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
 The xy plane is divided into four quadrants as shown in diagram 1.
 A positive angle is measured in an anticlockwise direction from the horizontal axis to the ray (a line that runs through the origin) as in diagram 1 and diagram 2.
 A negative angle is a clockwise sweep from the horizontal axis to the ray as in diagram 3.
This means a 330^{0} would look exactly like a 30^{0}.
 An angle that is between 0^{0} and 90^{0} 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.
 An angle that is greater than 90^{0} but less than 180^{0} is called an obtuse angle. An example is depicted in diagram 2 above.
Unit circle and trigonometry.
 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
 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.
 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.
 tan θ = sin θ / cos θ
Tan θ is not defined when cos θ is zero. For example, cos 90^{o}
= cos 270^{o} = 0.
Thus, tan 90^{o} and tan 270^{o} are undefined.
Let us refer to the rightangle 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
sin^{2} θ + cos ^{2}θ = 1
Note: sin^{2} θ = [sin θ]^{2}.

Extras.
cosecant: cscθ = 1/sinθ
secant: secθ = 1/cosθ
cotangent: cotθ=1/tanθ = cosθ/sinθ
If we divide
sin^{2} θ + cos ^{2}θ = 1through by sin^{2}θ then we obtain
1 + cot^{2}θ = csc^{2}θ
If we divide the original expression through by cos^{2} then we obtain
1 + tan^{2}θ = sec^{2}θ


Exploration A.
Below are some simulations for trigonometric functions from http://www.ies.co.jp/math/java/trig/index.html.
 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.
 Identify the range and domain for
 y =sin θ
 y =cos θ
 y =tan θ
 Complete the table below:
θ (degree) 
0^{o} 
30^{o} 
45^{o} 
60^{o} 
90^{o} 
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.
 What are the maximum and minimum values of
 y =sin θ
 y =cos θ
 y =tan θ
 Complete the following table with "+" for positive values and
"" for negative values for θ angle at different quadrants.

1st quadrant 
2nd quadrant 
3rd quadrant 
4th quadrant 
Sin θ
 + 



Cosθ
 + 



Tan θ
 + 



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
 Solve the following equations for 0^{o} ≤ θ ≤ 360^{o}.
 Example 1.
tan θ = 1
Solution: θ = arctan (1)
θ = 45^{o} This angle is in the fourth quadrant. The positive
angle is thus 360^{o}45^{o} = 315^{o}. 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 180^{o}45^{o} = 135^{o}. Answers: 135^{o},
315^{o}.
 Example 2.
Solve sin ( x + 20^{o}) = 0.5
Solution: ( x + 20^{o})
= arcsin (0.5)
x + 20^{o} = 30^{o}
Here we treat (x+20^{o}) as an item like θ and we should
not proceed to minus 20^{o} 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 360^{o}30^{o} = 330^{o} and 180^{o}+30^{o}
= 210^{o}.
That is x + 20^{o} = 210^{o} or 330^{o}.
Hence, x = 190^{o} or 310^{o}.
 Example 3.
Solve
2cos^{2}θ  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 .
Please remember when the question is framed in radian
then provide answers in radian.
If the question is framed in degree then provide answers in degree.

