Two special triangles Two families of angles often show up in - - PowerPoint PPT Presentation

SLIDE 1

Elementary Functions

Part 4, Trigonometry Lecture 4.2a, Two Special Triangles

• Dr. Ken W. Smith

Sam Houston State University

2013

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Two special triangles

Two families of angles often show up in computations on the unit circle and we use them to practice our understanding of trigonometry. In this presentation we examine the 30-60-90 triangle and the 45-90-45 triangle (and all triangles on the unit circle related to these.) The 30-60-90 triangle The 30-60-90 triangle has a right angle (90◦) and two acute angles of 30◦ and 60◦. We assume our triangle has hypotenuse of length 1 and draw it

• n the unit circle:

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The 30 − 60 − 90 triangle

Anytime we consider a 30-60-90 triangle, we imagine that triangle as half

• f an equilateral triangle.

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The 30 − 60 − 90 triangle

For example, we reflect the 30-60-90 triangle about the x-axis and so we imagine an equilateral triangle with one vertex at the origin and the other two vertices to the right of the origin, on the unit circle symmetrically spaced about the x-axis.

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SLIDE 2

The 30 − 60 − 90 triangle

This triangle is an equilateral triangle with three equal sides of length 1 unit. Since this equilateral triangle is symmetric about the x-axis, and since each side has length 1, then the y-coordinates of the two vertices on the unit circle are ± 1

2.

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The 30 − 60 − 90 triangle

By the Pythagorean Theorem, x2 + ( 1

2)2 = 12

so x2 = 3

4

and so the x-value of the points on the unit circle must be x =

√ 3 2 .

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The 30 − 60 − 90 triangle

The two points on the unit circle are P(

√ 3 2 , 1 2) and Q( √ 3 2 , − 1 2). From this

we see that the cosine of 30◦ is

√ 3 2 and the sine of 30◦ is 1 2.

(Also cos(−30◦) =

√ 3 2 and sin(−30◦) = − 1 2.)

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The 30 − 60 − 90 triangle

A similar argument with equilateral triangles works for an angle of 60◦. (We should always think of a 30 − 60 − 90 triangle as half of an equilateral triangle!) In the figure drawn below we see that the cosine of 60◦ is 1

2 and

the sine of 60◦ is

√ 3 2 .

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SLIDE 3

The 30 − 60 − 90 triangle

Any angle on the unit circle with a reference angle of 30◦ or 60◦ will have coordinates that involve the numbers 1

2 and √ 3 2 .

These angles include 30◦, 60◦, 120◦, 150◦, 210◦, 240◦, 300◦ 330◦ and so

• n....

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The 30 − 60 − 90 triangle and its siblings

x y 30◦

π 6

√

3 2 , 1 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 60◦

π 3

• 1

2, √ 3 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 90◦

π 2

(−1, 0) (1, 0) (0, −1) (0, 1)

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SLIDE 4

The 30 − 60 − 90 triangle and its siblings

x y 120◦

2π 3

• − 1

2, √ 3 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 150◦

5π 6

• −

√ 3 2 , 1 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 180◦ π (−1, 0) (1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 210◦

7π 6

• −

√ 3 2 , − 1 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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SLIDE 5

The 30 − 60 − 90 triangle and its siblings

x y 240◦

4π 3

• − 1

2, − √ 3 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 270◦

3π 2

(−1, 0) (1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 300◦

5π 3

• 1

2, − √ 3 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 330◦

11π 6

√

3 2 , − 1 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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SLIDE 6

The 30 − 60 − 90 triangle and its siblings

x y 360◦ 2π (−1, 0) (1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 390◦

13π 6

√

3 2 , 1 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 420◦

7π 3

• 1

2, √ 3 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 30 − 60 − 90 triangle and its siblings

x y 450◦

5π 2

(−1, 0) (1, 0) (0, −1) (0, 1)

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SLIDE 7

cos π

6 = √ 3 2 , sin π 6 = 1 2

x y 30◦

π 6

√

3 2 , 1 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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cos π

6 = − √ 3 2 , sin π 6 = 1 2

x y 150◦

5π 6

• −

√ 3 2 , 1 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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cos π

6 = − √ 3 2 , sin π 6 = −1 2

x y 210◦

7π 6

• −

√ 3 2 , − 1 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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cos π

