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University of Strathclyde Department of Mathematics & Statistics Circular Functions — Trigonometry Summer School Lecture Notes 1 Basic Definitions OAB is a triangle with a right angle at A. b =θ AOB AB opposite = OB hypotenuse B OA adjacent = OB hypotenuse ( ) AB opposite sin θ tan θ := = = OA adjacent cos θ use n ote hyp cos θ := cot θ := O 1 tan θ REMINDER: SOH CAH TOA 1 θ adjacent opposite sin θ := A 2 Summer School: Circular Functions 2 Circular Functions Y circle, radius r From the basic definitions, sin θ = opp y = hyp r B(x,y) ϕ adj x cos θ = = hyp r ( ) opp y sin θ tan θ = = = adj x cos θ r y θ x O A x y , cos φ = so we have r r x sin(90◦ − θ) = sin φ = = cos θ r Now φ = 90◦ − θ and sin φ = cos(90◦ − θ) = cos φ = y = sin θ. r These are Complementary Angles. Also, tan(90◦ − θ) = tan φ = 3 x = cot θ y Elementary Angles When θ = 0◦ : y = 0, x = r sin(0◦ ) = y = 0, r cos(0◦ ) = x y = 1, tan(0◦ ) = = 0 r x and using complementary angles sin(90◦ ) = sin(90◦ − 0◦ ) = cos(0◦ ) = 1 cos(90◦ ) = cos(90◦ − 0◦ ) = sin(0◦ ) = 0 tan(90◦ ) = tan(90◦ − 0◦ ) = 1 = undefined 0 X 3 Summer School: Circular Functions 3.1 Isosceles Triangle B b + OBA b = 90◦ AOB 45 b = OBA b AOB o 2a a a A b = OBA b = 45◦ ⇒ AOB OB 2 = OA2 + AB 2 = 2a2 √ ⇒ OB = 2 a 45 o O a 1 cos 45◦ = sin 45◦ = √ =√ 2a 2 tan 45◦ = 1 3.2 Equilateral Triangle B OB = BC = CO = 2l o 30 OA = AC = l Then OB 2 = OA2 + AB 2 ⇒ 4l = l + AB ⇒ AB = 2 2 2 √ 2l 3l 3l b = OBC b = B CO b = 60◦ C OB o b = ABC b = 30◦ OBA 60 O l Hence 1 = cos 60◦ 2 √ 3 = sin 60◦ cos 30◦ = 2 sin 30◦ = 1 tan 30◦ = √ 3 √ tan 60◦ = 3 A C 4 Summer School: Circular Functions 3.3 Table of Values (IMPORTANT) 0◦ 30◦ 45◦ 60◦ 90◦ √ 1 1 3 √ sin θ 0 1 2 2 2 √ 3 1 1 √ cos θ 1 0 2 2 2 √ 1 tan θ 0 √ 1 3 − 3 θ 3.4 3.4.1 Extension to other angles 90◦ < θ ≤ 180◦ Now x < 0, y > 0 and r > 0. Y We define sin θ = cos θ = y >0 r B x <0 r y r θ tan θ = y <0 x A x O b = 180◦ − θ, so We also have AOB sin(180◦ − θ) = sin θ cos(180◦ − θ) = − cos θ X 5 Summer School: Circular Functions 3.4.2 180◦ < θ ≤ 270◦ Y Here x < 0, y < 0, r > 0 sin θ = y <0 r cos θ = x <0 r A θ x X O y tan θ = > 0 x y r B 3.4.3 270◦ < θ ≤ 360◦ Y Here x > 0, y < 0, r > 0 sin θ = y <0 r cos θ = x >0 r tan θ = y <0 x O θ x r A y B X 6 Summer School: Circular Functions 3.5 Rule of Signs Y What’s positive? This shows that st nd 1 Quad 2 Quad sin θ > 0 in the 1st and 2nd quadrants sin all cos θ > 0 in the 1st and 4th quadrants X tan tan θ > 0 in the 1st and 3rd quadrants 3.6 cos rd th 3 Quad 4 Quad Negative Angles Angles are positive when measured anti-clockwise and negative when Y measured clockwise. y sin(−θ) = − = − sin θ r x = cos θ r y tan(−θ) = − = − tan θ x cos(−θ) = r θ −θ y X x r −y 7 Summer School: Circular Functions 4 Degrees and Radians The angle θ in radians is θ= length of arc AB r B (This does not depend