Download Trigonometry II Harder Exact Values, Trig Equations and Trig Curves.

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°