 Example 4.
We will resolve exaple 1 with
a different domain 180^{o} ≤ θ ≤ 180^{o}.
In particular tan θ = 1 with 180^{o} ≤ θ ≤ 180^{o}.
Solution: θ = arctan (1)
θ = 45^{o} 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 180^{o} then we start from the "third
quadrant" move in the anticlockwise direction to find our first answer.
We find the first angle in the "fourth quadrant" with a measure
of 45^{o}. The next angle is 180^{o}  45^{o} = 135^{o}
in the second quadrant.
Thus, θ = 45^{o} or θ = 135^{o}.
 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 (2ππ/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)
Solving Trigonometric Equations with simple Calculator
 Example 6.
cos θ = 0.35 for 0^{o} ≤ θ ≤ 360^{o}.
Solution: θ = arccos^{} 0.35
θ ≈ 69.5^{o} (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 360^{o}69.5^{o} = 290.5^{o}.
Answers: 69.5^{o} (1 d.p.), 290.5^{o} (1 d.p.).
 Example 7.
sin θ = 0.35 for 0^{o} ≤ θ ≤ 360^{o}.
Solution: θ = arcsin 0.35
θ ≈ 20.5^{o} (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 180^{o}20.5^{o} = 159.5^{o}. Answers: 20.5^{o}
(1 d.p.), 159.5^{o} (1 d.p.).
 Example 8.
Solve cos 2θ = 0.7 for 0^{o} ≤ x ≤ 360^{o}
Solution: 2θ= arccos 0.7
2θ ≈ 45.57^{o}
Since we have 2θ ≈ 45.57^{o}, 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
45.57^{o}, (360^{o}45.57^{o}) 314.43^{o},
(360^{o}+45.57^{o}) 405.57^{o} & (720^{o}45.57^{o})
674.43^{o}.
Thus, θ ≈ 22.785^{o}, 157.215^{o}, 202.785^{o},
& 337.215^{o}. 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.
 Example 9. Solve cosθ + 5 sinθcosθ = 0 for 0^{o}
≤ θ ≤ 360^{o}.
Solution: cos θ [ 1 + 5sinθ] = 0
cos θ = 0 or [ 1 + 5sinθ] = 0
θ = 90^{o}, 270^{o} 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 360^{o}11.5^{o} = 348.46^{o}
According to CAST, the other angle where its sin is negative will be in the
third quadrant. 180^{o}+11.54^{o} = 191.54^{o}.
Answers: 90^{o}, 191.54^{o}, 270^{o},& 348.46^{o}
(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. 
 Example 10. Solve 6 sin^{2}θ  sin θ  1 = 0
for values of 0 ≤ θ ≤ 2π.
Solution: Let u = sin θ
6
u^{2}  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).
 Example 11.Solve 2 sin^{2}θ  3 cosθ = 0 for
values of π ≤ θ ≤ π.
Solution:
2 [1cos^{2}θ]  3cosθ = 0
2  2 cos^{2}θ  3cosθ = 0
2  3cosθ  2 cos^{2}θ = 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 sin^{2}θ
+ cos^{2}θ = 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.
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 tan^{2}x + 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(x20^{o})+sinx = 3 for 180^{o}≤
x &le 180^{o}.
Working with GDC: Y1 = tan(x20)+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 ≈ 84.1^{o} or 83.5^{o}(3s.f.).


Double angle formulae.
 sin 2x = 2 sinxcosx
 cos 2x = cos^{2}x  sin^{2}x
cos 2x = 2cos^{2}x 1
cos 2x = 1  2sin^{2}x
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 [2cos^{3}x sin2xsinx]/cos x.
Solution:
[2cos^{3}x sin2xsinx]/cos
x = [2cosxcos^{2}x (2sinxcosx)sinx]/cos
x
[2cos^{3}x sin2xsinx]/cos x.= { 2cosx[cos^{2}x
(sinx)sinx] }/cos x
[2cos^{3}x sin2xsinx]/cos x.= 2[cos^{2}x sin^{2}x
]
[2cos^{3}x 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 √ (3^{2}
 1^{2}) = √(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 = cos^{2}x sin^{2}x
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 √ (5^{2}  3^{2}) = √(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= 2cos^{2}x1
cos2θ= 2(3/5)^{2} 1
cos2θ= 2(9/25) 1
cos2θ= (1825)/25
cos2θ= 7/25.
Let us review some basic concepts before we show that the two double
angle formulae above are true. For any rightangle 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 rightangle
triangle OAC.
Sin 2x = AC
= AB + BC, but ED = BC
= AB + ED,
Note that OE is the base of rightangle triangle OAE. Thus, OE is just
cosx. AE which is the opposite to rightangle 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 rightangle triangle BAE with hypotenus AE which
is sinx. AB = (sinx)cosx.
OED is also a rightangle 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 = cos^{2}x  sin^{2}x

Exercises Without Calculator
 Exercises With Calculator