Radians
What is a radian?
radius
arc=radius
o
Angle = 1
radian
radius
Why do mathematicians use radians instead of degrees?
Exact angle measure. It allows us to give exact numbers
rather than use decimals which are (sometimes)
inaccurate.
How many times does the radius divide into the
circumference?
There are 2
1 radian =
radians in a circle.
= 57.3o
Radians
Degrees
Degrees
Radians
Convert each angle in
radians to degrees.
Convert each angle in
degrees to radians.
1. 2c
114.6o
5c
286.5o
2.
540o
3.
120o
90o
4.
180o
2.
3. 3
4.
5.
c
c
c
240o
1. 65o
5.
200o
1.13c
3.49c
330o
Give 3,4,5 as exact numbers.
Two important formula using radians
Length of an arc using radians
Area of a sector using radians
A
Using the formulae
1. Find the length of the minor arc
AB.
3. Find the area of the minor sector
AOB.
AB  1.2  4.8
AB  5.76cm
3.5 2  0.7
AOB 
2
2
AOB  4.29cm
 the length of the minor
2. Given that
4. Given that the shaded area is 48
arc AB is 13.8 cm, find the missing
cm2, find the radius of the circle.
angle AOB.

6 13.8
  2.3
c
r 1.5
 48
2
2
r  64cm
2
r  8cm
The Unit Circle
The table so far…
General Angles – Unit Circle
x + y =1
2
2
(0,1)
(-1, 0)
(1, 0)
(0, -1)
Consider the circle shown
opposite.
The Unit Circle
y
Make a right angled triangle.
An expression for x:
(x,y)
1
y
x
x
An expression for y:
And now use tan:
General Angles – Unit Circle
x + y =1
2
2
(0,1)
(x, y)
θ
(-1, 0)
(1, 0)
(0, -1)
General Angles – Unit Circle
x + y =1
2
2
(0,1)
(-x, y)
(180 - q )
(x, y)
θ
θ
(-1, 0)
(1, 0)
(0, -1)
General Angles – Unit Circle
x + y =1
2
2
(0,1)
(x, y)
(180 + q )
θ
(-1, 0)
θ
(1, 0)
(-x,-y)
(0, -1)
General Angles – Unit Circle
x + y =1
2
2
(0,1)
(360 - q )
(x, y)
(-1, 0)
θ
θ
(1, 0)
(x,-y)
(0, -1)
Reference Angles
Quad II
θ’ = 180° – θ
θ’ = π – θ
θ’ = 180°+θ
θ’ = π +θ
Quad III
Quad I
θ’ = θ
θ’ = θ
θ’ = 360° – θ
θ’ = 2 π – θ
Quad IV
General Angles – Unit Circle
x + y =1
2
2
(0,1)
(-x, y)
(-1, 0)
θ
θ
θ
θ
(-x,-y)
(cosq,sinq )
(x, y)
(1, 0)
(x,-y)
(0, -1)
All Students Take Calculus.
Quad II
Quad I
cos(A)<0
cos(A)>0
sin(A)>0
sin(A)>0
tan(A)<0
tan(A)>0
cos(A)<0
cos(A)>0
sin(A)<0
sin(A)<0
tan(A)>0
tan(A)<0
Quad III
Quad IV
General Angles – Unit Circle
a = reference
Quadrant
sinq = sin a
θ in Quadrant I & II
sinq = -sin a
θ in Quadrant III & IV
cosq = cosa
θ in Quadrant I & IV
cosq = -cosa
θ in Quadrant II & III
tanq = tan a
θ in Quadrant I & III
tanq = -tan a
θ in Quadrant II & IV
General Angles – Unit Circle
Example
4
Given that cosq = and
5
Evaluate
tanq
and
180 < q < 270
o
o
sinq
Quadrant III
5
3
tanq =
4
3
θ
4
3
sin q = 5
General Angles – Unit Circle
x + y =1
2
2
(0,1)
(x, y)
θ
(-1, 0)
(1, 0)
(0, -1)
Other Identities
(x, y)
cos(-q ) = cosq
sin(-q ) = -sinq
tan(-q ) = -tanq
θ
-θ
(x,-y)
Special Triangles
450
300
√2
2
1
√3
450
600
1
2

