Smith (SHSU) Elementary Functions 2013 4 / 70 The 30 − 60 − 90 triangle The 30 − 60 − 90 triangle This triangle is an equilateral triangle with three equal sides of length 1 unit. By the Pythagorean Theorem, x2 + ( 12 )2 = 12 Since this equilateral triangle is symmetric about the x-axis, and since each side has length 1, then the y-coordinates of the two vertices on the unit circle are ± 12 . Smith (SHSU) Elementary Functions 2013 5 / 70 The 30 − 60 − 90 triangle √ (Also cos(−30◦ ) = Smith (SHSU) √ 3 2 3 4 √ and so the x-value of the points on the unit circle must be x = Smith (SHSU) Elementary Functions 3 2 . 2013 6 / 70 The 30 − 60 − 90 triangle The two points on the unit circle are P ( we see that the cosine of 30◦ is so x2 = √ 3 2 3 1 2 , 2) √ 3 1 2 , − 2 ). 30◦ is 12 . and Q( and the sine of From this and sin(−30◦ ) = − 21 .) Elementary Functions A similar argument with equilateral triangles works for an angle of 60◦ . (We should always think of a 30 − 60 − 90 triangle as half of an equilateral triangle!) In the figure drawn below we see that the cosine of 60◦ is 12 and the sine of 60◦ is 2013 7 / 70 Smith (SHSU) √ 3 2 . Elementary Functions 2013 8 / 70 The 30 − 60 − 90 triangle The 30 − 60 − 90 triangle and its siblings y (0, 1) √ Any angle on the unit circle with a reference √angle of 30◦ or 60◦ will have coordinates that involve the numbers 21 and 23 . These angles include on.... 30◦ , 60◦ , 120◦ , 150◦ , 210◦ , 240◦ , 300◦ 330◦ 3 1 2 ,2 30◦ and so π 6 (−1, 0) (1, 0) x (0, −1) Smith (SHSU) Elementary Functions 2013 9 / 70 The 30 − 60 − 90 triangle and its siblings Smith (SHSU) Elementary Functions 10 / 70 2013 12 / 70 The 30 − 60 − 90 triangle and its siblings y y (0, 1) (0, 1) √ 3 1 , 2 2 π 2 π 3 60◦ (−1, 0) 90◦ (1, 0) (−1, 0) (1, 0) x x (0, −1) Smith (SHSU) 2013 Elementary Functions (0, −1) 2013 11 / 70 Smith (SHSU) Elementary Functions The 30 − 60 − 90 triangle and its siblings − 12 , √ 3 2 The 30 − 60 − 90 triangle and its siblings y y (0, 1) (0, 1) 2π 3 √ 3 1 2 ,2 − 120◦ 5π 6 150◦ (−1, 0) (1, 0) (−1, 0) (1, 0) x x (0, −1) Smith (SHSU) (0, −1) Elementary Functions 2013 13 / 70 The 30 − 60 − 90 triangle and its siblings (−1, 0) Smith (SHSU) Elementary Functions y (0, 1) (0, 1) (1, 0) (−1, 0) 16 / 70 x √ − 3 1 2 , −2 (0, −1) Elementary Functions 2013 (1, 0) x 7π 6 Smith (SHSU) 14 / 70 The 30 − 60 − 90 triangle and its siblings y π 180◦ 2013 210◦ (0, −1) 2013 15 / 70 Smith (SHSU) Elementary Functions The 30 − 60 − 90 triangle and its siblings The 30 − 60 − 90 triangle and its siblings y y (0, 1) (0, 1) (−1, 0) (1, 0) (−1, 0) (1, 0) x x 240◦ 270◦ 4π 3 Smith (SHSU) − 12 , − √ 3 2 3π 2 (0, −1) (0, −1) Elementary Functions 2013 17 / 70 The 30 − 60 − 90 triangle and its siblings Smith (SHSU) Elementary Functions y (0, 1) (0, 1) (1, 0) (−1, 0) 300◦ (0, −1) 11π 6 √ 5π 3 Elementary Functions 20 / 70 x 330◦ Smith (SHSU) 2013 (1, 0) x 18 / 70 The 30 − 60 − 90 triangle and its siblings y (−1, 0) 2013 3 1 2 , −2 √ 3 1 , − 2 2 (0, −1) 2013 19 / 70 Smith (SHSU) Elementary Functions The 30 − 60 − 90 triangle and its siblings The 30 − 60 − 90 triangle and its siblings y y (0, 1) (0, 1) √ 3 1 2 ,2 (−1, 0) 360◦ 2π (1, 0) (−1, 0) (1, 0) x (0, −1) Elementary Functions 2013 21 / 70 The 30 − 60 − 90 triangle and its siblings Smith (SHSU) Elementary Functions 2013 22 / 70 2013 24 / 70 