## Unit Circle & Trigonometric Functions

Aims: By the end of this chapter, you will be able to
(i) define a negative angle, sine, cosine and tangent using a unit circle,
(ii) employ some of the basic identities to solve problems, and
(iii) solve trigonometric questions by hand and with the help of calculator.

### Angles

1. The x-y plane is divided into four quadrants as shown in diagram 1.   2. A positive angle is measured in an anti-clockwise direction from the horizontal axis to the ray (a line that runs through the origin) as in diagram 1 and diagram 2.
3. A negative angle is a clockwise sweep from the horizontal axis to the ray as in diagram 3.
This means a 3300 would look exactly like a -300.
4. An angle that is between 00 and 900 is called an acute angle. The angle depicted in diagram 1 is an acute angle. The angle in diagram 3 is a "negative" acute angle.
5. An angle that is greater than 900 but less than 1800 is called an obtuse angle. An example is depicted in diagram 2 above.

### Unit circle and trigonometry.

1. For any right angle triangle with hypotenus denoted as h the side opposite the given angle (θ in diagram at the side) is simply h sinθ and the adjacent is h cosθ
2. A unit circle is a circle with a radius one. Using what we have learned above, we can define the vertical side of the inscribed right angle triangle as sine θ and the horizontal base as the cosine θ.
3. The y = sin θ is thus the corresponding height of the vertical side for various angles θ. Similarly, y = cos θ is the corresponding base for various angles θ.
4. tan θ = sin θ / cos θ
Tan θ is not defined when cos θ is zero. For example, cos 90o = cos 270o = 0.
Thus, tan 90o and tan 270o are undefined.
 Let us refer to the right-angle triangle above. Let us allow the hypotenus to equal to 1. Thus, the opposite and adjacent are sin θ and cos θ respectively. Applying Pythagoras theorem, we obtain sin2 θ + cos 2θ = 1 Note: sin2 θ = [sin θ]2. Extras. cosecant: cscθ = 1/sinθ secant: secθ = 1/cosθ cotangent: cotθ=1/tanθ = cosθ/sinθ If we divide sin2 θ + cos 2θ = 1through by sin2θ then we obtain 1 + cot2θ = csc2θ If we divide the original expression through by cos2 then we obtain 1 + tan2θ = sec2θ
5. ### Exploration A.

Below are some simulations for trigonometric functions from http://www.ies.co.jp/math/java/trig/index.html.

1. Click on all the functions below. Increse the angle slowly and watch how the value of the function changes with the size of the angle.
 Sine Curve Cosine Curve Tangent Curve 2. Identify the range and domain for
1. y =sin θ
2. y =cos θ
3. y =tan θ

3. Complete the table below:
 θ (degree) 0o 30o 45o 60o 90o θ (radian) 0 π/6 p/2 Sin θ √0/2 = 0 √1/2 = 1/2 √2/2 = 1/√2 √3/2 √4/2 = 1 Cos θ √3/2 √1/2 = 1/2 √0/2 = 0 Tan θ 0 1 undefined
Can you identify any patterns from the table above?
In IB, you are required to memorize the table above.
4. What are the maximum and minimum values of
1. y =sin θ
2. y =cos θ
3. y =tan θ
5. Complete the following table with "+" for positive values and "-" for negative values for θ angle at different quadrants.
Summary: Since all three trigonometric functions in the first quadrant yield positive values then the quadrant is assigned symbol A for ALL. Sine is also positive in second quadrant and Tangent in third quadrant. Cosine is also positive in the fourth quadrant. For short, I refer to this summary as "CAST" or as "All Science Teachers are Crazy" which was the way I learned it.

## Solving Trigonometric Equations without Calculator

1. Solve the following equations for 0o ≤ θ ≤ 360o.
1. Example 1.
tan θ = -1 Solution: θ = arctan (-1)
θ = -45o This angle is in the fourth quadrant. The positive angle is thus 360o-45o = 315o. By CAST, we know the angle must be in the second quadrant for the value of tan to be negative (-1 in the question) (see diagram). Thus, the other angle is 180o-45o = 135o. Answers: 135o, 315o.
2. Example 2.
Solve sin ( x + 20o) = -0.5 Solution: ( x + 20o) = arcsin (-0.5)
x + 20o = -30o
Here we treat (x+20o) as an item like θ and we should not proceed to minus 20o from the right hand side. Since the value of sine is negative then then relevant angles are in third and four quadrants as drawn here.
The answers are 360o-30o = 330o and 180o+30o = 210o.
That is x + 20o = 210o or 330o.
Hence, x = 190o or 310o.

