Elementary Functions Two special triangles The 30-60-90 ......The 30-60-90 triangle The 30-60-90...

Elementary FunctionsPart 4, Trigonometry

Lecture 4.2a, Two Special Triangles

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 70

Two special triangles

Two families of angles often show up in computations on the unit circleand we use them to practice our understanding of trigonometry. In thispresentation 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 triangleThe 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 iton the unit circle:


The 30− 60− 90 triangle

Anytime we consider a 30-60-90 triangle, we imagine that triangle as halfof an equilateral triangle.



For example, we reflect the 30-60-90 triangle about the x-axis and so weimagine an equilateral triangle with one vertex at the origin and the othertwo vertices to the right of the origin, on the unit circle symmetricallyspaced about the x-axis.



This triangle is an equilateral triangle with three equal sides of length 1unit.

Since this equilateral triangle is symmetric about the x-axis, and sinceeach side has length 1, then the y-coordinates of the two vertices on theunit circle are ±1

2 .



By the Pythagorean Theorem,

x2 + (12)2 = 12

so x2 = 34

and so the x-value of the points on the unit circle must be x =√32 .



The two points on the unit circle are P (√32 ,

12) and Q(

√32 ,−

12). From this

we see that the cosine of 30◦ is√32 and the sine of 30◦ is 1

2 .

(Also cos(−30◦) =√32 and sin(−30◦) = −1

2 .)



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 equilateraltriangle!) In the figure drawn below we see that the cosine of 60◦ is 1

2 and

the sine of 60◦ is√32 .



Any angle on the unit circle with a reference angle of 30◦ or 60◦ will have

coordinates that involve the numbers 12 and

√32 .

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


The 30− 60− 90 triangle and its siblings

x

y

30◦π6

(√32 ,

12

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

60◦

π3

(12 ,√32

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

90◦

π2

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

120◦

2π3

(−1

2 ,√32

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

150◦5π6

(−√32 ,

12

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

180◦π(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

210◦7π6(

−√32 ,−

12

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

240◦4π3

(−1

2 ,−√32

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

270◦

3π2

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

300◦5π3

(12 ,−

√32

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

330◦11π6 (√

32 ,−

12

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

360◦ 2π(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

390◦13π6

(√32 ,

12

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

420◦

7π3

(12 ,√32

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

450◦

5π2

(−1, 0) (1, 0)

(0,−1)

(0, 1)


cos π6 =√32 , sin π

6 = 12

x

y

30◦π6

(√32 ,

12

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)


cos π6 = −√32 , sin π

6 = 12

x

y

150◦5π6

(−√32 ,

12

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)


cos π6 = −√32 , sin π

6 = −12

x

y

210◦7π6(

−√32 ,−

12

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



6 = −12

x

y

330◦11π6 (√

32 ,−

12

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)


cos π6 = 12 , sin π

6 =√32

x

y

60◦

π3

(12 ,√32

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)


cos π6 = −12 , sin π

6 =√32

x

y

120◦

2π3

(−1

2 ,√32

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)


cos π6 = −12 , sin π

6 = −√32

x

y

240◦4π3

(−1

2 ,−√32

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)


cos π6 = 12 , sin π

6 = −√32

x

y

300◦5π3

(12 ,−

√32

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



We collect this information in a table. The first two columns give theangle, first in degrees and then in radians. The last two columns give thex and y coordinates.

θ θ cos(θ) sin(θ)

30◦ π6

√32

12

60◦ π3

12

√32

120◦ 2π3 −1

2

√32

150◦ 5π6 −

√32

12

210◦ 7π6 −

√32 −1

2

240◦ 4π3 −1

2 −√32

300◦ 5π3

12 −

√32

330◦ 11π6

√32 −1

2

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


The 45-90-45 triangle

Another important triangle is the right triangle with both acute anglesequal to 45◦. Since this triangle has two equal angles then it is isoceles; italso has two equal sides. We may assume that the hypotenuse has length1 and so draw this triangle on the unit circle.



Since the triangle is isoceles, its height y is equal to the base x. By thePythagorean theorem,

x2 + y2 = 1 =⇒ x2 + x2 = 1 =⇒ 2x2 = 1.

=⇒ x2 = 12 .

=⇒ x = 1√2.

Since y = x then

cos(45◦) = sin(45◦) = 1√2.



cos(45◦) = sin(45◦) =1√2.

If we wish to get rid of the radical in the denominator, we can multiplynumerator 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.



x

y

45◦

π4

(√22 ,√22

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

135◦

3π4

(−√22 ,√22

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

225◦5π4(

−√22 ,−

√22

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

315◦7π4(√

22 ,−

√22

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

405◦

9π4

(√22 ,√22

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



x

y

495◦

11π4

(−√22 ,√22

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



4 =√22

x

y

45◦

π4

(√22 ,√22

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)


cos 3π4 = −

√22 , sin π

4 =√22

x

y

135◦

3π4

(−√22 ,√22

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)


cos 5π4 = −

√22 , sin π

4 = −√22

x

y

225◦5π4(

−√22 ,−

√22

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)


cos 7π4 =

√22 , sin π

4 = −√22

x

y

315◦7π4(√

22 ,−

√22

)

(−1, 0) (1, 0)

(0,−1)

(0, 1)



We summarize this in a table.

θ θ cos(θ) sin(θ)

45◦ π4

√22

√22

135◦ 3π4 −

√22

√22

225◦ 5π4 −

√22 −

√22

315◦ 7π4

√22 −

√22


Six Functions on the Unit Circle

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





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

(End)



Lecture 4.2b, Six Functions on the Unit Circle

Dr. Ken W. Smith


2013



A central angle θ determines a point P (x, y) on the unit circle. Thex-coordinate is cos θ and the y-coordinate is sin θ. There are four othertrig functions, based on this point.