The 30 − 60 − 90 triangle and its siblings y y (0, 1) √ 3 1 2, 2 (0, 1) 5π 2 7π 3 420◦ (−1, 0) 450◦ (1, 0) (−1, 0) (1, 0) x x (0, −1) Smith (SHSU) 13π 6 x (0, −1) Smith (SHSU) 390◦ Elementary Functions (0, −1) 2013 23 / 70 Smith (SHSU) Elementary Functions cos π 6 √ = 3 2 , sin π 6 1 2 = cos π 6 √ =− 3 2 , sin π6 = 1 2 y y (0, 1) (0, 1) √ 3 1 2 ,2 30◦ (−1, 0) √ − 3 1 2 ,2 π 6 5π 6 150◦ (1, 0) (−1, 0) (1, 0) x x (0, −1) Smith (SHSU) cos π 6 √ =− 3 2 , (0, −1) Elementary Functions sin π 6 = 2013 25 / 70 − 12 Smith (SHSU) cos π 6 √ = 3 2 , Elementary Functions y (0, 1) (0, 1) (1, 0) (−1, 0) √ − 3 1 2 , −2 210◦ 330◦ 11π 6 √ 3 1 2 , −2 Elementary Functions 28 / 70 x (0, −1) Smith (SHSU) 2013 (1, 0) x 7π 6 26 / 70 sin π6 = − 12 y (−1, 0) 2013 (0, −1) 2013 27 / 70 Smith (SHSU) Elementary Functions cos π 6 = 1 2, sin √ π 6 = 3 2 cos π 6 = − 12 , sin π 6 √ 3 2 = y y (0, 1) √ 3 1 2, 2 − 12 , √ 3 2 π 3 60◦ (−1, 0) (0, 1) 2π 3 120◦ (1, 0) (−1, 0) (1, 0) x x (0, −1) Smith (SHSU) cos π 6 = − 12 , (0, −1) Elementary Functions sin π 6 2013 29 / 70 Smith (SHSU) √ =− 3 2 cos π 6 = 1 2, sin Elementary Functions π 6 =− (0, 1) (0, 1) (−1, 0) x 240◦ Smith (SHSU) − 12 , − √ 300◦ 4π 3 5π 3 3 2 (0, −1) Elementary Functions 32 / 70 (1, 0) x 2013 3 2 y (1, 0) 30 / 70 √ y (−1, 0) 2013 (0, −1) 2013 31 / 70 Smith (SHSU) Elementary Functions √ 3 1 , − 2 2 The 30 − 60 − 90 triangle The 45-90-45 triangle We collect this information in a table. The first two columns give the angle, first in degrees and then in radians. The last two columns give the x and y coordinates. Another important triangle is the right triangle with both acute angles equal to 45◦ . Since this triangle has two equal angles then it is isoceles; it also has two equal sides. We may assume that the hypotenuse has length 1 and so draw this triangle on the unit circle. θ θ cos(θ) sin(θ) 30◦ 60◦ 120◦ 150◦ 210◦ 240◦ 300◦ 330◦ π 6 π 3 2π 3 5π 6 7π 6 4π 3 5π 3 11π 6 3 2 1 2 −√12 − √23 − 23 − 12 1 √2 3 2 1 √2 3 √2 3 2 1 2 −√12 − √23 − 23 − 12 √ I wouldn’t try to memorize this but instead be able to recreate it by proper use of the 30-60-90 triangle. Smith (SHSU) Elementary Functions 2013 33 / 70 The 45-90-45 triangle Smith (SHSU) Elementary Functions 2013 34 / 70 The 45-90-45 triangle Since the triangle is isoceles, its height y is equal to the base x. By the Pythagorean theorem, 1 cos(45◦ ) = sin(45◦ ) = √ . 2 x2 + y 2 = 1 =⇒ x2 + x2 = 1 =⇒ 2x2 = 1. =⇒ x2 = 12 . =⇒ x = √1 . 2 If we wish to get rid of the radical √ in the denominator, we can multiply numerator and denominator by 2 so that the equation becomes √ 2 ◦ ◦ cos(45 ) = sin(45 ) = . 2 Since y = x then cos(45◦ ) = sin(45◦ ) = √1 . 2 Any angle θ√in which the reference angle is 45◦ will have cosine and sine 2 equal to ± . 2 Here are some examples. Smith (SHSU) Elementary Functions 2013 35 / 70 Smith (SHSU) Elementary Functions 2013 36 / 70 The 45 − 90 − 45 triangle and its siblings The 45 − 90 − 45 triangle and its siblings y y (0, 1) (0, 1) √ √ 2 2 , 2 2 √ − 2 2 2 , 2 π 4 3π 4 45◦ (−1, 0) √ 135◦ (1, 0) (−1, 0) (1, 0) x x (0, −1) Smith (SHSU) (0, −1) Elementary Functions 2013 37 / 70 The 45 − 90 − 45 triangle and its siblings Smith (SHSU) Elementary Functions y (0, 1) (0, 1) (1, 0) (−1, 0) 315◦ 5π 4 7π 4 √ √ 2 2 , − 2 2 √ √ 2 2 , − 2 2 (0, −1) Smith (SHSU) 40 / 70 x 225◦ − 2013 (1, 0) x 38 / 