AMB11F Trigonometry Notes Trigonometry is a study of measurements of sides of triangles linked to the angles, and the application of this theory. Let ABC be right-angled so that angles A and B are acute (less than 90 o ). A c b C θ a B Let ABC = θ. We define (i) the sine of θ as sinθ = b = AC = opposite c AB hypotenuse (ii) the cosine of θ as cos θ = a = CB = adjacent c AB hypotenuse and (iii) the tangent of θ as tan θ = b = AC = opposite. a CB adjacent The three functions defined above are called trigonometric functions. We can deduce from (iii) that tan θ = sinθ. In fact all other trigonometric functions cosθ are defined in terms of sine and cosine only. The other three trig functions are defined as follows: (iv) the cosecant of θ is csc θ = 1 sinθ, (v) the secant of θ is sec θ = 1 cosθ, (vi) the cotangent of θ is sec θ = cos θ sin θ. Now sin θ+cos θ = b +a = c = 1; thus sin = 1 cos θ and cos = 1 sin θ. c c Also sin(90 o θ) = a/c = cos θ and cos(90 o θ) = b/c = sinθ. If and y are acute angles such that + y = 90 o then they are called complimentary angles. In such a case sin = cos y and cos = siny. Eample : If cos = 1/ and and y are complimentary angles, find siny. Soln. siny = cos = 1/. Some special (acute) angles (in degrees) viz. 0; 30; 45; 60 90 (i) sin0 o = 0, cos 0 o = 1. Since 0 and 90 o are complimentary angles, we conclude that sin90 o = cos0 o = 1 and cos 90 o = sin0 o = 0. (ii) Consider a right-angled isosceles triangle as shown: 1
A 45 o c a C 45 a o B Then sin45 o = a/c = cos45 o. Also a + a = c implies c = a. Thus sin45 o = 1/ = cos 45 o. Hence tan 45 o = 1. (iii) Consider the right-angled triangle with acute angles of 30 o and 60 o and a = c/ as shown: A 30 o c b C 60 a o B Then b = c (c/) = (3/4)c, implying b = 4c/. Thus sin60 o = 3 = cos30 o since 60 + 30 = 90. Also b = 4c/. Thus cos 60 o = 1 = sin30o since 60 + 30 = 90. Trig functions for angles > 90 o The y-plane is divided into 4 regions called quadrants: the 1st quadrant is formed by the +ve -ais and the +ve y-ais; the nd quadrant is formed by the +ve y-ais and the -ve -ais; the 3rd quadrant is formed by the -ve -ais and the -ve y-ais; the 4th quadrant is formed by the -ve y-ais and the +ve -ais; see diagram below:
y nd quad. 1st quad. 3rd quad. 4th quad. With this in mind, an acute angle is usually referred to as an angle in the first quad; an obtuse angle (between 90 o and 180 o ) is usually referred to as an angle in the nd quad; an angle between 180 o and 70 o is usually referred to as an angle in the 3rd quad; and an angle between 70 o and 360 o is usually referred to as an angle in the 4th quad. Let P(, y) be a point such that OP makes an angle θ with the +ve -ais. Denote the length of OP by r. Then we define (i) sinθ = y ; (ii) cos θ = ; (iii) tan θ = y. The other trig functions r r are defined as before. y P(, y) P(, y) r r θ r θ r P(, P(, y) y) Observe that r = + y in each quadrant. We say an angle is positive if it is measured counterclockwise. It is negative when measured clockwise. Now consider the above diagram: (1) sinθ = y/r 3
