notes- trig, evaluating trigonometric functions · terminal side and the negative x-axis and it has...

1

Calculus Evaluating Trigonometric Functions [Evaluating Trigonometric Functions With Special Angles 0 < ! < "

2 ]

(Ex1.) ! ="6

Step 1: First convert the given angle, which is measured in radians, into degree

! radians6

"180!

! radians=

180!

6= 30!

Step 2: Recall the 30! - 60! - 90! special right triangle

Step 3: Since we are interested in ! = 30! , the opp = 1 , adj = 3 and hyp = 2 . Therefore

sin !6 =

opphyp = 1

2 csc !6 =

hypopp = 2

1 = 2

cos !6 =

adjhyp =

32 sec !

6 =hypadj = 2

3

tan !6 =

oppadj = 1

3 cot !6 =

adjopp =

31 = 3

(Ex2.) ! ="4


! radians4

"180!

! radians=

180!

4= 45!


Step 3: Since we are interested in ! = 45! , the opp = 1 , adj = 1 and hyp = 2 . Therefore

sin !4 =

opphyp = 1

2 csc !

4 =hypopp =

21 = 2

cos !4 =

adjhyp = 1

2 sec !

4 =hypadj =

21 = 2

tan !4 =

oppadj = 1

1 = 1 cot !4 =adjopp = 1

1 = 1

1

3

2

30!

60!

12

45!

45!

1

2

1

32

30!

60!

(Ex3.) ! ="3


! radians3

"180!

! radians=

180!

3= 60!


Step 3: Since we are interested in ! = 60! ,

the opp = 3 , adj = 1 and hyp = 2 . Therefore

sin !3 =

opphyp =

32 csc !

3 =hypopp = 2

3

cos !3 =

adjhyp = 1

2 sec !3 =

hypadj = 2

1 = 2

tan !3 =

oppadj =

31 = 3 cot !3 =

adjopp = 1

3

[Evaluating Trigonometric Functions With Quadrantal Angles ! = 0, "2 , " , 3"

2 ]

Recall: If ! is an angle in standard position and (x, y)

is the point of intersection of the terminal side and the

unit circle (a circle with radius r = 1), then

cos! = x and sin! = y

(Ex4.) ! = 0

The terminal side of ! = 0 intersects the unit circle at (1,0)

*Note: “undef” means “undefined”

sin 0 = y = 0 csc 0 = 1sin0 = 1

0 = undef

cos0 = x = 1 sec 0 = 1cos0 = 1

1 = 1

tan 0 = sin0cos0 =

yx = 0

1 = 0 cot 0 = cos0sin0 = 1

0 = undef

(x, y) = cos!,sin!( )

terminal side

(1,0)0

!2

!

3!2

3

(Ex5.) ! = "2

The terminal side of ! = "2 intersects the unit circle at (0,1)

sin !2 = y = 1 csc !

2 = 1sin!2

= 11 = 1

cos !2 = x = 0 sec !

2 = 1cos!2

= 10 = undef

tan !2 =

sin!2cos!2

= 10 = undef cot !2 =

cos!2sin!2

= 01 = 0

(Ex6.) ! = "

The terminal side of ! = " intersects the unit circle at (–1,0)

sin! = yr = 0

1 = 0 csc! = 1sin! = 1

0 = undef

cos! = xr = "1

1 = "1 sec! = 1cos! = 1

"1 = "1

tan! = sin!cos! = 0

"1 = 0 cot! = cos!sin! = "1

0 = undef

(Ex7.) ! = 3"2

The terminal side of ! = 3"2 intersects the unit circle at (0,–1)

sin 3!

2 = y = "1 csc 3!2 = 1sin 3!2

= 1"1 = "1

cos 3!2 = x = 0 sec 3!2 = 1cos 3!2

= 10 = undef

tan 3!2 = sin 3!2

cos 3!2= "1

0 = undef cot 3!2 = cos 3!2sin 3!2

= 0"1 = 0

0

!2

!

