Download Trigonometry Notes on Right Triangle Trigonometry.

Trigonometry
Notes on Right Triangle Trigonometry.
Definitions of the Trigonometric Functions for Right Triangles: Historically, trigonometry was the study of right triangles, but today trigonometry has expanded to include more than just right
triangles, as we’ll see later. For now, we’ll start with the classical definitions.
Definition of the Trigonometric Functions for Right Triangles
Cosecant
opposite
y
hypotenuse h
sin(θ) =
=
csc(θ) =
=
hypotenuse h
opposite
y
Cosine
Secant
adjacient
x
hypotenuse h
cos(θ) =
=
sec(θ) =
=
hypotenuse h
adjacient
x
Tangent
Cotangent
opposite y
adjacient x
tan(θ) =
=
cot(θ) =
=
adjacient x
opposite y
Sine
Example #1: Find all 6 trigonometric functions for the right triangle at the right.
►
8
17
sin(θ) =
csc(θ) =
17
8
15
17
cos(θ) =
sec(θ) =
17
15
8
15
tan(θ) =
cot(θ) =
15
17
□
Notice that the 3 on the right are just reciprocals of the ones on the left. This leads us to our 1st set of trigonometric identities.
Remember that an identity is an equation that is true everywhere it is defined.
Reciprocal Identities for the Trigonometric Functions
1
1
sin(θ) =
csc(θ) =
csc(θ)
sin(θ)
cos(θ) =
1
sec(θ)
sec(θ) =
1
cos(θ)
tan(θ) =
1
cot(θ)
cot(θ) =
1
tan(θ)
This is why sine, cosine and tangent are referred to as the 3 basic trigonometric functions.
And, cosecant, secant and cotangent are sometimes called the reciprocal trigonometric functions.
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Notes on Right Triangle Trigonometry.
Page #1 of 6
Further, note that tangent and cotangent are the ratios of sine and cosine. Using the results from example #1:
8
8 17 sin(θ)
15 15 17 cos(θ)
tan(θ) =
=
=
, and cot(θ) =
=
=
15 15 17 cos(θ)
8
8 17 sin(θ)
This leads us to the next set of identities.
Quotient Identities for the Trigonometric Functions
sin(θ)
tan(θ) =
cos(θ)
cot(θ) =
cos(θ)
sin(θ)
Because, the basic trigonometric function tangent can be found from sine and cosine, they are sometimes referred to as the 2
fundamental trigonometric functions.
Example #2: If sin(θ) =
12
5
, and cos(θ) =
, find the other 4 trigonometric functions.
13
13
►
tan(θ) =
sin(θ) 12 13 12
=
=
cos(θ) 5 13
5
csc(θ) =
1
1
13
=
=
sin(θ) 12 13 12
sec(θ) =
1
1
13
=
=
cos(θ) 5 13 5
cot(θ) =
cos(θ) 5 13
5
□
=
=
sin(θ) 12 13 12
Actually, since we are right now only working with right triangles and therefore acute angles for θ, we only really need is just 1 of
the 6 trigonometric functions to find all of the rest.
Example #3: If sec(θ) =
18
and 0 < θ < 90° , find the other 5 trigonometric functions.
7
►
Since we know that secant is hypotenuse over adjacent we can work with the triangle to the right.
First, we need to find the opposite side, y, by using the Pythagoean theorem.
7 2 + y 2 = 182
49 + y 2 = 324
y 2 = 275
y = 275
y = 5 11
Thus,
5 11
18
7
cos(θ) =
18
sin(θ) =
tan(θ) =
5 11
7
csc(θ) =
18
=
18 11
5 11 5 11 11
18
sec(θ) = :Given
7
cot(θ) =
7
5 11
=
7 11
5 11 11
=
18 11 18 11
=
5(11)
55
=
7 11 7 11
=
5(11)
55
□
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Notes on Right Triangle Trigonometry.
Page #2 of 6
The Trigonometric Functions for Basic Acute Angles: π π
π
The basic acute angles are 30°, 45°, and 60°; or in radians , , and .
6 4
3
π
, the triangle would be a isoceles right triangle. Thus, opposite and adjacient sides
4
would be the same as shown in the triangle at the right. Therefore,
x2 + x2 = h2
For θ = 45° =
2x 2 = h 2
2x 2 = h
x 2=h
Thus,
While for both θ = 30° =
x
1
2
⎛π⎞
sin(45°) = sin ⎜ ⎟ =
=
=
2
2
⎝4⎠ x 2
⎛π⎞ x 2
csc(45°) = csc ⎜ ⎟ =
= 2
x
⎝4⎠
x
1
2
⎛π⎞
cos(45°) = cos ⎜ ⎟ =
=
=
2
2
⎝4⎠ x 2
⎛π⎞ x
tan(45°) = tan ⎜ ⎟ = = 1
⎝4⎠ x
⎛π⎞
sec(45°) = sec ⎜ ⎟ =
⎝4⎠
⎛π⎞
cot(45°) = cot ⎜ ⎟ =
⎝4⎠
x 2
= 2
x
x
=1
x
π
π
, and θ = 60° = , we need to start with an equilateral triangle and split it into 2 right triangles as shown
6
3
to the right.
