Review for Midterm Exam 2
tan θ =
sin θ
,
cos θ
sin2 θ + cos2 θ = 1,
csc θ =
1
,
sin θ
tan2 θ + 1 = sec2 θ,
cos(α + β) = cos α cos β − sin α sin β
cos(α − β) = cos α cos β + sin α sin β
sin(α + β) = sin α cos β + cos α sin β
sin(α − β) = sin α cos β − cos α sin β
tan α + tan β
1 − tan α tan β
tan α − tan β
tan(α − β) =
1 + tan α tan β
tan(α + β) =
sin(2θ) = 2 sin θ cos θ
cos(2θ) = cos2 θ − sin2 θ
r
µ ¶
1 − cos θ
θ
sin
=±
2
2
r
µ ¶
θ
1 + cos θ
cos
=±
2
2
c2 = a2 + b2 − 2ab cos γ
sin β
sin γ
sin α
=
=
a
b
c
sec θ =
1
,
cos θ
cot θ =
cot2 θ + 1 = csc2 θ
1
tan θ
Chapter 6 (sections 6.3, 6.4, 6.5, 6.7, 6.8)
Problem 1. Find the exact value of...
(a) cos 5π
sin 7π
− cos 7π
sin 5π
= sin( 7π
−
12
12
12
12
12
q
45◦
2
(b) sin 22.5◦ = sin
=
q
1−cos 45◦
2
r
(c) tan 7π
= tan
8
(d)
cos(− π8 )
7π
4
=
2
= cos
sin
cos
π
8
7π
4
2
7π
4
2
= cos
=
r
2
q
=
1+cos 7π
4
2
1+cos
2
π
4
= sin 2π
= sin π6 =
12
p
√
1− 22
2
1−cos 7π
4
2
−
π
4
=
5π
)
12
=
r
=−
q
=
√
2−
2
√
1− 22
√
1+ 22
2
s
= −
p
√
1+ 22
2
2+
2
=
(e) tan(θ − π4 ), sin(2θ), and cos( 2θ ) given that sin θ =
Use the fundamental identity to find that cos θ =
1
4
1
2
√
2− 2
√
2+ 2
√
2
and θ is in quadrant II
√
− 415 .
Thus tan θ = − √115 , so
√
− √115 − 1
tan θ − tan π4
π
−1 − 15
tan(θ − ) =
=
= √
1
4
1 + tan θ · tan π4
1 − √15·1
15 − 1
√
√
15
1 − 15
sin(2θ) = 2 sin θ cos θ = 2 · ·
= −
4
4
8
s
s
√
r
√
1 − 415
θ
1 + cos θ
4 − 15
cos( ) =
=
=
2
2
2
8
Note that cos( 2θ ) is positive since
θ
2
is in quadrant I ( π2 < θ < π ⇒
5
(f) tan(α − β), given that tan α = − 12
,π<α<
3π
2
and sin β = − 13 ,
π
4
3π
12
<
θ
2
< π2 )
< α < 2π
We need to know tan β, thus we need to find cos β (using fundamental identity):
cos β =
√
8
.
3
Thus tan(α − β) =
(g) tan(cos−1 45 + sin−1
Let α = cos−1
4
5
tan α−tan β
1+tan α·tan β
=
5
− 12
+ √1
8
5 √1
·
1+ 12
8
√
12 − 5 8
√
=
12 8 + 5
5
)
13
then cos α = 45 , therefore tan α =
triangle to see that).
Let β = sin−1
5
13
So tan(α + β) =
then sin β =
3
5
+ 12
4
3 5
1− 4 · 12
=
5
,
13
56
.
33
thus tan β =
5
.
12
3
4
(label sides of the right
(h) sec(2 tan−1 34 )
Let θ = tan−1
So sec(2θ) =
3
4
then tan θ =
1
cos(2θ)
=
3
4
thus cos θ =
1
cos2 θ−sin2 θ
=
1
9
16
− 25
25
=
4
5
and sin θ = 35 .
1
=
7
25
25
7
Problem 2. Establish each identity:
(a)
1 − cot2 θ
+ 2 cos2 θ = 1
1 + cot2 θ
Let’s start with the left hand side:
1 − cot2 θ
1 − cot2 θ
2
+
2
cos
θ
=
+ 2 cos2 θ = sin2 θ(1 − cot2 θ) + 2 cos2 θ
1 + cot2 θ
csc2 θ
= sin2 θ − cos2 θ + 2 cos2 θ = sin2 θ + cos2 θ = 1
and since we have arrived at the right hand side, the identity is proved.