cos( 45 ) 
2
2

sin( 45 ) 
2
tan( 45  )  1
0
60
1
3
cos(30 ) 
2
1

sin( 30 ) 
2
3

tan( 30 ) 
3

1
cos(60 ) 
2
3

sin( 60 ) 
2
tan( 60  )  3

Trig. ratios in the 1st
quadrant (1)
1. Find the missing side, giving your
answer as a surd.
2. Find an expression for sin .
2
3. Find an expression for cos .
1
4. Find an expression for tan .
5. Find the value of the angle, using
any of 2-4.
6. Fill in the appropriate column of
your table
Trig. ratios in the 1st
quadrant (2)
1. Find the missing side, giving your
answer as a surd.
2. Find an expression for sin .
1
1
3. Find an expression for cos .
4. Find an expression for tan .
5. Find the value of the angle, using
any of 2-4.
6. Fill in the appropriate column of
your table
Trig. ratios in the 1st
quadrant (3)
1. Find the missing side, giving your
answer as a surd.
2. Find an expression for sin .
2
1
3. Find an expression for cos .
4. Find an expression for tan .
5. Find the value of the angle, using
any of 2-4.
6. Fill in the appropriate column of
your table
Trig. ratios in the 2nd
quadrant
1. Find the missing side, giving your
answer as a surd.
2. Find an expression for sin .
2
-1
3. Find an expression for cos .
4. Find an expression for tan .
5. Find the value of the angle, using
any of 2-4.
6. Fill in the appropriate column of
your table
Use your answers from previous slides to fill in all the angles trig.
values in the second quadrant.
Trig. ratios in the 3rd
quadrant
1. Find the missing side, giving your
answer as a surd.
2. Find an expression for sin .
3. Find an expression for cos .
-1
4. Find an expression for tan .
-1
5. Find the value of the angle, using
any of 2-4.
6. Fill in the appropriate column of
your table
Use your answers from previous slides to fill in all the angles trig.
values in the third quadrant.
Trig. ratios in the 4th
quadrant
1. Find the missing side, giving your
answer as a surd.
2. Find an expression for sin .
3. Find an expression for cos .
4. Find an expression for tan .
2
-1
5. Find the value of the angle, using
any of 2-4.
6. Fill in the appropriate column of
your table
Use your answers from previous slides to fill in all the angles trig.
values in the second quadrant.
Investigation – using…
The completed table
All you need to remember
The Sine and Cosine rules
Area of a Triangle
Deriving the cosine rule
Find a2 in terms of b, c and A.
Make 2 right-angled triangles by
drawing a perpendicular from B.
Let AM=x.
Make 2 different equations from BM.
BM  c  x
2
2
x
2
BM  a  (b  x)
2
2
2
2
2

Using triangle ABM:
BM 2  a 2  b 2  2bx  x 2
 c  x  a  b  2bx  x
2
2
2
a  b  c  2bx
2
2
b-x
2
BM  a  b  2bx  x
2
M
2
2
x
cos A   x  c cos A
c
a  b  c  2bc cos A
2
2

2
2
When do we use the cosine rule?
a  b  c  2bc cos A
2
2
2
b c a
cos A 
2bc
2

2
2
When we have 3 sides and
 an unknown angle or 2
sides with the angle between them and we are looking
for the third side.
Using the cosine rule
Find the value of x in the triangle
above.
a  b  c  2bc
2
2
2
2
x  41.34
x  6.43 cm
b c a
cos A 
2bc
2
x  7.5  10  2  7.5 10cos 40
2
Find the value of the missing angle,
theta, in the triangle above.
2
2

2
2
7.5  13 10
cos A 
2 13  7.5
cos A  0.6423
A  50
2
2
2
Deriving the sine rule
Find sinC and sinA.
Make 2 right-angled triangles by
drawing a perpendicular from B.
BM
sin A 
c
BM
sin C 
a
M
 csin A  asin C
sin A  sin C

a
c

a
c
or

sin A sin C
sin A sin B sin C


a
b
c

a
b
c


sin A sin B sin C
When do we use the sine rule?
sin A sin B sin C


a
b
c

a
b
c


sin A sin B sin C

When we have 2 sides, 1 angle and an unknown angle
or 2 angles, 1 side and an unknown side.
Using the sine rule
Find the value of x in the triangle
above.
sin  sin 29

8.5
5.1
8.5  sin 29
sin  
5.1
180  (54  81)  45
x
9.5

sin 45 sin 81
9.5  sin 45
x
sin 81
x  6.8 cm
Find the value(s) of the missing
angle, theta, in the triangle above.

sin   0.808

  53.9,126.1

A problem in context
A yacht sails from a port, P, on a bearing of 100o. It sails for 3.2 km and
meets with ship, S. It then sails on a bearing of 330o towards a buoy, B, 6
km away. At the buoy the yacht turns and sails back to the Port.
Find the bearing and distance from the buoy to the port.
Draw a diagram.
150o
BP 2  3.2 2  6 2  (2  3.2  6cos50)
BP 2  21.557
BP  4.64 km
Angle inside the triangle at B.
sin B sin 50

3.2 4.6429
sin B  0.528
B  31.87,148.13
(ignore the second value as it will fit in the triangle)
Bearing:150  32 182
50o
The Unit Circle
- Pythagoras
Make a right angled triangle.
Make up an equation using
Pythagoras:
Using these relationships
There are 2 possible answers for
cosx. As x is obtuse, the angle is in
the 2nd quadrant so,
sin x
b) Use: tan x 
cos x
1. Given that
obtuse, find,
a) cosx
b) tanx
a) Use:
and x is