Pulling out just the right triangle, we get the 2nd triangle on the right.
Thus,
2
⎛x⎞
y2 + ⎜ ⎟ = x 2
⎝2⎠
x2
y2 +
= x2
4
3x 2
y2 =
4
60°
30°
3x 2
y=
4
y=
x 3
2
60°
Therefore,
3
⎛π⎞ x 3 2
sin(60°) = sin ⎜ ⎟ =
=
3
x
2
⎝ ⎠
⎛π⎞ x 2 1
cos(60°) = cos ⎜ ⎟ =
=
x
2
⎝3⎠
x
1
⎛π⎞
=
=2
sec(60°) = sec ⎜ ⎟ =
⎝3⎠ x 2 1 2
⎛π⎞ x 3 2
tan(60°) = tan ⎜ ⎟ =
= 3
x 2
⎝3⎠
x 2
1
3
⎛π⎞
cot(60°) = cot ⎜ ⎟ =
=
=
3
3
⎝3⎠ x 3 2
x
2
2 3
⎛π⎞
csc(60°) = csc ⎜ ⎟ =
=
=
3
3
x
3
2
3
⎝ ⎠
Similarly using the top angle we get,
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⎛π⎞ x 2 1
=
sin(30°) = sin ⎜ ⎟ =
x
2
⎝6⎠
x
1
⎛π⎞
=
=2
csc(30°) = csc ⎜ ⎟ =
6
x
2
1
2
⎝ ⎠
3
⎛π⎞ x 3 2
=
cos(30°) = cos ⎜ ⎟ =
6
x
2
⎝ ⎠
x
2
2 3
⎛π⎞
sec(30°) = sec ⎜ ⎟ =
=
=
6
3
3
⎝ ⎠ x 3 2
x 2
1
3
⎛π⎞
tan(30°) = tan ⎜ ⎟ =
=
=
3
3
⎝6⎠ x 3 2
⎛π⎞ x 3 2
cot(30°) = cot ⎜ ⎟ =
= 3
x 2
⎝6⎠
Notes on Right Triangle Trigonometry.
Page #3 of 6
Therefore, the complete table of the trigonometric function for the 3 basic acute angles is:
The Trigonometric Functions for the 3 Basic Acute Angles
sin(θ) cos(θ) tan(θ) csc(θ) sec(θ) cot(θ)
30° =
π
6
1
2
3
2
45° =
π
4
2
2
60° =
π
3
3
2
2
2
1
2
3
3
2
1
2
3
2 3
3
2 3
3
3
2
1
2
3
3
The Cofunctions Identities: Notice how the columns for sine and cosine, for tangent and cotangent, and for secant and cosecant are the reverse of each other.
Also, you may have noticed that for sine, secant and tangent there is a corresponding function with almost the same name except that
it begins with “co”, cosine, cosecant, and cotangent. This is the cofunction relation. These functions are related though complementary
angles. For example, using sine, cosine and the triangle to the right.
opposite side from θ y adjacient side to ( 90° − θ )
90° − θ
sin(θ) =
= =
= cos ( 90° − θ )
hypotenuse
h
hypotenuse
From the same process we get:
θ
The Cofunctions Identities
In Degrees
In Radians
⎛π
⎞
sin(θ) = cos ( 90° − θ ) sin(θ) = cos ⎜ − θ ⎟
⎝2
⎠
π
⎛
⎞
cos(θ) = sin ( 90° − θ ) cos(θ) = sin ⎜ − θ ⎟
⎝2
⎠
⎛π
⎞
tan(θ) = cot ( 90° − θ ) tan(θ) = cot ⎜ − θ ⎟
2
⎝
⎠
⎛π
⎞
csc(θ) = sec ( 90° − θ ) csc(θ) = sec ⎜ − θ ⎟
2
⎝
⎠
⎛π
⎞
sec(θ) = csc ( 90° − θ ) sec(θ) = csc ⎜ − θ ⎟
⎝2
⎠
π
⎛
⎞
cot(θ) = tan ( 90° − θ ) cot(θ) = tan ⎜ − θ ⎟
⎝2
⎠
Example #4: If tan ( 28° ) = 0.53171 , find cot ( 62° ) . Round to the 5th decimal place.
►
First, notice that 28˚ and 62˚ are a pair of complementary angles, 28˚+62˚ = 90˚. Thus,
cot ( 62° ) = tan ( 28° ) = 0.53171 .□
⎛π⎞
⎛ 2π ⎞
Example #5: If csc ⎜ ⎟ = 1.05146 , find cos ⎜ ⎟ . Round to the 5th decimal place.