(b)
sin(α + β)
= tan α + tan β
cos α · cos β
Start with the right hand side:
tan α + tan β =
sin α
sin β
sin α · cos β + sin β · cos α
sin(α + β)
+
=
=
cos α cos β
cos α · cos β
cos α · cos β
and since we have arrived at the left hand side, the identity is proved.
(c)
cos4 θ − sin4 θ = cos(2θ)
Start with the left hand side:
cos4 θ − sin4 θ = (cos2 θ − sin2 θ)(cos2 θ + sin2 θ) = cos2 θ − sin2 θ
and since cos2 θ − sin2 θ = cos(2θ) is the double identity, our identity is proved.
(d)
1
sin3 θ + cos3 θ
sin(2θ) =
2
sin θ + cos θ
Multiply both sides by sin θ + cos θ:
1−
(sin θ + cos θ)(1 −
1
sin(2θ)) = sin3 θ + cos3 θ
2
Now work on the left hand side:
(sin θ + cos θ)(1 −
1
sin(2θ)) = (sin θ + cos θ)(1 − sin θ cos θ)
2
= sin θ − sin θ cos2 θ + cos θ − sin2 θ cos θ
= sin θ(1 − cos2 θ) + cos θ(1 − sin2 θ)
= sin θ sin2 θ + cos θ cos2 θ
= sin3 θ + cos3 θ
which is the right hand side, thus the identity is proved.
Problem 3. Show that
(a)
v
1 − v2
We know that sin(sin−1 v) = v so label the right triangle (hypotenuse is 1, the
tan(sin−1 v) = √
opposite side (to the angle θ = sin−1 v is v and the adjacent side is, therefore,
√
1 − v 2 ), thus
v
tan(sin−1 v) = √
1 − v2
(b)
sin−1 v + cos−1 v =
π
2
Take cosine of both sides:
cos(sin−1 v + cos−1 v) = cos
π
2
Work with the left hand side:
√
√
cos(sin−1 v+cos−1 v) = v sin(cos−1 v)−v cos(sin−1 v) = v 1 − v 2 −v 1 − v 2 = 0
and since the right hand side is cos π2 = 0 we have established the identity.
Note, that the same argument could ”prove” that sin−1 v + cos−1 v =
example, since cos
5π
2
5π
2
is also 0. This would be very strange given that sin
for
−1
v
and cos−1 v are functions and thus have to have a unique output. The catch is
in their ranges:
−
π
π
< sin−1 v <
2
2
and
0 < cos−1 v < π
thus
π
3π
< sin−1 v + cos−1 v <
2
2
and so the only angle in this range which zero cosine is π2 .
−
Problem 4. Solve each equation on the interval 0 ≤ θ < 2π
(a)
sin θ + sin 2θ = 0
sin θ + 2 sin θ cos θ = 0
sin θ · (1 + 2 cos θ) = 0
Thus either sin θ = 0 or 1 + 2 cos θ = 0. The former equation has solutions
θ = 0 and θ = π. The latter equation:
1 + 2 cos θ = 0
1
cos θ = −
2
2π
4π
θ=
and θ =
3
3
So the answer is: θ = 0,
2π
4π
, π,
3
3
(b)
2 cos2 θ + cos θ − 1 = 0
Let x = cos θ then we have
2x2 + x − 1 = 0
thus x =
1
2
or x = −1.
So we need to solve two equations: cos θ =
So the answer is: θ =
sin θ −
Divide both sides by 2:
1
2
and cos θ = −1.
π
5π
, π,
3
3
(c)
Since cos π3 =
1
2
√
3 cos θ = 2
√
1
3
sin θ −
cos θ = 1
2
2
and sin π3 =
cos
√
3
2
we have:
π
π
sin θ − sin cos θ = 1
3
3
π
sin(θ − ) = 1
3
π
π
θ− =
3
2
5π
θ=
6
Chapter 7 (sections 7.1, 7.2, 7.3)
Problem 5. Find the exact value of 1 − cos2 20◦ − cos2 70◦
Since cos 20◦ = sin 70◦ we have
1 − cos2 20◦ − cos2 70◦ = 1 − sin2 70◦ − cos2 70◦ = 1 − (sin2 70◦ + cos2 70◦ ) = 1 − 1 = 0
Problem 6. Solve the triangle. (In cases d, e, f determine whether the given information results in one, two or no triangles)
(a) a = 6,
α = 40◦ ,
c=4
Use Law of Sines:
sin α
sin γ
=
a
c
⇒
c · sin α
4 sin 40◦
sin γ =
=
' 0.429
a
6
⇒
γ ' 25◦
Then since α + β + γ = 180◦ , we find that β = 115◦ .