The tangent function tan θ is equal to yx = sin θ

cos θ .

If the ratio yx 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 ofthe line joining O to P .



There are three more trig functions each of which is the reciprocal of oneof the first three.

1 The secant of θ is the reciprocal of cosine; sec θ = 1x .

2 The cosecant of θ is the reciprocal of sine; csc θ = 1y .

3 The cotangent of θ is the reciprocal of tangent; cot θ = xy .

The reciprocal of a trig function with the syllable “sine” in it (sine andcosine) 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.



Since we may write the six trig functions in terms of x and y on the unitcircle then we may also write the six trig functions in terms of cosine andsine.We may also write all six trig functions as ratios of the sides of the righttriangle with short legs of length x and y and hypotenuse 1.



Here is a summary of our results.

1 The cosine function cos θ is equal to x1 .

2 The sine function sin θ is equal to y1 .

3 The tangent function tan θ is equal to yx = sin θ

cos θ .4 The secant function sec θ is equal to 1

x = 1cos θ , the reciprocal of

cosine.5 The cosecant function csc θ is equal to 1

y = 1sin θ , the reciprocal of

sine.6 The cotangent function cot θ is equal to x

y = cos θsin θ , the reciprocal of

tangent.


The six trig functions on the right triangle

If we draw an angle in general position, intersecting the unit circle atP (x, y), vertices C(0, 0), P (x, y) and R(x, 0) form a right triangle with aright angle on the x-axis at R(x, 0).We can define the six trig functions in terms of the sides of the triangle4CPR.We follow tradition: call the side CR as the “adjacent” side (close to theangle θ), RP as the “opposite” side (far from the angle θ), and CP is the“hypotenuse.”


The six trig functions on the right triangle

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



Thousands of trig students have memorized the first three trig functionsas:

1 Sin θ = OH

2 Cos θ = AH

3 Tan θ = OA

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







A side storyI first learned of SOH-CAH-TOA at Central Michigan University when anumber of calculus students wrote the letters at the top of their exams.

Since the Saginaw Chippewa tribe is located near the university and theuniversity had a course in the Ojibwe language, I wondered if Soh-cah-toawas an Ojibwe phrase or prayer.

But one of the students explained that Soh-cah-toa was merely an Englishabbreviation for the trig functions. (But they didn’t deny saying a shortprayer as they wrote it at the top of the exam!)


Central Angles and Arcs

We will look more closely at these right triangle definitions of our trigfunctions in the next presentation.

(End)



Lecture 4.2c, Practicing With Six Trig Functions

Dr. Ken W. Smith


2013



Worked problems.

1 Find all six trig functions of the angle θ = 13π3 .

Solution. Draw the unit circle and find the point on the circle givenby the angle 13π

3 radians. Since 133 = 4+ 1

3 then 13π3 = 4π+ π

3 and sothe angle 13π

3 gives the same point on the unit circle as π3 . This point

is

P (12 ,√32 ).

Therefore cos 13π3 = 1

2 , sin13π3 =

√32 and tan 13π

3 =

√3

212

=√3. The

values of secant, cosecant and tangent are reciprocals of these so thevalues of all six trig functions are:

cos(13π3 ) =√32

sin(13π3 ) = 12

tan(13π3 ) =√3

sec(13π3 ) = 2√3

csc(13π3 ) = 2

cot(13π3 ) = 1√3


http://en.wikipedia.org/wiki/Ojibway


2 Suppose sin θ = −√32 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 theangle −60◦ = −π

3 .)So

cos(5π3 ) = 12 , sin(5π3 ) = −

√32 , tan(5π3 ) = −

√3.

The reciprocals of these are

sec(5π3 ) = 2, csc(5π3 ) = − 2√3= −2

√3

3 , cot(5π3 ) = − 1√3= −

√33 .


Complementary angles

Given an angle θ we will talk about the angle we need to add to θ to makeup 180◦ or 90◦. The angle we need to add to θ to create a 180◦ angle isthe “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. Inthe figure below, with central angle θ (in black) the complementary angleof θ is drawn in blue.



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 isAH then the sine of the angle π

2 − θ is also AH ; 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”.


The Pythagorean Theorem on the unit circle

Since the points P (x, y) are on the unit circle then, by the PythagoreanTheorem, 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” oftrigonometry since they all depend on the Pythagorean theorem on theunit circle.


A Worked Problem.

Given the cosine or sine of θ we should (by the Pythagorean Theorem) beable 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 andsin(θ) = 2

3 .Solution. If sin(θ) = 2

3 then by the Pythagorean Theorem

cos(θ)2 + (23)2 = 1

=⇒ cos(θ)2 = 1− (23)2 = 9

9 −49 = 5

9 .

Since cos(θ)2 = 59 then cos(θ) = ±

√53 . 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) = −√53 . The point on the unit circle corresponding to the

angle θ is therefore P (−√53 ,

23) and with this information we can compute

the remaining four trig functions of θ.

cos(θ) = −√53

sin(θ) = 23

tan(θ) = − 2√5= −2

√5

5

sec(θ) = − 3√5= −3

√5

5

csc(θ) = 32

cot(θ) = −√52


Central Angles and Arcs

In the next presentation, we will look in depth at the sine function andtransformations of it.

(End)


Elementary Functions Two special triangles The 30-60-90 ......The 30-60-90 triangle The 30-60-90...

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