C1: Chapters 8 & 10
Trigonometry
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
www.drfrostmaths.com
Last modified: 1st September 2015
Sin Graph
What does it look like?
-360
-270
-180
-90
?
90
180
270
360
Sin Graph
What do the following graphs look like?
-360
-270
-180
-90
90
180
270
Suppose we know that sin(30) = 0.5. By thinking about symmetry in the graph,
how could we work out:
sin(150) = 0.5?
sin(-30) = -0.5?
sin(210) = -0.5
?
360
Cos Graph
What do the following graphs look like?
-360
-270
-180
-90
?
90
180
270
360
Cos Graph
What does it look like?
-360
-270
-180
-90
90
180
270
Suppose we know that cos(60) = 0.5. By thinking about symmetry in the graph,
how could we work out:
cos(120) = -0.5
?
cos(-60) = 0.5?
cos(240) = -0.5
?
360
Tan Graph
What does it look like?
-360
-270
-180
-90
?
90
180
270
360
Tan Graph
What does it look like?
-360
-270
-180
-90
90
180
270
Suppose we know that tan(30) = 1/√3. By thinking about symmetry in the
graph, how could we work out:
tan(-30) = -1/√3
?
tan(150) = -1/√3
?
360
Laws of Trigonometric Functions
We saw for example sin(30) = sin(150) and cos(30) = cos(330). It’s also easy to
see by looking at the graphs that cos(40) = sin(50). What laws does this give us?
!
sin(x) = sin(180? – x)
cos(x) = cos(360? – x)
sin and cos repeat every 360?
tan repeats every 180?
sin(x) = cos(90?– x)
Bro Tip: These 5 things are pretty much the only
thing you need to learn from this Chapter!
Practice
Find all the values in the range 0 to 360 for which sin/cos/tan
will be the same.
1
?
sin(30) = sin(150)
?
cos(70) = cos(290)
2
?
cos(30) = cos(330)
3
sin(-10) = sin(190)
? = sin(350)
?
4
cos(-40) = cos(40)
? = cos (320)
5
sin(20) = cos(70)
?
6
sin(80) = sin(100)
?
7
8
11
cos(-25) = cos(25)
? = cos(335)
?
9
10
cos(80) = sin(10)
?
?
sin(15) = sin(165)
sin(-60) = sin(240)
? = sin(300)
?
12
?
tan(80) = sin(260)
Dr Frost’s technique for remembering trig values
(once described by a KGS tutee of mine as ‘the Holy Grail of teaching’)
I literally picture this table in my head when
I’m trying to remember my values.
0
45
All the values in this square are
over 2.
90
30
60
sin
0
_1_
√2
1
_1_
2
√3
2
cos
1
_1_
√2
0
√3
2
_1_
2
tan
0
1
_1_
√3
√3
All the surds in this block are √2
The diagonals
starting from the
top left are
rational. The
other values in
the square are
not.
All the surds in this block are √3
I remember that out of tan(30) and tan(60), one is 1/√3 and
the other √3. However, by considering the graph of tan, clearly
tan(30) < tan(60), so tan(30) must be the smaller one, 1/√3
Practice
0
45
90
30
60
sin
0
?
_1_
?
√2
1
?
_1_
?2
√3
?2
cos
?1
_1_
?
√2
?0
√3
?2
_1_
?2
tan
?
0
?
_1_
?
√3
√3
1
?
‘Magic Triangles’
You can easily work out sin(45), cos(45), sin(30), tan(30) etc. if
you were ever to forget.
?
√2
1
45
1
_1_
sin(45) = ?
√2
30
2
?
√3
60
1?
_√3_
cos(30) = ?
2
Angle quadrants
Given that sin α = 2/5, and that α is obtuse, find
(without a calculator) the exact value of cos α.
21
?
cos 𝜙 =
5
5?
2?
ф
?
√21
Imagine working instead with the
acute angle ф such that sin ф = 2/5
Therefore thinking about the
graph of cos:
21
?
cos 𝛼 = −
5
Since by thinking about the
graph of cos, we can see that
cos 180 − 𝑥 = − cos 𝑥
Angle quadrants
1
Given that tan α = 5/12, and that α is acute, find
the exact value of sin α and cos α.
?
?
sin α = 5/13,
cos α = 12/13
2
Given that cos α = -3/5, and that α is obtuse, find
the exact value of sin α and tan α.
? tan α = -4/3
?
sin α = 4/5,
3
Given that tan α = -√3, and that α is reflex, find the
if tan α is negative, then is our reflex
exact value of sin α and tan α. Hint:
angle between 180 and 270, or 270 and 360?
?
sin α = -√3/2,
cos α = 1/2?
Onwards to Chapter 10...
The only 2 identities you need this chapter...
r
? 
y = r sin

? 
x = r cos
1
2
sin  = y/r and cos  = x/r
and tan  = y/x
Pythagoras gives
you...
sin  = tan? 
cos 
? 2=1
sin2  + cos
Examples of use
1
Simplify sin2 3 + cos2 3
=1
?
Simplify 5 – 5sin2 
= 5cos2?
3
Given that p = 3 cos 
and q = 2 sin , show that
4p2 + 9q2 = 36.
2
Show that:
cos 4 𝜃 − sin4 𝜃
2
≡
1
−
tan
𝜃
2
cos 𝜃
This box is
intentionally left
blank.
Solving Trigonometric Equations
Edexcel May 2013 ()
 = 123.44, ?176.57
Bro Tips for solving:
1. If 0 ≤  < 180, then what range does 2 – 30 have?
2. Immediately after the point at which you do sin-1 of both sides, list out the
other possible angles in the above adjusted range. Recall that
sin(x) = sin(180-x) and that sin repeats every 360.
Solving Trigonometric Equations
Edexcel June 2010
a
b
tan  = 0.4?
tan 2x = 0.4
0 ≤ 2x < 720
2x = 21.801, 201.801,? 381.801, 561.801,
x = 10.9, 100.9, 190.9, 280.9
Solving Trigonometric Equations
Edexcel Jan 2010
(2sin x – 1)(sin x + 3) = 0
sin x = 0.5 or? sin x = -3
x = 30°, 150°
Bro Tip: In general, when you have sin and cos, and
one is squared, change the squared term to be
consistent with the other.
Exercises
Edexcel Jan 2009
Edexcel Jun 2009
?
?
𝜃 = −45°, 135°, 23.6°, 156.4° 𝑥 = 41.4°, 318.6°
Edexcel Jun 2008
284.5,
435.5,
?
644.5
Edexcel Jan 2008
65, 155
?
40 80 160 ?
200 280 320
Edexcel Jan 2013
41.2, 85.5,
161.2
?
θ = 230.785, 309.23152,
50.8, 129.2
?
Things to remember
1
If you square root both sides, don’t forget the . You’ll probably lose
2 marks otherwise.
sin2 3𝑥 =
1
2
⇒ sin 3𝑥 = ±
1
2
2
Don’t forget solutions. If you have sin, you’ll always be able to get an extra
solution by using 180 – x. If you have cos you can get an extra one using 360-x.
3
Remember that tan repeats every 180, sin/cos every 360.
4
If you had sin2x and cos x, you’d replace the sin2 x with 1 – cos2 x.
You’d then have a quadratic in terms of cos x which you can factorise.
5
Check whether the question expects you to give your answers in degrees or
radians. If they say 0 ≤ 𝑥 ≤ 𝜋, then clearly they want radians.