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Trigonometry II
Harder Exact Values and Simple Trig Equations.
By Mr Porter
Harder Exact Value Questions.
1) Given that tan θ = ⅗ and sin θ < 0 ,
find the exact value of cos θ .
3
a) Construct a right angle triangle, label
Opposite (3) and Adjacent (5).
c =3 +5
2
2
2
c = 34
2
c) Construct TRIG CIRCLE.
c = 34
90
°
Sin(+)
180
°
d) tan θ is positive (1st and
3rd Quad.) and sin θ is
negative (3rd and 4th Quad.)
Hence, θ is in 3rd Quadrant.
cos θ is negative.
All (+)
180° —
θ°
180° +
θ°
Tan(+)
θ
0°(36
°
360° — 0°)
θ°
Cos(+)
270
°
θ
5
c2 = a 2 + b2
b) Use Pythagoras Thm
to find missing side.
34
By definition
A
H
-5
cos q = 34
cos q =
Harder Exact Value Questions.
2) Given that sin θ = -⅔ and cos θ < 0 ,
find the exact value of cot θ .
3
2
a) Construct a right angle triangle, label
Opposite (2) and Hypotenuse (3).
b) Use Pythagoras Thm
to find missing side.
5
a 2 + b2 = c2
a +2 =3
2
2
2
By definition
a 2 = 32 - 2 2
c) Construct TRIG CIRCLE.
a = 5
90
°
Sin(+)
d) sin θ is negative (3rd and 4th 180
Quad.) and cos θ is negative (2nd °
and 3rd Quad.)
Hence, θ is in 3rd Quadrant.
cot θ is positive (same as tan θ).
θ
All (+)
180° —
θ°
180° +
θ°
Tan(+)
θ
0°(36
°
360° — 0°)
θ°
Cos(+)
270
°
1
tan q
A
cot q = O
5
cot q =
2
cot q =
Harder Exact Value Questions.
3) Given that cos θ = -⅜ and tan θ < 0 ,
find the exact value of cosec θ .
8
55
a) Construct a right angle triangle, label
Adjacent (3) and Hypotenuse (8).
3
a 2 + b2 = c2
b) Use Pythagoras Thm
to find missing side.
a +3 =8
2
2
2
By definition
a 2 = 82 - 32
c) Construct TRIG CIRCLE.
a = 55
90
°
Sin(+)
(2nd
3rd
d) cos θ is negative
and
Quad.) and tan θ is negative (2nd
and 4th Quad.)
Hence, θ is in 2nd Quadrant.
cosec θ is positive (same as sin
θ).
180
°
θ
All (+)
180° —
θ°
180° +
θ°
Tan(+)
θ
0°(36
°
360° — 0°)
θ°
Cos(+)
270
°
1
sin q
H
cosec q = O
8
cosec q =
55
cosec q =
Exercise 1
a) Given that cos θ = ¼ and sin θ < 0, find the
exact value of tan θ.
ans : tanq = - 15
b) Given that tan θ = -⅕ and sin θ > 0, find
the exact value of sec θ.
ans : secq =
- 26
5
c) Given that sin θ = -⅘ and cos θ > 0, find the
exact value of cot θ.
ans : cot q =
-3
4
d) Given that cot θ = -⅞ and sec θ < 0, find
the exact value of cosec θ.
ans : cosecq =
113
8
Simple Trigonometric Equations & Exact Values.
{Hints: Exact values triangles and trig circle.}
90
°
Sin(+)
a) Solve 2sin α = 1, for 0° ≤ α ≤ 360°.
180
°
[These normally have 2 answers]
180° —
θ°
180° +
θ°
Tan(+)
Rearrange the equation, making sinα, the subject.
sin a =
1
2
30°
1st Quad answer
sin a = sin30°
a = 30°
45
°
2
2
1
3
1
answers are possible.
θ
0°(36
°
360° — 0°)
θ°
Cos(+)
270
°
Construct the trig circle and exact values triangles.
From the triangles, sin30° =
All (+)
60
°
1
45
°
1 , and RHS is positive, indicating a 1st and 2nd Quadrant
2
2nd Quad answer (180 – α)
sin a = sin(180 - 30°)
sin a = sin150°
a = 150°
Simple Trigonometric Equations & Exact Values.
90
°
Sin(+)
b) Solve 2 cos α + 1 = 0, for 0° ≤ α ≤ 360°.
180
°
Rearrange the equation, making cos α, the subject.
All (+)
180° —
θ°
180° +
θ°
Tan(+)
θ
0°(36
°
360° — 0°)
θ°
Cos(+)
45
°
2
1
1
45
°
270
°
1
cos a = - 2
Construct the trig circle and exact values triangles.
30°
2
3
1
60
°
1
From the triangles, cos 45° =
, and RHS is negative, indicating a 2nd and 3rd Quadrant
2
answers are possible.
2nd Quad answer (180 – α)
cosa = cos(180° - 45°)
cos a = cos135°
a = 135°
3rd Quad answer (180 + α)
cosa = cos(180 + 45°)
cosa = cos225°
a = 225°
Simple Trigonometric Equations & Exact Values.
c) Solve sin α = cos α, for 0° ≤ α ≤ 360°.
Rearrange the equation, making single trig ratio (divide by cos α).
Sin(+)
180
°
tan a = 1
Construct the trig circle and exact values triangles.
From the triangles, tan 45° = 1 , and RHS is positive,
1
All (+)
180° —
θ°
180° +
θ°
Tan(+)
θ
0°(36
°
360° — 0°)
θ°
Cos(+)
45
°
30°
2
3
270
°
indicating a 1st and 3rd Quadrant answers are possible.
tan a = tan 45°
a = 45°
2
1
90
°
sin a
cos a
= =1
cos a
cos a
1st Quad answer (α)
45
°
3rd Quad answer (180 + α)
tan a = tan(180 + 45°)
tan a = tan 225°
a = 225°
1
60
°
Exercise :The domain of θ is 0° ≤ θ ≤ 360°
a) Find all solution to 2cosq = -1
b) Find all solution to
3 tanq +1 = 0
ans : q = 120° or q = 240°
ans : q = 150° or q = 330°
c) Find all solution to 2sinq - 3 = 0
ans : q = 60° or q = 120°
d) Find all solution to sinq - 3 cosq = 0
ans : q = 60° or q = 240°
e) Find all solution to 2sin 2 q - sinq = 0
ans : q = 0°,180°,360° or q = 30°,150°
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