on r) r θ The circumference is 2πr. Hence a A r ◦ complete rotation (360 ) is 2πr = 2π radians r ( ◦ 2π radians ≡ 360 ⇒ 1 radian ≡ 180 π )◦ π radians ≡ 1◦ 180 4.1 Table of degrees and radians Degrees 0◦ 30◦ 45◦ 60◦ 90◦ 120◦ 135◦ 150◦ Radians 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 Degrees 180◦ 210◦ 225◦ 240◦ 270◦ 300◦ 330◦ Radians π 315◦ 7π/6 5π/4 4π/3 3π/2 5π/3 7π/4 11π/6 Now look back at the table of values and make sure you know exact values for radians!!! NOTE: Be careful to include ◦ when referring to degrees. 8 Summer School: Circular Functions 5 Pythagoras’ Theorem Pythagoras’ Theorem says x2 + y 2 = r 2 Dividing by r2 , we get r 2 y 2 x y + 2 =1 2 r r θ and using the definition of sin θ and cos θ: x cos2 θ + sin2 θ = 1 6 Graphs The graph of sin θ has a maxi- sin(θ) mum value of 1 when θ takes the 1 values ··· −3π π 5π , , ··· 2 2 2 0.5 It takes the minimum value of −1 when θ takes the values ··· −π 3π , ,··· 2 2 0 −0.5 and is equal to 0 when θ takes −1 the values −6 0, ±π, ±2π, · · · The graph of sin θ has a period of 2π. −4 −2 0 θ 2 4 6 9 Summer School: Circular Functions The graph of cos θ has a maxicos(θ) mum value of 1 when θ takes the values 1 0, ±2π, ±4π · · · 0.5 It takes the minimum value of −1 when θ takes the values ±π, ±3π, · · · 0 −0.5 and is equal to 0 when θ takes the values −1 −6.2832 π 3π 5π ± ,± ,± ,··· 2 2 2 −3.1416 0 θ 3.1416 The graph of cos θ has a period of 2π. Note the similarity between the graphs of sin θ and cos θ. tan(θ) 6 The graph of tan θ takes the value 0 when θ takes the values 0, ±π, ±2π, ±3π · · · It is undefined when θ takes the values π 3π 5π ± ,± ,± ,··· 2 2 2 The graph of tan θ has a period of π. 4 2 0 −2 −4 −6 −4.7124 −1.5708 θ 1.5708 4.7124 6.2832 10 Summer School: Circular Functions 7 Compound Angles P α L α N Q β α O K M b is a right angle, so LN bQ = α Triangles OQK, P QN are similar, so QPbN = α. N LP PK P L + LK LK + P L NM + P L = = = PO PO PO PO N M ON PL PN = · + · ON P O P N P O = sin α cos β + cos α sin β sin(α + β) = Hence sin(α + β) = sin α cos β + cos α sin β It can also be shown that cos(α + β) = cos α cos β − sin α sin β sin(α − β) = sin α cos β − cos α sin β cos(α − β) = cos α cos β + sin α sin β tan(α + β) = tan α + tan β 1 − tan α tan β tan(α − β) = tan α − tan β 1 + tan α tan β Summer School: Circular Functions Example 7.1 Find the value of sin(75◦ ) sin(75◦ ) = sin(45◦ + 30◦ ) = sin 45◦ cos 30◦ + cos 45◦ sin 30◦ √ √ 1 1 1+ 3 1 3 = √ +√ = √ 2 2 22 2 2 Example 7.2 Find the value of cos(15◦ ) cos(15◦ ) = cos(45◦ − 30◦ ) = cos(45◦ ) cos(30◦ ) + sin(45◦ ) sin(30◦ ) √ √ 1 3 1 1 1+ 3 = √ +√ = √ 2 2 22 2 2 Example 7.3 Find the value of tan(105◦ ) tan(105◦ ) = tan(60◦ + 45◦ ) tan(60◦ ) + tan(45◦ ) = 1 − tan(60◦ ) tan(45◦ ) √ √ √ 3+1 3+1 1+ 3 √ = √ × √ = 1− 3 1− 3 1+ 3 √ √ 3+3+1+ 3 √ √ = 1+ 3− 3−3 √ √ 2 3+4 = =− 3−2 −2 8 Double Angles From Section 7, we have sin(2a) = 2 sin a cos a cos(2a) = cos2 a − sin2 a We can use cos2 a + sin2 a = 1 to obtain alternative formulae for cos(2a): cos(2a) = cos2 a − sin2 a = cos2 a − (1 − cos2 a) = 2 cos2 a − 1 cos(2a) = cos2 a − sin2 a = (1 − sin2 a) − sin2 a = 1 − 2 sin2 a 11 12 Summer School: Circular Functions ( ) 1◦ Example 8.1 Find the value of sin 22 . 