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Trigonometric Equations I
By Mr Porter
Basic Knowledge 1
1: Students need a basic understanding of solving
a) Linear type equations: 3x + 1 = - 17
b) Quadratic Equations:
2x2 – 3x + 1 = 0
2: Students need a basic understanding of trigonometric identities:
cos2 q + sin 2 q = 1
cos (a - b ) = cos a cos b + sin a sin b
1+ tan 2 q = sec2 q
sin (a - b ) = sin a cos b - sin b cos a
cot 2 q +1 = cosec2 q
tan (a - b ) =
cos (a + b ) = cos a cos b - sin a sin b
sin (a + b ) = sin a cos b + sin b cos a
tan a + tan b
tan (a + b ) =
1- tan a tan b
tan a - tan b
1+ tan a tan b
tan 2a =
2 tan a
1- tan 2 a
sin2a = 2sin a cos a
cos2a = cos2 a - sin 2 a
Basic Knowledge 2
1: Students must know the exact values triangles used in trigonometry.
30°
2
3
1
1
60
°
45
°
2
1
45
°
2: Students must know the acute angle equivalent of angles of any magnitude in
trigonometry
90
°
Sin(+)
180
°
All (+)
180° —
θ°
180° +
θ°
Tan(+)
θ
0°(36
°
360° — 0°)
θ°
Cos(+)
270
°
Radian Angles
Radian Angles are part of the HSC course, but some questions are written in radians.
Radian Conversion
Degrees (θ) to Radians (x)
x =q ´
30° :
p
q = x´
180
x = 30 ´
x=
π radians = 180°
Radian (x) to Degrees (θ)
p
5p
:
6
p
180
2
1
4
2
p
p
4
3
1
5p 180
´
6
p
q = 150°
q=
p
2
Sin(+)
p
p
6
p
6
Exact Values Triangles in Radians
3
180
p
All (+)
π—x
Note the following angles:
x
0 or 2π
π +x
2π – x
Tan(+)
Cos(+)
1
3p
2
30° =
p
6
45° =
p
4
60° =
p
3
If you have a single
trigonometric function, make it
the subject of the equation.
Example 1:
Solve 2 sin θ – 1 = 0, for 0° ≤ θ ≤ 360°
Solution:
2sinq -1 = 0
2sinq = 1
1
sinq =
2
So, the solutions are
We can use the degree exact
values triangles (or calculator)
to find an angle that gives ½.
Sin(+)
2
1
sin A =
60
°
1
2
æ 1ö
A = sin -1 ç ÷
è 2ø
A = 30°
θ = 30° or
θ = 150°
90
°
30°
3
Degrees should be used in the solution.
180
°
All (+)
180° —
θ°
180° +
θ°
Tan(+)
θ
0°(36
°
360° — 0°)
θ°
Cos(+)
270
°
The trigonometric angle circle, tells us there
are two possible POSITIVE sine angle ratios
A and 180 - A
A = 30° or
A = 180° - 30°
A = 150°
Always check your answer(s)
using a calculator!
If you have a single
trigonometric function, make it
the subject of the equation.
Example 2:
Solve 3 cos θ + 1 = 0, for 0° ≤ θ ≤ 360°
Solution:
3cosq +1 = 0
3cosq = -1
1
cosq = 3
So, the solutions are
Note:
1/
3 givens us the acute angle
required.
– tells us which Quadrant to
look in.
1
cos A =
3
90
°
Sin(+)
æ 1ö
A = cos -1 ç ÷
è 3ø
A = 70°32 '
Degrees should be used in the solution.
180
°
θ = 109°28’
or θ =
250°32’
All (+)
180° —
θ°
180° +
θ°
Tan(+)
θ
0°(36
°
360° — 0°)
θ°
Cos(+)
270
°
The trigonometric angle circle, tells us there
are two possible NEGATIVE cosine angle
ratios 180 – A and 180 + A
A = 180° - 70°32' or
A = 109°28'
A = 180° + 70°32'
A = 250°32'
Always check your answer(s)
using a calculator!