6 = √ 3 2 , sin π 6 = −1 2

x y 330◦

11π 6

√

3 2 , − 1 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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SLIDE 8

cos π

6 = 1 2, sin π 6 = √ 3 2

x y 60◦

π 3

• 1

2, √ 3 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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cos π

6 = −1 2, sin π 6 = √ 3 2

x y 120◦

2π 3

• − 1

2, √ 3 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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cos π

6 = −1 2, sin π 6 = − √ 3 2

x y 240◦

4π 3

• − 1

2, − √ 3 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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cos π

6 = 1 2, sin π 6 = − √ 3 2

x y 300◦

5π 3

• 1

2, − √ 3 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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SLIDE 9

The 30 − 60 − 90 triangle

We collect this information in a table. The first two columns give the angle, first in degrees and then in radians. The last two columns give the x and y coordinates. θ θ cos(θ) sin(θ) 30◦

π 6 √ 3 2 1 2

60◦

π 3 1 2 √ 3 2

120◦

2π 3

− 1

2 √ 3 2

150◦

5π 6

−

√ 3 2 1 2

210◦

7π 6

−

√ 3 2

− 1

2

240◦

4π 3

− 1

2

−

√ 3 2

300◦

5π 3 1 2

−

√ 3 2

330◦

11π 6 √ 3 2

− 1

2

I wouldn’t try to memorize this but instead be able to recreate it by proper use of the 30-60-90 triangle.

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The 45-90-45 triangle

Another important triangle is the right triangle with both acute angles equal to 45◦. Since this triangle has two equal angles then it is isoceles; it also has two equal sides. We may assume that the hypotenuse has length 1 and so draw this triangle on the unit circle.

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The 45-90-45 triangle

Since the triangle is isoceles, its height y is equal to the base x. By the Pythagorean theorem, x2 + y2 = 1 = ⇒ x2 + x2 = 1 = ⇒ 2x2 = 1. = ⇒ x2 = 1

2.

= ⇒ x =

1 √ 2.

Since y = x then cos(45◦) = sin(45◦) =

1 √ 2.

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The 45-90-45 triangle

cos(45◦) = sin(45◦) = 1 √ 2. If we wish to get rid of the radical in the denominator, we can multiply numerator and denominator by √ 2 so that the equation becomes cos(45◦) = sin(45◦) = √ 2 2 . Any angle θ in which the reference angle is 45◦ will have cosine and sine equal to ± √ 2 2 . Here are some examples.

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SLIDE 10

The 45 − 90 − 45 triangle and its siblings

x y 45◦

π 4

√

2 2 , √ 2 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 45 − 90 − 45 triangle and its siblings

x y 135◦

3π 4

• −

√ 2 2 , √ 2 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 45 − 90 − 45 triangle and its siblings

x y 225◦

5π 4

• −

√ 2 2 , − √ 2 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 45 − 90 − 45 triangle and its siblings

x y 315◦

7π 4

√

2 2 , − √ 2 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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SLIDE 11

The 45 − 90 − 45 triangle and its siblings

x y 405◦

9π 4

√

2 2 , √ 2 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 45 − 90 − 45 triangle and its siblings

x y 495◦

11π 4

• −

√ 2 2 , √ 2 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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cos π

4 = √ 2 2 , sin π 4 = √ 2 2

x y 45◦

π 4

√

2 2 , √ 2 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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cos 3π

4 = − √ 2 2 , sin π 4 = √ 2 2

x y 135◦

3π 4

• −

√ 2 2 , √ 2 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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SLIDE 12

cos 5π

4 = − √ 2 2 , sin π 4 = − √ 2 2

x y 225◦

5π 4

• −

√ 2 2 , − √ 2 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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cos 7π

4 = √ 2 2 , sin π 4 = − √ 2 2

x y 315◦

7π 4

√

2 2 , − √ 2 2

• (−1, 0)

(1, 0) (0, −1) (0, 1)

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The 45-90-45 triangle

We summarize this in a table. θ θ cos(θ) sin(θ) 45◦

π 4 √ 2 2 √ 2 2

135◦

3π 4

−

√ 2 2 √ 2 2

225◦

5π 4

−

√ 2 2

−

√ 2 2

315◦

7π 4 √ 2 2

−

√ 2 2

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Six Functions on the Unit Circle

All of this information has been provided for us (calculator free!) by our understanding of two basic triangles. Make sure you can recover these results just by drawing the appropriate triangle in the appropriate location.