2 Let a = 22 9 1◦ then consider cos(2a). 2 1 √ = cos(45◦ ) 2 ( ) 1◦ 2 ⇒ 2 sin 22 2 ( ) 1◦ 2 ⇒ sin 22 2 ( ) 1◦ ⇒ sin 22 2 ( ) 1◦ = 1 − 2 sin 22 2 √ 1 2−1 =1− √ = √ 2 2 √ 2−1 = √ 2 2 √√ 2−1 √ = 2 2 2 Secant, Cosecant We define sec θ = 1 cos θ and csc θ = 1 sin θ Recall that cos2 θ + sin2 θ = 1. Divide by cos2 θ to get 1 + tan2 θ = sec2 θ Divide by sin2 θ to get cot2 θ + 1 = csc2 θ 10 Identities Example 10.1 Show that cos θ 1 − sin θ = 1 + sin θ cos θ cos θ(1 − sin θ) cos θ(1 − sin θ) cos θ(1 − sin θ) 1 − sin θ cos θ = = = = 2 2 1 + sin θ (1 + sin θ)(1 − sin θ) cos θ cos θ 1 − sin θ Example 10.2 Show that 1 + sin(2θ) = (cos θ + sin θ)2 1 + sin(2θ) = cos2 θ + sin2 θ + 2 sin θ cos θ = cos2 θ + 2 cos θ sin θ + sin2 θ = (cos θ + sin θ)2 13 Summer School: Circular Functions Example 10.3 Show that sin(2θ) − cos(2θ) + 1 = tan θ sin(2θ) + cos(2θ) + 1 sin(2θ) − cos(2θ) + 1 2 sin θ cos θ − (1 − 2 sin2 θ) + 1 = sin(2θ) + cos(2θ) + 1 2 sin θ cos θ + (2 cos2 θ − 1) + 1 2 sin θ cos θ + 2 sin2 θ = 2 sin θ cos θ + 2 cos2 θ sin θ(cos θ + sin θ) = cos θ(sin θ + cos θ) sin θ = = tan θ. cos θ 11 Pythagorean Triples A set of three positive integers x, y, z such that x2 + y 2 = z 2 is called a Pythagorean Triple. Examples are {3, 4, 5}, {5, 12, 13}, {7, 24, 25}, {8, 15, 17}, {9, 40, 41}, {11, 60, 61}, {12, 35, 37}, {13, 84, 85}, {15, 112, 113}, {16, 63, 65}, {17, 144, 145}, {19, 180, 181}, {20, 21, 29}, {20, 99, 101}. Example 11.1 If sin θ = 15 , what is 17 B cos θ? Hint: draw a triangle! From diagram BC = 15, AB = 17 17 15 Hence AC 2 = 172 −152 = 289−225 = 64 = 82 AC 8 cos θ = = AB 17 A θ 8 C 14 Summer School: Circular Functions Example 11.2 If sin α = 5 24 and cos β = , find tan(α − β) 13 25 13 25 5 α 7 β 24 12 tan α − tan β 1 + tan α tan β 7 5 − 12 24 = 5 × 24 − 7 × 12 = 5 7 12 × 24 + 5 × 7 1+ · 12 24 120 − 84 36 = = 288 + 35 323 tan(α − β) = 12 Product Formulae Recall that sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β ⇒ sin(α + β) + sin(α − β) = 2 sin α cos β p+q p−q Put α + β = p, α − β = q ⇒ α = ,β = to get 2 2 ( ) ( ) p+q p−q sin p + sin q = 2 sin cos 2 2 ( cos p + cos q = 2 cos ( cos p − cos q = 2 sin Example 12.1 p+q 2 p+q 2 ( ◦ ◦ sin 105 + sin 15 ) ( cos ) ( sin p−q 2 q−p 2 ) ) ) ( ◦) 120◦ 90 = 2 sin cos 2 2 ◦ ◦ = 2 sin 60 cos√ 45 √ 3 1 3 √ = = 2 2 2 2 15 Summer School: Circular Functions Example 12.2 sin 105◦ − sin 15◦ = sin 105◦ + sin(−15◦ ) ( ◦) ( ) 120◦ 90 = 2 sin cos 2 2 ◦ ◦ = 2 sin 45 cos 60 1 1 1 = 2√ =√ 22 2 Example 12.3 ( ◦ ◦ cos 15 − cos 105 ) ( ◦) 120◦ 90 = 2 sin sin 2 2 ◦ ◦ = 2 sin 60 sin 45 √ √ 3 1 3 √ =√ = 2 2 2 2 Example 12.4 ( ) ( ) 5x + x 5x − x 2 sin cos sin(5x) + sin x 2 2 ( ) ( ) = 5x + x 5x − x cos(5x) + cos x 2 cos cos 2 2 sin(3x) = = tan(3x) cos(3x) Product formulae can also be re-written in the form 1 sin α cos β = {sin(α + β) + sin(α − β)} 2 1 cos α cos β = {cos(α + β) + cos(α − β)} 2 1 sin α sin β = {cos(α − β) − cos(α + β)} 2 Example 12.5 ( ) ( ◦) ) ( )] [ ( 1◦ 1 1 1◦ 1◦ 1◦ 1◦ cos 37 cos 7 = + cos 37 − 7 cos 37 + 7 2 2 2 2 2 2 2 1 = {cos(45◦ ) + cos(30◦ )} 2[ √ ] √ √ 1 1 3 2+ 3 √ + = = 2 2 4 2 16 Summer School: Circular Functions Example 12.6 ) ) ) ( ◦) ( ( 1◦ 1◦ 1◦ 1 1◦ 1◦ 2 sin 37 cos 7 = sin 37 + 7 + sin 37 − 7 2 2 2 2 2 2 ◦ ◦ = sin(45 ) + sin(30 ) √ 1 1 2+1 = √ + = 2 2 2 ( 13 Trigonometric Equations Question: Given a value c of sin(θ), what is θ? 1 Example 13.1 Suppose sin(θ) = : 2 sin(θ) 1 0.5 0 −0.5 −1 −6.2832 −3.1416 0 3.1416 θ 6.2832 9.4248 12.5664 1 There are infinitely many values of θ for which sin(θ) = , for example 2 θ= π 5π π 5π , , 2π + , 2π + ,... 6 6 6 6 In general θ= π + 2kπ, 6 θ= 5π + 2kπ, 6 k = 0, ±1, ±2, ±3, . . . Because of these endless answers we are usually asked to find values of θ within a certain range. 17 Summer School: Circular Functions Solving for θ ∈ [0, 2π] 13.1 Given we want to solve, for θ ∈ [0, 2π], trig f unction(θ) = c we first solve trig f unction(θ̂) = |c| i.e. for positive c. Here trig f unction is sin, cos or tan. Y nd 2 st Quad 1 Quad sin all π−θ rd 3 π ). We 2 then refer to our ’Rules of Signs’ from This gives an acute angle (θ̂ < section 3.5. So all of our functions are θ π+θ 2π − θ tan cos positive in two quadrants and negative in X two quadrants. This gives, at most, two values for θ ∈ [0, 2π]. This process is th Quad 4 Quad best seen in an example. cos(θ) Example 13.2 Find all the solutions for x ∈ [0, 2π] of 1 2 cos2 x − 3 cos x + 1 = 0 0.5 Factorise: (2 cos x − 1)(cos x − 1) = 0 0 1st 1 cos x = , 2 π x̂ = , 3 π x= , 3 cos x = 1 −0.5 x̂ = 0 1st x=0 −1 0 4th x = 2π− 5π π = , 3 3 4th x = 2π−0 = 2π 1.5708 3.1416 θ 4.7124 6.2832 18 Summer School: Circular Functions Example 13.3 Find x ∈ [0, 2π] for which cos(2x) − 3 cos x + 2 = 0 Put cos(2x) = 2 cos2 x − 1 to obtain 2 cos2 x − 3 cos x + 1 = 0 See previous example. Example 13.4 Find x ∈ [0, 2π] for which sin(θ) 1 sin x − 2 cos2 x + 2 = 0 0.5 Put cos2 x = 1 − sin2 x: sin x − 2(1 − sin2 x) + 2 = 0 2 sin2 x + sin x = 0 1 sin x(sin x + ) = 0 2 sin x = 0, x̂ = 0, 1st x = 0 or 2π, 2nd x = π−0 = π, sin x = − 1 2 π x̂ = 6 3rd 4th π 7π = 6 6 π 11π x = 2π− = 6 6 x = π+ 0 −0.5 −1 0 1.5708 3.1416 θ 4.7124 6.2832 4 x 10 19 Summer School: Circular Functions Example 13.5 Solve, for x ∈ [0, 2π] cos2 x − sin2 x − cos x + 1 = 0 cos(θ) 1 Put sin2 x = 1 − cos2 x to obtain cos2 x − (1 − cos2 x) − cos x + 1 = 0 0.5 2 cos2 x − cos x = 0 0 cos x(2 cos x − 1) = 0 cos x = 0, x̂ = 1st 4th 14 x= π , 2 cos x = x̂ = π , 2 x = 2π− π 3π = , 2 2 1 2 −0.5 π 3 −1 π 3 1st x= 4th x = 2π− 0 1.5708 3.1416 θ 4.7124 π 5π = 3 3 Inverse Functions We have seen in the previous section that there are an infinite number of solutions, x, to equations such as sin x = c where x is an angle and c is a constant. [ π π] with sin(x) = c: This unique If we restrict c to c ∈ [−1, 1] there is a unique x ∈ − , 2 2 value is the inverse sine of c. We write x = sin−1 c or x = arcsin c Similarly, for each value of c ∈ [−1, 1] there is a unique x ∈ [0, π] with cos(x) = c: This x is the inverse cosine function x = cos−1 c or x = arccos c ( π π) And again, for each c ∈ (−∞, ∞) there is a unique x ∈ − , with tan(x) = c: This x 2 2 is the inverse tangent function x = tan−1 c or x = arctan c 6.2832 20 Summer School: Circular Functions Note that sin−1 (sin α) = α and sin(sin−1 α) = α etc. We can use these inverse functions to help find values of some circular functions. ( ) ( ) 4 5 −1 Example 14.1 If θ = sin + sin find sin θ. 5 13 ( ) ( ) 4 5 −1 −1 Let sin = α, sin =β ⇒θ =α+β 5 13 −1 sin θ = sin(α + β) = sin α cos β + cos α sin β α 13 5 5 3 β 4 12 3 4 cos α = , sin α = 5 5 cos β = 12 , 13 sin β = 5 13 48 + 15 63 4 12 3 5 · + · = = 5 13 5 13 65 65 ( ) ( ) 12 24 Example 14.2 If θ = sin−1 + tan−1 find cos θ. 13 7 sin θ = ( Let sin −1 12 13 ) ( −1 = α, tan 24 7 ) =β ⇒θ =α+β cos θ = cos α cos β − sin α sin β β α 13 25 12 cos α = 5 , 13 7 5 24 sin α = 12 13 cos θ = cos β = 7 , 25 sin β = 5 7 12 24 35 − 288 253 · − · = =− 13 25 13 25 325 325 24 25 21 Summer School: Circular Functions 15 Polar Coordinates Y Let P have Cartesian coordinates (x, y). OP has length r ≥ 0. P(x,y) b = θ. X OP Then P has polar coordinates (r, θ). We have r x = r cos θ, x2 + y 2 = r 2 , y y = r sin θ r= √ x2 + y 2 θ y tan θ = x X x O θ is usually restricted to −π < θ ≤ π or 0 ≤ θ < 2π. (The later in our case). We use a similar method to find θ as we did for solving trigonometric equations. That is, firstly solve y tan θ̂ = x Then we check which quadrant we are in and use the same relations as before to find θ. Example 15.1 (x, y) = (1, r= √ x2 + y 2 = tan(θ̂) = √ √ √ 3) Y (1, 3) 1+3=2 3 ⇒ θ̂ = π 3 π The point is in the 1st quad so θ = is ( π )3 fine and Polar coordinates are 2, . 3 θ X 22 Summer School: Circular Functions √ Example 15.2 (x, y) = (−1, − 3) r= √ Y 1+3=2 θ √ − 3 √ π = 3 ⇒ θ̂ = tan(θ̂) = −1 3 X The point is in the 3rd quad so π 4π = 3 3 ( ) 4π and Polar coordinates are 2, . 3 θ=π+ (−1,− 3) Example 15.3 (x, y) = (4, −4) r= √ x2 + y 2 = √ 16 + 16 = √ √ 32 = 4 2 Y θ 4 π tan(θ̂) = = 1 ⇒ θ̂ = 4 4 X The point is in the 4th quad so π 7π = 4 4 ( ) √ 7π and Polar coordinates are 4 2, . 4 √ Example 15.4 (x, y) = (− 3, 3) θ = 2π − r= √ 3+9= √ (4,−4) √ 12 = 2 3 Y √ 3 π tan(θ̂) = √ = 3 ⇒ θ̂ = 3 3 (− 3, 3) The point is in the 2nd quad so 2π π = 3 3 ( ) √ 2π and Polar coordinates 2 3, . 3 θ=π− θ THE END X