2. Example 3. Solve 2cos2θ - cos θ - 1 = 0 for 0 ≤ θ ≤ 2 π.
Solution: Since this is a quadratic equation then we will try first to factorize the equation.
(2 cos θ + 1)(cos θ - 1) = 0.
cos θ = -1/2 or cos θ = 1
When cos θ = -1/2 then the relevant angle in the CAST diagram is in the second and third quadrants where values of cosine is actually negative. Thus, θ = π - π/3 or θ = π +π/3
θ = 2π/3 or θ = 4π/3
When cos θ = 1 then θ = 0 or 2π .
Thus, the complete answer is θ = 2π/3 or θ = 4π/3 or 0 or 2π .

3. Example 4. We will resolve exaple 1 with a different domain -180o ≤ θ ≤ 180o.
In particular tan θ = -1 with -180o ≤ θ ≤ 180o.
Solution: θ = arctan (-1)
θ = -45o This angle is in the fourth quadrant. By CAST, we know the angle must be in the second quadrant for the value of tan to be negative (-1 in the question) (see diagram). Now we need to take the domain into consideration. Since the domain starts from -180o then we start from the "third quadrant" move in the anti-clockwise direction to find our first answer. We find the first angle in the "fourth quadrant" with a measure of -45o. The next angle is 180o - 45o = 135o in the second quadrant.
Thus, θ = -45o or θ = 135o.
4. Example 5. cos ( (x/2) - (π /3)) = 1/2 for 0 ≤ x ≤ 2π .
Solution.
( (x/2) - (π /3)) = arccos (1/2)
(x/2) - (π /3) = -π/3 or π/3 or 5π/3
Here we draw a cast diagram with the relevant angle in quadrant 1 and quadrant 4 where cosine values are positive. The possible relevant angle in this case is -π/3 (fourth quadrant), π/3 and 5π/3 (2π-π/3 ).
x/2 = -π/3 + (π /3) or x/2 = π/3 + (π /3) or x /2 = 5π/3 +(π /3)
x/2 = 0 or x/2 = (2π /3) or x/2 = 2π
x = 0 or x = (4π /3) or x = 4π but the last answer is outside the given domain.
Thus, x = 0 or x = (4π /3)
5. ## Solving Trigonometric Equations with simple Calculator

6. Example 6.
cos θ = 0.35 for 0o ≤ θ ≤ 360o. Solution: θ = arccos 0.35
θ ≈ 69.5o (1 d.p.)
Since the value of the cost is 0.35 which is positive then the angle must be in either the first or the fourth quadrant (see diagram). Thus, the other angle (in fourth angle) is 360o-69.5o = 290.5o.
Answers: 69.5o (1 d.p.), 290.5o (1 d.p.).

7. Example 7.
sin θ = 0.35 for 0o ≤ θ ≤ 360o. Solution: θ = arcsin 0.35
θ ≈ 20.5o (1 d.p.)
Since the value of the sin is 0.35 which is positive then the angle must be in either the first or the second quadrant (see diagram). The other angle is 180o-20.5o = 159.5o. Answers: 20.5o (1 d.p.), 159.5o (1 d.p.).

8. Example 8.
Solve cos 2θ = 0.7 for 0o ≤ x ≤ 360o Solution: 2θ= arccos 0.7
2θ ≈ 45.57o
Since we have 2θ ≈ 45.57o, we need to consider two revolutions on the unit circles. A revolution is a complete sweep of a circle. Since the value of this cos function is 0.7 which is positive then the angle is either in first or fourth quadrants. All possible angles for 2θ are
45.57o, (360o-45.57o) 314.43o, (360o+45.57o) 405.57o & (720o-45.57o) 674.43o.
Thus, θ ≈ 22.785o, 157.215o, 202.785o, & 337.215o. If you arrive at the stage where (1/2)θ = 0.5 then we need to consider only half of a revolution. (1/2)θ should be treated as an object as in the example above. All possible angles associated with this object calls (1/2)θ should be found. Then, we calculate θ by multiplying all the obtained angles by 2. Similar method is used to solve all kθ=x where k is a rational number and x a real number between -1 and 1.

9. Example 9. Solve cosθ + 5 sinθcosθ = 0 for 0o ≤ θ ≤ 360o. Solution: cos θ [ 1 + 5sinθ] = 0
cos θ = 0 or [ 1 + 5sinθ] = 0
θ = 90o, 270o or sin θ = -1/5
θ = arcsin (-0.2)
θ ≈ -11.54 (2 d.p.)o so the angle is in the fourth quadrant. The positive angle is thus 360o-11.5o = 348.46o
According to CAST, the other angle where its sin is negative will be in the third quadrant. 180o+11.54o = 191.54o.
Answers: 90o, 191.54o, 270o,& 348.46o (2 d.p.)

