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.

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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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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.

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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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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