(ii) Since 180 o θ is in the nd quad, sin(180 o θ) = y/r = sinθ by using the coordinates of P in the nd quad.; (iii) Since 180 o +θ is in the 3rd quad, sin(180 o +θ) = y/r = sinθ by using the coordinates of P in the 3rd quad.; (iv) Since 360 o θ is in the 4th quad, sin(360 o θ) = y/r = sinθ by using the coordinates of P in the 4th quad.; Similarly (v) cos θ = /r (vi) Since 180 o θ is in the nd quad, cos(180 o θ) = /r = cosθ by using the coordinates of P in the nd quad.; (vii) Since 180 o +θ is in the 3rd quad, cos(180 o +θ) = /r = cos θ by using the coordinates of P in the 3rd quad.; (viii) Since 360 o θ is in the 4th quad, cos(360 o θ) = /r = cos θ by using the coordinates of P in the 4th quad.; NOTE: From above we see that (a) both sine and cosine are +ve in the first quad (b) In the nd quad sine is +ve while cosine is -ve (c) In the 3rd quad both sine and cosine are -ve (d) In the 4th quad sine is -ve while cosine is +ve y sine +ve & cos -ve. sine +ve & cos +ve. sine -ve & cos -ve. sine -ve & cos +ve. Eamples: 1. Find sin10 o without a calculator. Soln. sin10 o = sin(180 o 60 o ) = sin60 o = 3 by (ii) above. Find cos5 o without a calculator. Soln. cos 5 o = cos(180 o + 45 o ) = cos45 o = 1 by (vii) above 4
3. Find cos300 o without a calculator. Soln. cos 300 o = cos(360 o 60 o ) = cos 60 o = 1 by (viii) above 4. Find sin315 o without a calculator. Soln. sin315 o = sin(360 o 45 o ) = sin45 o = 1 by (iv) above Note: sin(180 o ) = sin(180 o 0 o ) = sin0 o = 0 cos(180 o ) = cos(180 o 0 o ) = cos0 o = 1 sin(70 o ) = sin(180 o + 90 o ) = sin90 o = 1 cos(70 o ) = cos(180 o + 90 o ) = cos 90 o = 0 sin(360 o ) = sin(360 o 0 o ) = sin0 o = 0 cos(360 o ) = cos(360 o 0 o ) = cos0 o = 1 Periodicity of sine and cosine Sine and cosine are periodic with period 360 o since after 360 O they start to repeat themselves, i.e. sin(+360 o ) = sin and cos(+360 o ) = cos. Hence after any multiple of 360, sine and cosine repeat themselves, i.e. sin(+360n o ) = sin and cos( + 360n o ) = cos where n is an integer, e.g. 3 or -3. Eamples: 1. Find sin765 o without a calculator. Soln. sin765 o = sin( 360 o + 45 o ) = sin45 o = 1. Find sec θ = 5/, find cos θ, sin θ and cot θ without a calculator. Soln. sec θ = 5/ implies cosθ = /5. Therefore r = 5 and =. Therefore y = r = 5 4 = 1 by the Theorem of Pythagoras. Thus y = 1. Since this cosine is +ve, the angle θ is either in the 1st or 4th quad. See diagram below: 5
y P(, 1) 5 θ 5 P(, 1) Case θ is in the first quad: sinθ = 1 and cotθ = 5 1. Case θ is in the 4th quad: sinθ = 1 and cot θ = 5 1. Negative Angles An angle is negative when measured clockwise. y P(, y) r θ θ r P(, y) sin( θ) y/r = sinθ cos( θ)/r = cos θ Eamples: 1. cos( 45 o ) = cos(45 o ) = 1. sin( 690 o ) = sin(690 o ) = sin(70 o 30 o ) [ sin30 o ] = sin30 o = 1 6
Trigonometric Identities Recall: 1. sin + cos = 1 (1), for any angle. This is an eample of a trig identity.. Divide eqn (1) by cos to get tan + 1 = sec (a), and divide eqn (1) by sin to obtain 1 + cot = csc (b) We give the following identities without proof: 3. sin( + y) = sincosy + cos siny (3) 4. cos( + y) = cos cosy sinsiny (4) We derive the following: 5. In (3), letting = y we get sin = sin cos (5) 6. In (4), letting = y we get cos = cos sin (6) Eample: (a) Find sin15 o cos 15 o Soln. By (5) above sin 15 o cos15 o = sin(15o ) = sin(30o ) = 1/4 (b) Find cos 105 o sin 105 o Soln. By (6) above cos 105 o sin 105 o = cos(105 o ) = cos 10 o = cos(180 o + 30 o ) = cos30 o = 3 (7) Now (6) implies cos = cos (1 cos ) = cos 1, implying cos 1+cos = (7), (double angle formula); 8) Also (6) implies cos = 1 sin sin = 1 sin, implying sin 1 cos = (8), (double angle formula); Eample: (a) Find sin15 o without a calculator. Soln. sin 15 o = 1 cos (15o ) = 1 3 = 3, thus sin 15 o 3 =. 4 (b) Find cos(, 5) o without a calculator. Soln. cos (, 5) o 1+cos 45o = = 1+ 1 = +1 +1, thus cos(, 5)o =. 9. Add the following eqns: sin( + y) = sincos y + cos siny (3) and sin( y) = sincos y cos siny (3) to get sin cos y = sin( + y) + sin( y) 9 10. Add the following eqns: cos( + y) = cos cos y sinsiny (4) and cos( y) = cos cosy + sin siny (4) to get cos cos y = cos( + y) + cos( y) 10 11. Subtracting in the above eqns we have sin siny = cos( y) cos( + y) 11 Eample: Find sin75 o cos 15 o = sin 90o +sin 60 o = + 3 by eqn (9) 4 More Eamples 7