 However, factorising can sometime lead to a factor that has no solution. An example is (2 cosθ - 1)(cosθ -2) = 0 Thus, cos θ = 1/2 or cos θ = 2. This second factor is rejected because the maximum value of cosine is 1. Thus, there is no solution to cos θ = 2.

10. Example 10. Solve 6 sin2θ - sin θ - 1 = 0 for values of 0 ≤ θ ≤ 2π.
Solution: Let u = sin θ 6 u2 - u - 1 = 0
(3u + 1)(2u -1 ) = 0
u = -1/3 or u = 1/2
sin θ = -1/3 or sin θ = 1/2
θ = arcsin (-1/3) or θ = arcsin (1/2)
From θ = arcsin (-1/3)
θ ≈ -0.33984 (I am using 5 significant figures here so that my final answer is accurate to 3 significant figures) Thus, the relavant angles are in third and fourth quadrants.
θ ≈ π + 0.33984 or θ ≈ 2π -0.33984
θ ≈ 3.48 (3s.f.) or θ ≈ 5.94 (3s.f.)
And for θ = arcsin (1/2)
θ = π/ 6 (first quadrant as in the CAST diagram here) or θ = π - π/ 6 (second quadrant)
θ = π/ 6 or θ = 5π/ 6
Answers are π/6, 5π/6, 3.48 (3 s.f.), or 5.94 (3 s.f).

11. Example 11.Solve 2 sin2θ - 3 cosθ = 0 for values of -π ≤ θ ≤ π. Solution: 2 [1-cos2θ] - 3cosθ = 0
2 - 2 cos2θ - 3cosθ = 0
2 - 3cosθ - 2 cos2θ = 0
(2 + cosθ)(1 -2cosθ) = 0
cosθ = -2 or cosθ = 1/2
The first factor is rejected because the range of cos θ is [-1,1].
From the second factor, we obtain θ= -π/3 (fourth quadrant) or π/3 (first quadrant). Please check your given domain for such question carefully.
 We will use the identity sin2θ + cos2θ = 1 when the question involves both sine and cosine and one of these functions is squared.

## Solving trigometric equations purely with GDC and a relevant sketch.

12. Example 12.
(a) Solve 3sin x - 2cos x = 0.65 for 0 ≤ x &le 4.
 Working with GDC: Y1 = 3sin(x)-2cos(x) ; Y2 = 0.65 Make sure your [MODE] is set at radian. In this case, you want your window to be Xmin=0, Xmax=4, Xscl=1, Ymin=-5, Ymax=5,Yscl=1 Press [2nd][TRACE] for calculate. Select Intersect and find the relevant intersection points. Click here to review GDC use. One of the intersection is shown on screenshot 1. As part of your working, make sure you sketch a graph similar to diagram 4. Clearly label all axes, all curves and end points (indicated by black dots) as there is a given domain in the question. Also clearly indicate all intersections as shown in diagram 4. The answers in this case x ≈ 0.769 or 3.55 (3s.f.).

(b) Solve tan2x + 3tanx - x = 2 for -2π≤ x &le 2π.
 Working with GDC: Y1 = (tan(x))2+3tan(x)-x; Y2 = 1.23 Make sure your [MODE] is set at radian. In this case, you want your window to be Xmin=-2π, Xmax=2π, Xscl=π/2, Ymin=-8, Ymax=8,Yscl=1 Press [2nd][TRACE] for calculate. Select Intersect and find the relevant intersection points. Click here to review GDC use. One of the intersection is shown on screenshot 2. As part of your working, make sure you sketch a graph similar to diagram 5. Clearly label all axes, all curves and end points (indicated by black dots) as there is a given domain in the question. Also clearly indicate all intersections as shown in diagram 5. Notice that I have indicated the vertical asymptotes with dotted lines. The answers in this case x ≈ -1.25 or 0.449 or 1.83 or 4.03 or 4.94 (3s.f.). ( c) Solve tan(x-20o)+sinx = 3 for -180o≤ x &le 180o.

 Working with GDC: Y1 = tan(x-20)+sinx; Y2 = 3 Make sure your [MODE] is set at Degree in this case. In this case, you want your window to be Xmin=-180, Xmax=180, Xscl=60, Ymin=-8, Ymax=8,Yscl=1 Press [2nd][TRACE] for calculate. Select Intersect and find the relevant intersection points. Click here to review GDC use. One of the intersection is shown on screenshot 3. As part of your working, make sure you sketch a graph similar to diagram 6. Clearly label all axes, all curves and end points (indicated by black dots) as there is a given domain in the question. Also clearly indicate all intersections as shown in diagram 6. Notice that I have indicated the vertical asymptotes with dotted lines. The answers in this case x ≈ -84.1o or 83.5o(3s.f.).