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SLIDE 13

Two special triangles

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Two special triangles

In the next presentation, we will look at the six trig functions. (End)

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Elementary Functions

Part 4, Trigonometry Lecture 4.2b, Six Functions on the Unit Circle

• Dr. Ken W. Smith

Sam Houston State University

2013

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Six Functions on the Unit Circle

A central angle θ determines a point P(x, y) on the unit circle. The x-coordinate is cos θ and the y-coordinate is sin θ. There are four other trig functions, based on this point. The tangent function tan θ is equal to y

x = sin θ cos θ.

If the ratio y

x looks familiar, this is the “rise” over “run” as we move from

the origin O(0, 0) out to the point P(x, y); that is, tan θ is the slope of the line joining O to P.

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SLIDE 14

Six Functions on the Unit Circle

There are three more trig functions each of which is the reciprocal of one

• f the first three.

1 The secant of θ is the reciprocal of cosine; sec θ = 1 x. 2 The cosecant of θ is the reciprocal of sine; csc θ = 1 y. 3 The cotangent of θ is the reciprocal of tangent; cot θ = x y.

The reciprocal of a trig function with the syllable “sine” in it (sine and cosine) will be a trig function involving “secant” (cosecant and secant.) Each reciprocal pair has exactly one “co” in the list, so: the reciprocal of sine is cosecant; the reciprocal of cosine is secant; the reciprocal of tangent is cotangent.

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Six Functions on the Unit Circle

Since we may write the six trig functions in terms of x and y on the unit circle then we may also write the six trig functions in terms of cosine and sine. We may also write all six trig functions as ratios of the sides of the right triangle with short legs of length x and y and hypotenuse 1.

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Six Functions on the Unit Circle

Here is a summary of our results.

1 The cosine function cos θ is equal to x 1. 2 The sine function sin θ is equal to y 1. 3 The tangent function tan θ is equal to y x = sin θ cos θ. 4 The secant function sec θ is equal to 1 x = 1 cos θ, the reciprocal of

cosine.

5 The cosecant function csc θ is equal to 1 y = 1 sin θ, the reciprocal of

sine.

6 The cotangent function cot θ is equal to x y = cos θ sin θ , the reciprocal of

tangent.

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The six trig functions on the right triangle

If we draw an angle in general position, intersecting the unit circle at P(x, y), vertices C(0, 0), P(x, y) and R(x, 0) form a right triangle with a right angle on the x-axis at R(x, 0). We can define the six trig functions in terms of the sides of the triangle △CPR. We follow tradition: call the side CR as the “adjacent” side (close to the angle θ), RP as the “opposite” side (far from the angle θ), and CP is the “hypotenuse.”

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SLIDE 15

The six trig functions on the right triangle

If we abbreviate the lengths “adjacent”, “opposite” and “hypotenuse” by their first letters, A, O and H, we might draw the triangle like this.

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Six Functions on the Unit Circle

Thousands of trig students have memorized the first three trig functions as:

1 Sin θ = O H 2 Cos θ = A H 3 Tan θ = O A

and put these together into a chant: SOH-CAH-TOA.

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Six Functions on the Unit Circle

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Six Functions on the Unit Circle

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SLIDE 16

Six Functions on the Unit Circle

A side story I first learned of SOH-CAH-TOA at Central Michigan University when a number of calculus students wrote the letters at the top of their exams. Since the Saginaw Chippewa tribe is located near the university and the university had a course in the Ojibwe language, I wondered if Soh-cah-toa was an Ojibwe phrase or prayer. But one of the students explained that Soh-cah-toa was merely an English abbreviation for the trig functions. (But they didn’t deny saying a short prayer as they wrote it at the top of the exam!)

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Central Angles and Arcs

We will look more closely at these right triangle definitions of our trig functions in the next presentation. (End)

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Elementary Functions

Part 4, Trigonometry Lecture 4.2c, Practicing With Six Trig Functions

• Dr. Ken W. Smith

Sam Houston State University

2013

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Six Functions on the Unit Circle

Worked problems.

