Download Precalculus: Trigonometric Functions of Acute Angles Practice

Precalculus: Trigonometric Functions of Acute Angles Practice Problems
Questions
1. Find the value of all six of the trigonometric functions of the angle θ given the following right angle triangle. Note the
triangle is not drawn to scale.
8
6
θ
2. Using the labeling of the triangle below, prove that if θ is an acute angle in any right triangle, then (sin θ)2 +(cos θ)2 = 1.
c
a
θ
b
3. Find the value of all six of the trigonometric functions of the angle θ given the following right angle triangle. Note the
triangle is not drawn to scale.
13
1
θ
x
7
4. Assume that θ is an acute angle in a right triangle which satisfies sec θ = . Evaluate the remaining trigonometric
6
functions of the angle θ.
5. What are the values of the following? I would recommend you know how to work them out from the special triangles,
but if you want to you can memorize them.
(a) sin(π/3) =
(b) tan(π/3) =
(c) csc(π/4) =
(d) sec(π/4) =
(e) cos(π/6) =
(f) cot(π/6) =
Page 1 of 6
Precalculus: Trigonometric Functions of Acute Angles Practice Problems
Solutions
1. Find the value of all six of the trigonometric functions of the angle θ given the following right angle triangle. Note the
triangle is not drawn to scale.
8
6
θ
First, we need to use the Pythagorean theorem to find the length of the side adjacent to θ in the triangle, which I have
labelled x:
8
6
θ
x
x2 + 62 = 82
⇒
x2 = 64 − 36 = 28
⇒
x=
√
√
28 = 2 7
We choose the positive root, not the negative root, since the angle is acute.
hyp=8
The triangle can be labeled as
opp=6
θ
√
adj=2 7
From the reference triangle, we get value of the six trig functions:
sin θ
=
cos θ
=
tan θ
=
csc θ
=
sec θ
=
cot θ
=
6
3
opposite
= =
hypotenuse
8
4
√
√
adjacent
2 7
7
=
=
hypotenuse
8
4
opposite
3
6
= √ =√
adjacent
2 7
7
hypotenuse
8
4
= =
opposite
6
3
hypotenuse
8
4
= √ =√
adjacent
2 7
7
√
√
adjacent
2 7
7
=
=
opposite
6
3
Page 2 of 6
Precalculus: Trigonometric Functions of Acute Angles Practice Problems
2. Using the labeling of the triangle below, prove that if θ is an acute angle in any right triangle, then (sin θ)2 +(cos θ)2 = 1.
c
a
θ
b
We need to work out sine and cosine of the angle θ:
sin θ
=
cos θ
=
a
opposite
=
hypotenuse
c
adjacent
b
=
hypotenuse
c
Now we need to show the following quantity reduces to 1 (break out the algebra!):
sin2 θ + cos2 θ
=
=
=
=
=
=
(sin θ)2 + (cos θ)2
a 2 b 2
+
c
c
2
2
b
a
+ 2
c2
c
a2 + b2
c2
c2
(by Pythagorean theorem, a2 + b2 = c2 )
c2
1
3. Find the value of all six of the trigonometric functions of the angle θ given the following right angle triangle. Note the
triangle is not drawn to scale.
13
1
θ
x
First, we need to use the Pythagorean theorem to find the length of the side adjacent to θ in the triangle, which I have
labelled x:
x2 + 12 = 132
⇒
x2 = 169 − 1 = 168
⇒
x=
√
√
168 = 2 42
Page 3 of 6
Precalculus: Trigonometric Functions of Acute Angles Practice Problems
hyp=13
opp=1
θ √
adj=2 42
The triangle can be labeled as
From the reference triangle, we get value of the six trig functions:
sin θ
=
cos θ
=
tan θ
=
csc θ
=
sec θ
=
cot θ
=
1
opposite
=
hypotenuse
13
√
adjacent
2 42
=
hypotenuse
13
1
opposite
= √
adjacent
2 42
hypotenuse
13
=
= 13
opposite
1
hypotenuse
13
= √
adjacent
2 42
√
√
adjacent
2 42
=
= 2 42
opposite
1
7
4. Assume that θ is an acute angle in a right triangle which satisfies sec θ = . Evaluate the remaining trigonometric
6
functions of the angle θ.
To begin, we need to draw the right triangle with θ in it, and then use that to help us evaluate the trigonometric functions.
The triangle is based on the relation we have been given, and use SOH-CAH-TOA to get determine how to label:
sec θ =
7
1
hypotenuse
=
=
.
6
cos θ
adjacent
hyp=7
opp=x
θ
adj=6
We use the Pythagorean theorem to determine the length of the opposite side.
72 = 62 + x2
⇒
x2 = 49 − 36 = 13
⇒
x=
√
13
Page 4 of 6
Precalculus: Trigonometric Functions of Acute Angles Practice Problems
hyp=7
Our triangle becomes:
√
opp= 13
θ
adj=6
From the reference triangle, we get value of the six trig functions:
sin θ
=
cos θ
=
tan θ
=
csc θ
=
sec θ
=
cot θ
=
√
13
opposite
=
hypotenuse
7
adjacent
6
=
hypotenuse
7
√
opposite
13
=
adjacent
6
hypotenuse
7
=√
opposite
13
hypotenuse
7
=
adjacent
6
6
adjacent
=√
opposite
13
5. What are the values of the following? I would recommend you know how to work them out from the special triangles,
but if you want to you can memorize them.
√
3
2
(a) sin(π/3) =
(b) tan(π/3) =
(c) csc(π/4) =
(d) sec(π/4) =
√
√
3
2
√
2
√
(e) cos(π/6) =
(f) cot(π/6) =
3
2
√
3
Start by writing down the two special triangles, and then use SOH-CAH-TOA and the reciprocal definitions to read off
the values. The details are on the following page.
Page 5 of 6
Precalculus: Trigonometric Functions of Acute Angles Practice Problems
A 45-45-90 Triangle
1
√
2
π/4
1
π/4
1
opp
1
=√
4
hyp
2
π
adj
1
=
cos
=√
4
hyp
2
π opp
1
=
tan
= =1
4
adj
1
sin
1
π
=
√
1
=
2
π =
4
sin 4
π
√
1
= 2
sec
=
π
4
cos 4
π
1
=1
=
cot
4
tan π4
csc
π
A 30-60-90 Triangle
√
opp
3
=
3
hyp
2
π
adj
1
=
=
cos
3
hyp
2
√
π opp
3 √
tan
=
=
= 3
3
adj
1
sin
π
opp
1
=
6
hyp
2
√
π
adj
3
cos
=
=
6
hyp
2
π opp
1
tan
=
=√
6
adj
3
sin
π
=
π
1
2
=√
=
3
sin π3
3
π
1
=2
sec
=
3
cos π3
π
1
1
=√
cot
=
3
tan π3
3
csc
=
csc
π
6
1
sin
π
3
=2
1
2
=√
π
6
cos 3
3
√
π
1
3 √
=
cot
=
= 3
π
6
1
tan 3
sec
π
=
=
Page 6 of 6