Use Law of Sines again to find the remaining side:
sin α
sin β
=
a
b
⇒
b=
a · sin β
'8
sin α
a = 6, b = 8, c = 4, α = 40◦ , β = 115◦ , γ = 25◦
(b) α = 70◦ ,
β = 60◦ ,
c=4
We find that γ = 180◦ − 70◦ − 60◦ = 50◦ .
Using the Law of Sines we get:
b=
c · sin β
' 4.5
sin γ
a=
c · sin α
' 4.9
sin γ
Note that this answer makes sense: since all angles are close to 60◦ the triangle
must be close to equilateral (all sides are equal).
a = 4.9, b = 4.5, c = 4, α = 70◦ , β = 60◦ , γ = 50◦
(c) α = 50◦ ,
γ = 20◦ ,
a=4
We find that β = 180◦ − 50◦ − 20◦ = 110◦ .
Using the Law of Sines we get:
a · sin β
' 4.9
sin α
a · sin γ
c=
' 1.8
sin α
b=
a = 4, b = 4.9, c = 1.8, α = 50◦ , β = 110◦ , γ = 20◦
(d) a = 2,
c = 1,
α = 120◦
Use Law of Sines:
√
sin α
3
sin γ
c · sin α
sin 120◦
=
⇒ sin γ =
=
=
⇒ γ ' 26◦ or γ ' 154◦
a
c
a
2
4
However, since α + β + γ = 180◦ , γ must be less than 180◦ − 120◦ = 60◦ and
therefore, we conclude that γ = 26◦ (i.e. there is only one triangle, that satisfies
the above conditions).
β = 180◦ − 120◦ − 26◦ = 34◦
By Law of Sines:
b=
a · sin β
' 1.3
sin α
a = 2, b = 1.3, c = 1, α = 120◦ , β = 34◦ , γ = 26◦
(e) b = 4,
c = 6,
β = 20◦
Use Law of Sines:
sin β
sin γ
=
⇒
b
c
sin γ =
c · sin γ
6 sin 20◦
=
' 0.513
b
4
⇒
γ ' 31◦ or γ ' 149◦
Note that both values can be possible (since β + γ < 180◦ ), therefore there are
two triangles that satisfy the above conditions.
α = 129◦ or α = 11◦
Use Law of Sines:
sin γ
sin α
=
a
c
⇒
a=
c · sin α
sin γ
⇒
a = 9.1, b = 4, c = 6, α = 129◦ , β = 20◦ , γ = 31◦
a = 2.2, b = 4, c = 6, α = 11◦ , β = 20◦ , γ = 149◦
a ' 9.1 or a ' 2.2
(f) b = 4,
c = 5,
β = 95◦ Use Law of Sines:
sin γ
sin β
=
b
c
⇒
c · sin γ
5 sin 95◦
=
' 1.245
b
4
sin γ =
However, there is no such angle (since range of sine is [−1, 1]), i.e. there is no
such triangle that satisfies the above conditions.
(g) a = 3,
b = 4,
γ = 40◦
Use Law of Cosines:
c2 = a2 + b2 − 2ab cos γ
⇒
c2 = 9 + 16 − 24 cos 40◦
⇒
c ' 2.6
Then by Law of Sines:
sin α =
a sin γ
' 0.658
c
⇒
α ' 41◦
and thus β = 180◦ − 41◦ − 40◦ = 99◦
a = 3, b = 4, c = 2.6, α = 41◦ , β = 99◦ , γ = 40◦
(h) a = 10,
b = 8,
c=5
Use Law of Cosines:
c2 = a2 + b2 − 2ab cos γ
⇒
cos γ =
c2 − a2 − b2
' 0.869
−2ab
Then by Law of Sines:
sin α =
a sin γ
' 0.991
c
⇒
α = 82.3◦
and thus β = 180◦ − 29.7◦ − 81.3◦ = 69◦
a = 10, b = 8, c = 5, α = 82.3◦ , β = 69◦ , γ = 29.7◦
⇒
γ ' 29.7◦