1 Find all six trig functions of the angle θ = 13π 3 .

• Solution. Draw the unit circle and find the point on the circle given

by the angle 13π

3

• radians. Since 13

3 = 4 + 1 3 then 13π 3

= 4π + π

3 and so

the angle 13π

3

gives the same point on the unit circle as π

3 . This point

is P( 1

2, √ 3 2 ).

Therefore cos 13π

3

= 1

2, sin 13π 3

=

√ 3 2 and tan 13π 3

=

√ 3 2 1 2 =

√

• 3. The

values of secant, cosecant and tangent are reciprocals of these so the values of all six trig functions are: cos( 13π

3 )

=

√ 3 2

sin( 13π

3 )

=

1 2

tan( 13π

3 )

= √ 3 sec( 13π

3 )

=

2 √ 3

csc( 13π

3 )

= 2 cot( 13π

3 )

=

1 √ 3

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SLIDE 17

Six Functions on the Unit Circle

2 Suppose sin θ = − √ 3 2 and cos θ is positive. Identify the angle θ and

then find all six trig functions of the angle θ.

• Solution. The angle is 5π

3 , equal to 300◦. (Or we could choose the

angle −60◦ = − π

3 .)

So cos( 5π

3 ) = 1 2, sin( 5π 3 ) = − √ 3 2 , tan( 5π 3 ) = −

√ 3. The reciprocals of these are sec( 5π

3 ) = 2, csc( 5π 3 ) = − 2 √ 3 = − 2 √ 3 3 , cot( 5π 3 ) = − 1 √ 3 = − √ 3 3 .

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Complementary angles

Given an angle θ we will talk about the angle we need to add to θ to make up 180◦ or 90◦. The angle we need to add to θ to create a 180◦ angle is the “supplement” of θ. So the supplementary angle of θ is 180◦ − θ or (in radians) π − θ. The angle we need to add to θ to “complete” a right angle is the “complement” of θ; the complementary angle of θ is 90◦ − θ or π

2 − θ. In

the figure below, with central angle θ (in black) the complementary angle

• f θ is drawn in blue.

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Six Functions on the Unit Circle

The term “complementary” explains the syllable “co-” in the terms cosine, cotangent and cosecant. The cosine of the angle θ is merely the “sine of the complement” of θ; co-sine is an abbreviation for “sine of the complement.” From the drawing we see that if the sine of θ is O

H and the cosine of θ is A H then the sine of the angle π 2 − θ is also A H ; the

“sine-of-the-complement” is merely the “co-sine.” In a similar way, “co-tangent” and “co-secant” stand for “tangent-of-the-complement” and “secant-of-the-complement”.

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The Pythagorean Theorem on the unit circle

Since the points P(x, y) are on the unit circle then, by the Pythagorean Theorem, x2 + y2 = 1. In terms of sine and cosine, this translates into (cos θ)2 + (sin θ)2 = 1. (1) Dividing by (cos θ)2 we have 1 + (tan θ)2 = (sec θ)2. (2) Dividing the top equation by (sin θ)2 we have (cot θ)2 + 1 = (csc θ)2. (3) These three identities are often called “The Pythagorean Identities” of trigonometry since they all depend on the Pythagorean theorem on the unit circle.

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SLIDE 18

A Worked Problem.

Given the cosine or sine of θ we should (by the Pythagorean Theorem) be able to find all the trig functions of θ. Here is an example. Find all trig functions of the angle θ if θ is in the second quadrant and sin(θ) = 2

3.

• Solution. If sin(θ) = 2

3 then by the Pythagorean Theorem

cos(θ)2 + ( 2

3)2 = 1

= ⇒ cos(θ)2 = 1 − ( 2

3)2 = 9 9 − 4 9 = 5 9.

Since cos(θ)2 = 5

9 then cos(θ) = ± √ 5 3 . Since θ is in the second quadrant

we know that the x-value of the point P(x, y) is negative and so x = cos(θ) = −

√ 5 3 . The point on the unit circle corresponding to the

angle θ is therefore P(−

√ 5 3 , 2 3) and with this information we can compute

the remaining four trig functions of θ. cos(θ) = −

√ 5 3

sin(θ) =

2 3

tan(θ) = − 2

√ 5 = − 2 √ 5 5

sec(θ) = − 3

√ 5 = − 3 √ 5 5

csc(θ) =

3 2

cot(θ) = −

√ 5 2

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Central Angles and Arcs

In the next presentation, we will look in depth at the sine function and transformations of it. (End)

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