C2: Trigonometrical Equations
Learning Objective: to be able to
solve simple trigonometrical
equations in a given range
Starter:
1. Calculate the area of the shaded
segment
A
O
π/4
B
15 cm
The three trigonometric ratios
The three trigonometric ratios, sine, cosine and tangent, can
be defined using the ratios of the sides of a right-angled
triangle as follows:
O
P
P
O
S
I
T
E
H
Y
P
O
T
E
N
U
S
E
θ
Opposite
Sin θ =
Hypotenuse
SOH
Adjacent
Cos θ =
Hypotenuse
CAH
Tan θ =
Opposite
Adjacent
TOA
ADJACENT
Remember: S O H C A H
TOA
The sine, cosine and tangent of any angle
These definitions are limited because a right-angled triangle
cannot contain any angles greater than 90°.
To extend the three trigonometric ratios to include angles
greater than 90° and less than 0° we consider the rotation of a
straight line OP of fixed length r about the origin O of a
coordinate grid.
Angles are then measured
anticlockwise from the
positive x-axis.
y
P(x, y)
θ
r
α
O
x
For any angle θ there is an
associated acute angle α
between the line OP and
the x-axis.
The graph of y = sin θ
The graph of y = cos θ
The graph of y = tan θ
Remember CAST
We can use CAST to remember in which quadrant each of
the three ratios is positive.
2nd quadrant
1st quadrant
S
Sine is positive
A
All are positive
3rd quadrant
4th quadrant
T
Tangent is positive
C
Cosine is positive
Task 1:
Write each of the following as trigonometric ratios of
positive acute angles:
•
sin 260°
•
cos 140°
•
tan 185°
•
tan 355°
•
cos 137°
•
sin 414°
•
sin (-194)°
•
cos (-336)°
•
tan 396°
•
tan 148°
Sin, cos and tan of 45°
A right-angled isosceles triangle has two acute angles of 45°.
Suppose the equal sides are of
unit length.
45°
2
1
Using Pythagoras’ theorem:
2
2
The hypotenuse  1  1
45°
 2
1
We can use this triangle to write exact values for sin, cos and
tan 45°:
1
sin 45° =
2
1
cos 45° =
2
tan 45° = 1
Sin, cos and tan of 30°
Suppose we have an equilateral triangle of side length 2.
60°
30°
2
2
3
60°
1
If we cut the triangle in half then we have
a right-angled triangle with acute angles
of 30° and 60°.
Using Pythagoras’ theorem:
2
2

2

1
60° The height of the triangle
 3
2
We can use this triangle to write exact values for sin, cos and
tan 30°:
1
sin 30° =
2
3
cos 30° =
2
1
tan 30° =
3
Sin, cos and tan of 60°
Suppose we have an equilateral triangle of side length 2.
60°
30°
2
3
60°
1
2
If we cut the triangle in half then we have
a right-angled triangle with acute angles
of 30° and 60°.
Using Pythagoras’ theorem:
2
2

2

1
60° The height of the triangle
 3
2
We can also use this triangle to write exact values for sin, cos
and tan 60°:
3
sin 60° =
2
1
cos 60° =
2
tan 60° = 3
Sin, cos and tan of 30°, 45° and 60°
Write the following ratios exactly:
1) cos 300° =
1
2
2) tan 315° =
–1
3) tan 240° =
3
4) sin –330° =
1
2
5) cos –30° =
3
2
6) tan –135° =
1
7) sin 210° =
1
2
8) cos 315° =
1
2
Task 2 :
Write down the value of the following leaving your answers
in terms of surds where appropriate:
1. sin 120°
2. cos 150°
3. tan 225°
4. cos 300°
5. sin (-30)°
6. cos (-120)°
7. sin 240°
8. sin 420°
9. cos 315°
Equations of the form sin θ = k
Equations of the form sin θ = k, where –1 ≤ k ≤ 1, have an
infinite number of solutions.
If we use a calculator to find sin–1 k the calculator will give a
value for θ between –90° and 90°.
There is one and only one solution in this range.
This is called the principal solution of sin θ = k.
Other solutions in a given range can be found by considering
the unit circle.
For example:
Solve sin θ = 0.7 for –360° < θ < 360°.
sin-1 0.7 = 44.4° (to 1 d.p.)
Equations of the form sin θ = k
We solve sin θ = 0.7 for –360° < θ < 360° by considering
angles in the first and second quadrants of a unit circle where
the sine ratio is positive.
Start by sketching the principal solution 44.4° in the first
quadrant.
Next, sketch the associated acute
angle in the second quadrant.
–224.4°
–315.6°
135.6°
44.4°
Moving anticlockwise from the
x-axis gives the second solution:
180° – 44.4° = 135.6°
Moving clockwise from the
x-axis gives the third and fourth
solutions:
–(180° + 44.4°) = –224.4°
–(360° – 44.4°) = –315.6°
Examples:
1. Solve for 0 ≤ x ≤ 360°, cos x = ½
2. Solve for 0 ≤ x ≤ 360°, sin x = - 0.685
Task 3: Solve for 0 ≤ x ≤ 360°
1.
2.
3.
4.
5.
6.
Sin x = 0.6
Cos x = 0.8
Tan x = 0.4
Sin x = -0.8
Cos x = -0.6
Tan x = -0.5
Task 4: Solve for 0 ≤ x ≤ 2π
1.
2.
3.
4.
5.
6.
sin x = 1/2
cos x = 1/ √2
tan x = - √3
sin x = √3 / 2
cos x = 1/2
cos x = - 1 / √2
Examples:
Solve for -180° ≤ x ≤ 180°,
tan 2x = 1.424
Solve for 0 ≤ x ≤ 360°,
sin (x + 30°) = 0.781
Task 5: Solve for 0 ≤ x ≤ 360°
1.
2.
3.
4.
5.
sin 3x = 0.7
sin (x / 3) = 2/3
tan 4x = 1/3
cos 2x = 0.63
cos (x + 72°) = 0.515
Solve for 0 ≤ x ≤ 2π
1. tan 2x = 1
2. sin (x / 3) = ½
3. sin (x + π/6) = √3 / 2