Eample 1. Let tan θ = 4/3, 70 o < θ < 360 o. Find sin θ; cos θ and tan θ. Soln. y θ r = 5 P(3, 4) sinθ = sin θ cosθ = ( 4/5)(3/5) = 4/5 cos θ = cos θ sin θ = (3/5) ( 4/5) = (9 16)/5 = 7/5 Proving given Identities In proving identities, take one side viz the right hand side (rhs) or the left hand side (lhs) and show that it equals the other side. It is advisable to take the more complicated side and simplify it until it gives the other side. Before giving eamples, we derive an identity involving tan θ: We want to show that tan(θ + α) = tanθ+tan α Soln. tan(θ + α) = sin(θ+α) cos(θ+α) = 1 tan θ sinθ cosα+cosθ sinα). cos θ cosα sinθ sinα cos θ cos α, then tan(θ + α) = tanθ+tan α 1 tan θ In the above eqn let θ = α, then tan θ = tanθ 1 tan θ. Eample: Find tan 75 o without a calculator. Soln. tan 75 o = tan(30 o + 45 o ) = tan30o +tan 45 o More Eamples: sin 1. Prove that = tan 1+cos Soln. lhs = sin = sin cos = 1+cos 1+cos sin sin cos = tan = rhs cos 1 sin. Prove that = sincos sin cos Soln. lhs = sin +cos sin cos sin cos = 1+ 3 1 tan30 o tan45 o 3 1 sincos cos +sin + cos sin = = (sin cos)(sin cos) sin cos 8 Now divide the rhs by sin cos cos = = sin cos = rhs
cot 3. Prove that = sin 1+cot Soln. lhs = cos / sin 1+cos /sin 4. Prove that 1 1+cos + 1 = sin cos sin +cos Soln. lhs = 1 cos+1+cos 1 cos 5. Prove that 1 cos = tan sin / cos = sin/1 = rhs = 1 cos csc = /sin = csc@ = rhs 1+tan Soln. rhs = = sin /1 = 1 cos = lhs (cos +sin )/cos csc 6. Prove that = sec 1+cot 1+tan 1/sin = 1 (sin+cos )/sin sin+cos Soln. lhs = lhs. 7. Prove that (sec tan ) = 1 sin 1+sin and rhs = 1/ cos (sin+cos)/cos = 1 sin cos = (1 sin) Soln. rhs = = 1 sin+sin = 1/cos sin /cos + (1+sin)(1 sin) 1 sin sin /cos = sec tan sec + tan = (sec tan) = lhs. 8. Prove that cot = sin 1 cos sin cos Soln. rhs = = cot = rhs. Trig Eqns = sincos 1 cos +sin sinsin Eamples : 1. Solve cos = 1, [0 o, 360 o ] without the use of a calculator. Soln. cos = 1/ is negative in nd and 3rd quadrants. Thus the soln is = 180 o 60 o = 10 o or = 180 o + 60 o = 40 o.. Solve cos = 1, [ 180 o, 360 o ] without the use of a calculator. Soln. cos = 1/ is negative in nd and 3rd quadrants. Thus the soln is = 180 o 60 o = 10 o or = 180 o + 60 o = 40 o or = 10 o. 3. Find a generalsolution of the eqn cos = 1, without the use of a calculator. Soln. cos = 1/ is negative in nd and 3rd quadrants. Thus the general soln is = 10 o + 360 o n or = 40 o + 360 o n where n is an integer. (NOTE: = 10 o + 360 o n is the same as = 40 o + 360 o n and = 40 o + 360 o n is the same as = 10 o + 360 o n) A general soln caters for all possible values of angle. 4. Find a general soln of sin sin 1 = 0, without the use of a calculator. Soln. (sin + 1)(sin 1) = 0 implies sin = 1/ or sin = 1. sin is negative in 3rd and 4th quadrants. Thus = 10 o +360 o n or = 330 o +360 o n 9