5
⎝ 10 ⎠
⎝ ⎠
►
1
1
⎛π⎞
⎛π π ⎞
⎛ 5π − π ⎞
⎛ 4π ⎞
⎛ 2π ⎞
cos ⎜ ⎟ = sin ⎜ − ⎟ = sin ⎜
=
= .95106 .□
⎟ = sin ⎜ ⎟ = sin ⎜ ⎟ =
2
π
10
2
10
10
10
5
1.05146
⎝ ⎠
⎝
⎠
⎝
⎠
⎝ ⎠
⎝ ⎠ csc ⎛ ⎞
⎜ ⎟
⎝ 5 ⎠
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Notes on Right Triangle Trigonometry.
Page #4 of 6
Solving Trigonometric Equations Involving Acute Angles: Example #6: Solve exactly cos(θ) =
3
for 0° < θ < 90° .
2
►
In this case, since
3
3
is a basic result for cosine, we can use the table above to that get that cos(30°) =
. Thus, θ = 30° .□
2
2
Note, that since the restriction on θ was given in degrees, the answer was expected back in degrees. Similarly if the problem is given
using radians, the answer would be expected back in radians.
Example #7: Solve exactly cot(θ) = 1 for 0 < θ <
π
.
2
►
π
⎛π⎞
In this case, since 1 is a basic result for cotangent, we can use the table above to that get that cot ⎜ ⎟ = 1 . Thus, θ = .□
4
⎝4⎠
Example #8: Solve exactly sin(θ) = 0.256 for 0° < θ < 90° . Round to the 4th decimal place.
►
Here 0.256 isn’t a basic result for sine. Thus, we need to use the calculator and the inverse sine function.
Remember that if y = f (x) and f(x) is an invertible function, then f −1 (y) = x where the inverse of f(x) is denoted by f −1 (y) .
We’ll learn more about the inverse trigonometric functions later, but for now it is safe to say that if θ is an acute angle and
y
y
⎛ y⎞
sin(θ) = , then θ = sin −1 ⎜ ⎟ . So in this case, the = 0.256 , and we get that θ = sin −1 ( 0.256 ) . Finally, we use the calculator in
h
h
h
⎝ ⎠
degree mode to get θ = sin −1 ( 0.256 ) = 14.8328° .□
Be careful not to confuse an inverse trigonometric function with a reciprocal trigonometric function, i.e. sin −1 ( z ) ≠ csc ( z ) . The -1 is
NOT an exponent on the function.
How to use your calculator with the trigonometric functions is shown in the calculator handouts on the website.
Example #9: Solve sec(θ) = 3.7468 for 0 < θ <
►
π
. Round answer to the 4th decimal place.
2
sec(θ) = 3.7468
1
= 3.7468
cos(θ)
1
= cos(θ)
3.7468
⎛ 1 ⎞
cos −1 ⎜
⎟=θ
⎝ 3.7468 ⎠
1.3006 ≈ θ
Make sure your calculator is in radian mode.
□
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Notes on Right Triangle Trigonometry.
Page #5 of 6
Applications of Right Triangle Trigonometry: Example #10: A person, who is standing 10ft from a flag pole, has to look up at an angle of 58˚ to see the top of the flag pole. How
tall is the flag pole, if the distance from the ground to their eyes is 6.1ft? Round the answer to the tenths.
►
First, notice that the right triangle gives us that
y
tan ( 58° ) =
10
10 tan ( 58° ) = y
And, the height of the flag pole is given by,
58°
height = y + 6.1
= 10 tan ( 58° ) + 6.1
≈ 22.1
Thus, the height of the flag pole is 22.1ft.□
Example #11: If a 12m long guy wire, attached to a building and the ground, makes an angle of 50° with the ground, how far from
building is the guy wire attached to the ground? Round answer to the hundreths.
►
x
cos ( 50° ) =
12
50°
12 cos ( 50° ) = x
x
7.71 ≈ x
Therefore, the distance is 7.71m.□
Example #12: If a rower paddles due north across a 20ft wide river that runs west to east, and finds himself 13ft down stream from
where he started, at what bearing, angle from north, did he actually travel? Round answer thousandths of a degree.
►
For B=bearing,
13
tan(B) =
20
⎛ 13 ⎞
B = tan −1 ⎜ ⎟
Make sure your calculator is in degree mode.
⎝ 20 ⎠
B ≈ 33.024°
Thus, the bearing is N 33.024° E .□
Example #13: If a simple pool has depths ranging from 3ft to 10ft and the bottom of the pool is 18ft, find the angle of depression of
the bottom of the pool. Round answer to the hundredths of a radian.
►
θ
First notice that the opposite side from θ is 10ft − 3ft = 7ft . Therefore,
7
sin ( θ ) =
18
⎛7⎞
θ = sin −1 ⎜ ⎟
⎝ 18 ⎠
θ ≈ 0.40
Thus, the angle of depression is 0.40 radians.□
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Notes on Right Triangle Trigonometry.
Page #6 of 6