If you have a single
trigonometric function, make it
the subject of the equation.
Example 3:
Radians should be used in the solution.
Solve 6 sin 2θ = 3, for 0 £ q £ 2p
Note: Original domain 0 ≤ θ ≤ 2π
then for 2θ, domain is 0 ≤ 2θ ≤ 4π
Solution:
6sin2q = 3
1
sin 2q =
2
1
sin A =
2
p
-1 æ 1 ö
6
A = sin ç ÷
3
è 2ø
A=
p
Note:
1/
2 givens us the acute angle
required.
+ tells us which Quadrant to
look in.
p
2
Sin(+)
2
p
p
π—x
x
0 or 2π
2π – x
Tan(+)
6
3p
2
p
6
or
A=p A=
5p
6
p
6
q=
6
,p -
p
, 2p +
p
, 3p -
p
6
p 5p 13p 17p
,
12 12
,
12
,
12
Cos(+)
The trigonometric angle circle, tells us there are two possible POSITIVE
sine angle ratios A and π – A in 0 ≤ A ≤ 2π.
A=
p
6
6
p 5p 13p 17p
= ,
,
,
6 6
6
6
So, the solutions are
All (+)
π +x
3
1
2q =
Always check your answer(s)
using a calculator!
If you have a single
trigonometric function, make it
the subject of the equation.
Example 4:
æx
ö
Solve 4 sin ç + 30°÷ = 3 , for 0° £ x £ 360°
è2
ø
Solution:
æx
ö
4 sin ç + 30°÷ = 3
è2
ø
æx
ö 3
sin ç + 30°÷ =
è2
ø 4
sin A =
90
°
3
4
æ 3ö
A = sin -1 ç ÷
è 4ø
Note:
3/
4 givens us the acute angle
required.
+ tells us which Quadrant to
look in.
Sin(+)
180
°
All (+)
180° —
θ°
180° +
θ°
Tan(+)
θ
0°(36
°
360° — 0°)
θ°
Cos(+)
270
°
A = 48°35 '
The trigonometric angle circle, tells us there are two possible POSITIVE
sine angle ratios A and 180° – A in 0° ≤ A ≤ 360°.
So, A = 48°35
or
A = 131°25’
Degrees should be used in the solution.
Note: Original domain 0° ≤ x ≤ 360°
then for ˆx/2, domain is 0° ≤ ˆx/2 ≤ 180°
x
+ 30° = 48°35, 131°25'
2
So, subtracting 30°
x
= 18°35, 101°25'
2
x = 37°10, 202°50'
Always check your answer(s)
using a calculator!
If you have a single
trigonometric function, make it
the subject of the equation.
Example 5:
1
, for
2
Solve cos 2x ³
0 £ x £ 2p
Solution: Solve for the equals case.
1
Note:
cos 2x =
1/
√2 givens us the acute angle
2
required.
+ tells us which Quadrant to
look in.
1
cos A =
2
æ 1 ö
A = cos -1 ç
è 2 ÷ø
A=
Note: Original domain 0 ≤ x ≤ 2π
then for 2x, domain is 0 ≤ 2x ≤ 4π
p 7p 9p 15p
2x = ,
,
,
4 4 4
4
p 7p 9p 15p
x= ,
,
,
8 8 8
8
To solve the inequality, you need to test in-between
the above boundary points plus x = 0 and x = 2π
OR use a neat sketch.
p
2
Sin(+)
p
1
p
Radians should be used in the solution.
4
2
p
p
4
4
1
All (+)
π—x
x
0 or 2π
π +x
2π – x
Tan(+)
Cos(+)
3p
2
The trigonometric angle circle, tells us there are two possible POSITIVE
cosine angle ratios A and 2π – A in 0 ≤ A ≤ 2π.
So, A =
p
4
or
A=
7p
4
Solutions are:
0£x£
p
8
,
7p
9p 15p
£x£
,
£ x £ 2p
8
8
8