70 The 45 − 90 − 45 triangle and its siblings y (−1, 0) 2013 Elementary Functions (0, −1) 2013 39 / 70 Smith (SHSU) Elementary Functions The 45 − 90 − 45 triangle and its siblings The 45 − 90 − 45 triangle and its siblings y y (0, 1) (0, 1) √ √ 2 2 , 2 2 √ √ 2 2 2 , 2 − 9π 4 405◦ 11π 4 495◦ (−1, 0) (1, 0) (−1, 0) (1, 0) x x (0, −1) Smith (SHSU) cos π 4 √ = 2 2 , (0, −1) Elementary Functions sin π 4 2013 41 / 70 Smith (SHSU) √ = 2 2 cos 3π 4 Elementary Functions √ =− 2 2 , sin 42 / 70 2013 44 / 70 √ π 4 = 2 2 y y (0, 1) (0, 1) √ √ 2 2 2 , 2 √ − 3π 4 45◦ (−1, 0) √ 2 2 2 , 2 π 4 135◦ (1, 0) (−1, 0) (1, 0) x x (0, −1) Smith (SHSU) 2013 Elementary Functions (0, −1) 2013 43 / 70 Smith (SHSU) Elementary Functions cos 5π 4 √ =− 2 2 , sin √ π 4 2 2 =− cos 7π 4 √ = 2 2 , sin π 4 √ =− 2 2 y y (0, 1) (0, 1) (−1, 0) (1, 0) (−1, 0) (1, 0) x x 225◦ 315◦ 5π 4 7π 4 √ − √ 2 2 , − 2 2 √ √ 2 2 , − 2 2 (0, −1) Smith (SHSU) (0, −1) Elementary Functions 2013 45 / 70 The 45-90-45 triangle Smith (SHSU) Elementary Functions 2013 46 / 70 Six Functions on the Unit Circle All of this information has been provided for us (calculator free!) by our understanding of two basic triangles. Make sure you can recover these results just by drawing the appropriate triangle in the appropriate location. We summarize this in a table. Smith (SHSU) θ θ cos(θ) sin(θ) 45◦ 135◦ 225◦ 315◦ π 4 3π 4 5π 4 7π 4 2 2√ 2 √2 2 2√ √ − √22 −√ 22 2 2 Elementary Functions √ − √22 − 22 2013 47 / 70 Smith (SHSU) Elementary Functions 2013 48 / 70 Two special triangles Two special triangles In the next presentation, we will look at the six trig functions. (End) Smith (SHSU) Elementary Functions 2013 49 / 70 Smith (SHSU) Elementary Functions 2013 50 / 70 Six Functions on the Unit Circle A central angle θ determines a point P (x, y) on the unit circle. The x-coordinate is cos θ and the y-coordinate is sin θ. There are four other trig functions, based on this point. Elementary Functions Part 4, Trigonometry Lecture 4.2b, Six Functions on the Unit Circle The tangent function tan θ is equal to Dr. Ken W. Smith y x = sin θ cos θ . If the ratio xy looks familiar, this is the “rise” over “run” as we move from the origin O(0, 0) out to the point P (x, y); that is, tan θ is the slope of the line joining O to P . Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 51 / 70 Smith (SHSU) Elementary Functions 2013 52 / 70 Six Functions on the Unit Circle Six Functions on the Unit Circle There are three more trig functions each of which is the reciprocal of one of the first three. 1 The secant of θ is the reciprocal of cosine; sec θ = x1 . 2 The cosecant of θ is the reciprocal of sine; csc θ = y1 . 3 The cotangent of θ is the reciprocal of tangent; cot θ = xy . Since we may write the six trig functions in terms of x and y on the unit circle then we may also write the six trig functions in terms of cosine and sine. We may also write all six trig functions as ratios of the sides of the right triangle with short legs of length x and y and hypotenuse 1. The reciprocal of a trig function with the syllable “sine” in it (sine and cosine) will be a trig function involving “secant” (cosecant and secant.) Each reciprocal pair has exactly one “co” in the list, so: the reciprocal of sine is cosecant; the reciprocal of cosine is secant; the reciprocal of tangent is cotangent. Smith (SHSU) Elementary Functions 2013 53 / 70 Smith (SHSU) Elementary Functions 2013 54 / 70 Six Functions on the Unit Circle The six trig functions on the right triangle Here is a summary of our results. x 1 The cosine function cos θ is equal to 1 . y 2 The sine function sin θ is equal to 1 . y sin θ 3 The tangent function tan θ is equal to x = cos θ . 