### Double angle formulae.

1. sin 2x = 2 sinxcosx
2. cos 2x = cos2x - sin2x
cos 2x = 2cos2x -1
cos 2x = 1 - 2sin2x
3. Note: sin 2x ≠ sin x + sin x.

Example 13. Solve without a calculator. Solve sinxcosx = 1/4 for values of 0 ≤ θ ≤ 2π.
Solution: Since sin2x = 2sinxcosx then (1/2)sin2x = sinxcosx.
Thus, sinxcosx =1/4 can be written as
(1/2)sin2x = 1/4
sin2x = 1/2
2x = arcsin (1/2)
2x = π/ 6 (first quadrant as in the CAST diagram here) or θ = π - π/ 6 (second quadrant)
2x = π/ 6 or 5π/ 6
x = π/ 12 or 5π/ 12

Example 14.
Simplify [2cos3x -sin2xsinx]/cos x.
Solution:
[2cos3x -sin2xsinx]/cos x = [2cosxcos2x -(2sinxcosx)sinx]/cos x
[2cos3x -sin2xsinx]/cos x.= { 2cosx[cos2x -(sinx)sinx] }/cos x
[2cos3x -sin2xsinx]/cos x.= 2[cos2x -sin2x ]
[2cos3x -sin2xsinx]/cos x.= 2cos2x.

Example 15. Given that x is acute and that sin x = 1/3,
evaluate (i) tan x, (ii) cosx, (iii) cos2x (iv) sin2x and (v) tan2x.
Solution: For this sort of question it is based to sketch a right angle triangle as shown here with the relevant information. Since the angle x is acute then the triagle is in the first quadrant. The base is calculated as √ (32 - 12) = √(8) using Pythagoras Theorem. Fill this information into the diagram as shown here.
Thus, (i) tan x = 1/√(8) ------reading off the completed diagram.
(ii) cosx = √(8)/3 ------reading off the completed diagram.

(iii) cos2x = cos2x -sin2x
cos2x = (√(8)/3)2 -(1/3)2
cos2x = (8/9) - (1/9)
cos2x = 7/9

(iv) sin2x = 2sinxcosx
sin2x = 2(1/3)(√(8)/3 )
sin2x = 2√(8)/9

(v) tan2x = sin2x/cos2x
tan2x = [ 2√(8)/9 ] / (7/9 )
tan2x = 2√(8)/7

Example 16.
Given than x is an obtuse angle and cos θ = -3/5 evaluate (i) sin2θ and (ii) cos2θ.
Solution: For this sort of question it is based to sketch a right angle triangle as shown here with the relevant information. Since the angle θ is obtuse then the triagle is actually in the second quadrant. Actually the right angle triangle in the second quadrant is with θ* but sin θ = sin θ* and cos θ= cos θ* and tanθ=tanθ*. Thus, very often we dispense with θ* and just label the angle in the second quadrant as θ. The height (opposite) is calculated as √ (52 - 32) = √(16) using Pythagoras Theorem. Fill this information into the diagram as shown here.
(i) sin2θ= 2sinθcosθ
From the completed diagram, we have
sin2θ= 2(4/5)(-3/5)
sin2θ= -24/25
(ii) cos2θ= 2cos2x-1
cos2θ= 2(-3/5)2 -1
cos2θ= 2(9/25) -1
cos2θ= (18-25)/25
cos2θ= -7/25.
 Let us review some basic concepts before we show that the two double angle formulae above are true. For any right-angle triangle as in this diagram the opposite is hsinθ and the adjacent (the base) is hcosθ. Let us consider a diagram as in the left. Sin 2x is thus AC which is the opposite to right-angle triangle OAC. Sin 2x = AC = AB + BC, but ED = BC = AB + ED, Note that OE is the base of right-angle triangle OAE. Thus, OE is just cosx. AE which is the opposite to right-angle triangle OAE is hence sinx. We can also easily establish angle BAE to be x using "the butterfly principle for angles." BAE is also a right- angle triangle. Thus, AB is the base of a right-angle triangle BAE with hypotenus AE which is sinx. AB = (sinx)cosx. OED is also a right-angle triangle with hypotenus OE which is cosx as calculated above. ED is the opposite to OED and thus (cosx)sinx. Sin 2x = AB + ED = (sinx)cosx + (cosx)sinx = 2 sinxcosx Similarly, 2cosx = OC which is OD -CD. You can use the method above to show that cos 2x = cos2x - sin2x