or = 90 o + 360 o n where n is an integer. 5. Solve sec =, [ 180 o, 180 o ] without the use of a calculator. Soln. cos = 1/ is +ve in 1st and 4th quadrants. Thus the soln is = 45 o or = 45 o or = 360 o 45 o = 315 o or = 45 o. 6. Solve cos = sin0 o, [0 o, 360 o ], without the use of a calculator. Soln. cos = cos 70 o > 0 since 0 + 70 = 90. Thus the soln is = 70 o or = 360 o 70 o = 90 o. Using a calculator If cos = 0.35 then to find a solution from a calculator, enter the number 0.35 and then press cos 1. Then adjust by 180 o or 360 o to find a correct answer. Eamples: 1. Find a general soln of the eqn 3cos + 5sin = 1, using a calculator. Soln. 3cos +5sin = 3(1 sin )+5sin = 1 implies 3sin 5sin = 0. Thus (3sin + 1)(sin ) = 0 Thus sin = 1/3 since the last bracket is impossible. Therefore = 19, 5 o + 180 o + 360 o n = 199, 5 o + 360 o n or = 360 o 19, 5 o + 180 o + 360 o n = 340, 5 o + 360 o n. Solve 4cot + 3csc = 0, [0 o, 360 o ] using a calculator. Soln. 4 cos + 3/sin = 0 implies cos = 3/4 is negative in nd and sin 3rd quadrant. Using a calculator we have = 180 o 41, 4 o = 138, 6 o or = 180 o + 41, 4 o = 1, 4 o. Graphs of Trigonometric Functions Here we look at graphs of y = sin ; y = cos and y = tan. Note that sin and cos are periodic with period 360 o while tan has period 180 o. The the graph of each trig function repeats itself after its period. 1. Graph of y = sin: Consider the table of points on the graph: -360-70 -180-90 0 90 180 70 360 y 0 1 0-1 0 1 0-1 0 Thus the graph of y = sin is as follows: 10
y 360 70 180 90 1 0 90 180 70 360 1 The maimum of sin = 1 and is called the amplitude of the graph or the function y = sin. In general the graph of y = k sin has amplitude = k. Secondly the graph of y = sin(n) has period = 360 o /n. Similarly for the cosine function; i.e. y = k cos n has amplitude = k and its graph has period = 360 o /n. y = k sin n has amplitude = k and period = 360 o n. Similarly for the cosine function.. Graph of y = 3sin : Its period = 360/ = 180 o, so the graph repeats itself after intervals of 180 o. Its amplitude = 3. Consider the table of points on the graph: -180-135 -90-45 0 45 90 135 180 y 0-3 0 3 0-3 0 3 0 Thus the graph of y = sin is as follows: 11
y 3 360 70 180 90 0 90 180 70 360 3 The maimum of sin = 1 and is called the amplitude of the graph or the function y = sin. In general the graph of y = k sin has amplitude = k. Secondly the graph of y = sin(n) has period = 360 o /n. 3. Graph of y = sin(/): Its period = 360 = 70 o, so the graph repeats itself after intervals of 70 o. Its amplitude = 3. Consider the table of points on the graph: -70-540 -360-180 0 180 360 540 70 y 0 0-0 0-0 Thus the graph of y = sin is as follows: 1
y 70 540 360 180 0 180 360 540 70 4. Graph of y = cos: Its period = 360 o, so the graph repeats itself after intervals of 360 o. Its amplitude = 1. Consider the table of points on the graph: -360-70 -180-90 0 90 180 70 360 y 0 1 0-1 0 1 0-1 0 Thus the graph of y = cos is as follows: 13
y 360 70 180 90 1 0 90 180 70 360 1 Graph of y = tan y = tan n has period = 180/n and NO amplitude is defined for it. 5. Graph y = tan. Soln. Period = 180 o. This graph has asymptotes at points such that cos = 0. Thus the asymptotes are the lines = ±90 o ; = ±70 o and further multiples of 90 o. Consider the table of points on the graph: -315-5 -135-45 0 45 135 180 5 315 360 y 1 0 1-1 0 1-1 0 1-1 0 Thus the graph of y = tan is as follows: 14
y 1 70 5 180 135 90 45 0 45 90 135 180 5 70 1 6. Graph y = tan. Soln. Period = 180 o / = 90 o. This graph has asymptotes at points such that cos = 0. Thus the asymptotes are the lines = ±45 o ; = ±135 o and further multiples of 45 o. Consider the table of points on the graph: -90-67.5-45 -.5 0.5 45 67.5 90 11.5 135 y -1 0 1 0-1 0 1-1 0 1 Thus the graph of y = tan is as follows: 15
y 1 135 11.5 90 67.5 45.5 0.5 45 67.5 90 11.5 135 1 16