1 1 4 The secant function sec θ is equal to x = cos θ , the reciprocal of cosine. 1 1 5 The cosecant function csc θ is equal to y = sin θ , the reciprocal of sine. x cos θ 6 The cotangent function cot θ is equal to y = sin θ , the reciprocal of tangent. If we draw an angle in general position, intersecting the unit circle at P (x, y), vertices C(0, 0), P (x, y) and R(x, 0) form a right triangle with a right angle on the x-axis at R(x, 0). We can define the six trig functions in terms of the sides of the triangle 4CP R. We follow tradition: call the side CR as the “adjacent” side (close to the angle θ), RP as the “opposite” side (far from the angle θ), and CP is the “hypotenuse.” Smith (SHSU) Elementary Functions 2013 55 / 70 Smith (SHSU) Elementary Functions 2013 56 / 70 The six trig functions on the right triangle Six Functions on the Unit Circle If we abbreviate the lengths “adjacent”, “opposite” and “hypotenuse” by their first letters, A, O and H, we might draw the triangle like this. Thousands of trig students have memorized the first three trig functions as: O 1 Sin θ = H A 2 Cos θ = H O 3 Tan θ = A and put these together into a chant: SOH-CAH-TOA. Smith (SHSU) Elementary Functions 2013 57 / 70 Six Functions on the Unit Circle Smith (SHSU) Elementary Functions Smith (SHSU) Elementary Functions 2013 58 / 70 2013 60 / 70 Six Functions on the Unit Circle 2013 59 / 70 Smith (SHSU) Elementary Functions Six Functions on the Unit Circle Central Angles and Arcs A side story I first learned of SOH-CAH-TOA at Central Michigan University when a number of calculus students wrote the letters at the top of their exams. Since the Saginaw Chippewa tribe is located near the university and the university had a course in the Ojibwe language, I wondered if Soh-cah-toa was an Ojibwe phrase or prayer. We will look more closely at these right triangle definitions of our trig functions in the next presentation. (End) But one of the students explained that Soh-cah-toa was merely an English abbreviation for the trig functions. (But they didn’t deny saying a short prayer as they wrote it at the top of the exam!) Smith (SHSU) Elementary Functions 2013 61 / 70 Smith (SHSU) Elementary Functions 2013 62 / 70 Six Functions on the Unit Circle Worked problems. 13π 1 Find all six trig functions of the angle θ = 3 . Solution. Draw the unit circle and find the point on the circle given 13 1 13π π by the angle 13π 3 radians. Since 3 = 4 + 3 then 3 = 4π + 3 and so 13π π the angle 3 gives the same point on the unit circle as 3 . This point is √ P ( 12 , 23 ). √ √ 3 √ 3 13π 1 13π 13π Therefore cos 3 = 2 , sin 3 = 2 and tan 3 = 21 = 3. The Elementary Functions Part 4, Trigonometry Lecture 4.2c, Practicing With Six Trig Functions Dr. Ken W. Smith 2 Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 63 / 70 values of secant, cosecant and tangent are reciprocals of these so the values of all six trig functions are: √ 3 cos( 13π ) = 3 2 sin( 13π = √12 