Simplify the following:
1.
2.
3.
4.
5.
Questions – Proving Identities
Prove
1
− cos(𝑥)
cos(𝑥)
LHS  RHS
or
RHS  LHS
Examples
More Examples
Compound angle identities



Compound angles -Summary
sin A  B  sin Acos B  cos Asin B
cosA  B  cos Acos B sin Asin B
tan A  tan B
tanA  B 
1 tan Atan B
Example
1
2
Given that sin a = and cosb = ,
3
3
and a+b is acute find,
a) sin(a+b),
b) cos(a - b),
2 2 2  1  5 
cosA  B  
    
 3 3  3  3 
4 2 5
cosA  B 
9
c) tan(a - b).
sin 2 a  cos 2 a  1
sin 2 b  cos 2 b  1
1
 cos2 a  1
9
8
cos2 a 
9
2 2
cos a  
3
sin 2 b 
4
1
9
5
2
sin b 
9
5
sin b  
3

a) sin A  B  sin Acos B  cos A sin B
1 2  2 2  5 
sin A  B     
 
3 3   3  3 
2  2 10
sin A  B 
9
b) cosA  B  cos A cos B  sin A sin B
tan A  tan B
c) tanA  B 
1 tan A tan B
1
5
tan a 
,tan b 
2
2 2
 1
5 



2 
2 2
tanA  B 
 1
5 
1 


2 
2 2
2  2 10
tanA  B 
4 2 5
Example 2
Find sin(a+b) in the diagram.
sin A  B  sin Acos B  cos Asin B
missing side (x)= 8 (pythagoras)
missing side (y)=
17 (pythagoras)
6
17
sin a = ,sin b 
10
9
8
8
cos a  ,cosb 
10
9
 6 8   8  17 
sin( a  b)      

10 9  10  9 
48  8 17
24 + 4 17
sin( a  b) 

90
45
x
y


Trigonometric equations
1
𝑜
cos 𝑥 = 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 360
2
What do you
need to consider
to solve this?
Trigonometric equations
Using a Graph
Using a GDC
Using a GDC
Using Algebra - Remember
Special Triangles
450
300
√2
2
1
√3
450
600
1
2

cos( 45 ) 
2
2

sin( 45 ) 
2
tan( 45  )  1
0
60
1
3
cos(30 ) 
2
1

sin( 30 ) 
2
3

tan( 30 ) 
3

1
cos(60 ) 
2
3

sin( 60 ) 
2
tan( 60  )  3

Solving Trigonometric Equations
1.
tan x  1  0
3 7
x
,
4 4
Using Algebra
Using Algebra
Solving Trigonometric Equations
Solve:
2cos x 1  0
Step 1: Isosolate cos x using algebraic skills.
2cos x  1
cos x  1
2
Step 2: Determine in which quadrants cosine is positive. Use the inverse
function to assist by finding the angle in Quad I first. Then use that angle
as the reference angle for the other quadrant(s).
QI
x
QIV
 5
3
,
3
Note: cosine is positive in
Quad I and Quad IV.
Note: The reference angle is /3.
Solving Trigonometric Equations
Solve:
Step 1:
tan x  1  0
2
tan x  1
2
tan x   1
tan x  1
2
Step 2:
Q1
x
QIV
QIII
QII
 3 5 7
,
4 4
,
4
,
4
Note: Since there is a  , all four quadrants
hold a solution with /4 being the reference
angle.
Using Algebra
Using Algebra
Solving Trigonometric Equations
Solve:
2sin 2 x  sin x  1  0
0 ≤ 𝑥 ≤ 2𝜋
Factor the quadratic equation.
 2sin x 1sin x 1  0
2sin x  1  0 or sin x  1  0
1
sin x  
2
7 11
x
,
6
6
sin x  1
x

2
Set each factor equal to zero.
Solve for sin x
Determine the correct
quadrants for the solution(s).
Solving Trigonometric Equations
Solve:
2sin 2 x  3cos x  3  0
0 ≤ 𝑥 ≤ 2𝜋
2 1  cos 2 x   3cos x  3  0
Replace sin2x with 1-cos2x
2  2 cos x  3cos x  3  0
Distribute
2 cos 2 x  3cos x  1  0
Combine like terms.
2 cos 2 x  3cos x  1  0
Multiply through by – 1.
 2cos x 1 cos x 1  0
Factor.
2
2 cos x  1  0 or cos x  1  0
1
cos x 
2
x
cos x  1
 5
3
,
3
x0
Set each factor equal to zero.
Solve for cos x.
Determine the solution(s).