3 ) tan( 13π = 3 3 ) 13π 2 √ sec( 3 ) = 3 csc( 13π ) = 2 3 13π Functions Smith (SHSU) Elementary ) 2013 64 / 70 cot( = √13 3 Six Functions on the Unit Circle Complementary angles √ 2 Suppose sin θ = − 23 and cos θ is positive. Identify the angle θ and then find all six trig functions of the angle θ. ◦ Solution. The angle is 5π 3 , equal to 300 . (Or we could choose the angle −60◦ = − π3 .) So √ √ 3 1 5π 5π cos( 5π ) = , sin( ) = − 3 2 3 2 , tan( 3 ) = − 3. Given an angle θ we will talk about the angle we need to add to θ to make up 180◦ or 90◦ . The angle we need to add to θ to create a 180◦ angle is the “supplement” of θ. So the supplementary angle of θ is 180◦ − θ or (in radians) π − θ. The angle we need to add to θ to “complete” a right angle is the “complement” of θ; the complementary angle of θ is 90◦ − θ or π2 − θ. In the figure below, with central angle θ (in black) the complementary angle of θ is drawn in blue. The reciprocals of these are √ 2 3 5π 5π √2 √1 sec( 5π 3 ) = 2, csc( 3 ) = − 3 = − 3 , cot( 3 ) = − 3 = − Smith (SHSU) Elementary Functions 2013 √ 3 3 . 65 / 70 Six Functions on the Unit Circle Smith (SHSU) Elementary Functions 2013 66 / 70 The Pythagorean Theorem on the unit circle The term “complementary” explains the syllable “co-” in the terms cosine, cotangent and cosecant. The cosine of the angle θ is merely the “sine of the complement” of θ; co-sine is an abbreviation for “sine of the complement.” O and the cosine of θ is From the drawing we see that if the sine of θ is H π A A then the sine of the angle − θ is also ; the 2 H H “sine-of-the-complement” is merely the “co-sine.” In a similar way, “co-tangent” and “co-secant” stand for “tangent-of-the-complement” and “secant-of-the-complement”. Since the points P (x, y) are on the unit circle then, by the Pythagorean Theorem, x2 + y 2 = 1. In terms of sine and cosine, this translates into (cos θ)2 + (sin θ)2 = 1. (1) Dividing by (cos θ)2 we have 1 + (tan θ)2 = (sec θ)2 . (2) Dividing the top equation by (sin θ)2 we have (cot θ)2 + 1 = (csc θ)2 . (3) These three identities are often called “The Pythagorean Identities” of trigonometry since they all depend on the Pythagorean theorem on the unit circle. Smith (SHSU) Elementary Functions 2013 67 / 70 Smith (SHSU) Elementary Functions 2013 68 / 70 A Worked Problem. Central Angles and Arcs Given the cosine or sine of θ we should (by the Pythagorean Theorem) be able to find all the trig functions of θ. Here is an example. Find all trig functions of the angle θ if θ is in the second quadrant and sin(θ) = 23 . Solution. If sin(θ) = 23 then by the Pythagorean Theorem In the next presentation, we will look in depth at the sine function and transformations of it. cos(θ)2 + ( 23 )2 = 1 =⇒ cos(θ)2 = 1 − ( 23 )2 = 9 9 − 4 9 = 59 . √ Since cos(θ)2 = 59 then cos(θ) = ± 35 . Since θ is in the second quadrant we know that the x-value of the point P (x, y) is negative and so √ 5 x = cos(θ) = − 3 . The point on the unit circle corresponding to the (End) √ angle θ is therefore P (− 35 , 23 ) and with this information we can compute the remaining four trig functions of θ. √ cos(θ) sin(θ) tan(θ) Smith (SHSU) = = = − 2 3 5 3 √ − √25 = − 2 5 5 √ Elementary Functions sec(θ) csc(θ) cot(θ) = = = − √35 = − 3 5 5 3 2√ − 5 2 2013 69 / 70 Smith (SHSU) Elementary Functions 2013 70 / 70