3!2

(0,!1)

0

!2

!

3!2

(!1,0)

0

!2

!

3!2

(0,1)

4

[Evaluating Trigonometric Functions With Special Angles !2 < " < 2! ]

(Ex8.) ! = 2"3

Step 1: Convert 2!3 into degree:

2!3

"180!

!= 120! , then draw the terminal side of ! = 120!

on the xy-coordinate plane

Step 2: Determine the reference angle (the acute angle formed by the terminal side and

either the positive x-axis or the negative x-axis). Here, the reference angle is formed by the

terminal side and the negative x-axis and it has 180! !120! = 60!

Step 3: Since the reference angle is 60! , we can draw the 30! - 60! - 90! special right

triangle and label the ratio of each side, attach the negative sign if the side is “to the left

of the y-axis” or “below the x-axis”. Here, the adjacent side of the reference angle is to

the left of the y-axis, thus it is negative.

Step 4: Looking at the reference angle and the special triangle, we have

opp = y = 3 , adj = x = !1, and hyp = r = 2

(Note: r is always positive). Therefore

sin 2!3 = opp

hyp =32 csc 2!3 = hyp

opp = 23

cos 2!3 = adjhyp = "1

2 sec 2!3 = hypadj = 2

"1 = "2

tan 2!3 = opp

adj =3

"1 = " 3 cot 2!3 = adjopp = "1

3

30!

3

!1

2

Step 1 Step 2 Step 3

5

(Ex9.) ! = 3"4


3!4

"180!



Step 2: Determine the reference angle, which is 180! !135! = 45!

Step 3: Since the reference angle is 45! , we can draw the 45! - 45! - 90! special right


of the y-axis” or “below the x-axis”. Here, the adjacent side of the reference angle is to

the left of the y-axis, thus it is negative.


opp = y = 1, adj = x = !1, and hyp = r = 2


sin 3!4 = opp

hyp = 12

csc 3!4 = hypopp =

21 = 2

cos 3!4 = adjhyp = "1

2 sec 3!4 = hyp

adj =2

"1 = " 2

tan 3!4 = opp

adj = 1"1 = "1 cot 3!4 = adj

opp = "11 = "1

45!1

!1

2


6

(Ex10.) ! = 7"6


7!6

"180!



Step 2: Determine the reference angle, Here, the reference angle is formed by the terminal

side and the negative x-axis and it has 210! !180! = 30!

Step 3: : Since the reference angle is 30! , we can draw the 30! - 60! - 90! special right


of the y-axis” or “below the x-axis”. Here, both adjacent and opposite sides of the

reference angle are negative.


opp = y = !1 , adj = x = ! 3 , and hyp = r = 2


sin 7!6 = opp

hyp = "12 csc 7!6 = hyp

opp = 2"1 = "2

cos 7!6 = adjhyp =

" 32 sec 7!6 = hyp

adj = 2" 3

tan 7!6 = opp

adj = "1" 3 =

13

cot 7!6 = adjopp =

" 3"1 = 3

!1

! 3

2


7

(Ex11.) ! = 5"3


5!3

"180!



Step 2: Determine the reference angle, Here, the reference angle is formed by the terminal

side and the positive x-axis and it has 360! ! 300! = 60!

Step 3: : Since the reference angle is 60! , we can draw the 30! - 60! - 90! special right


of the y-axis” or “below the x-axis”. Here, the opposite sides of the reference angle is

below the x-axis, thus it is negative.


opp = y = ! 3 , adj = x = 1, and hyp = r = 2


sin 5!3 = opp

hyp =" 32 csc 5!3 = hyp

opp = 2" 3

cos 5!3 = adjhyp = 1

2 sec 5!3 = hypadj = 2

1 = 2

tan 5!3 = opp

adj =" 31 = " 3 cot 5!3 = adj

opp = 1" 3

1

! 32


8

[Evaluating Trigonometric Functions With Negative Special Angles]

When evaluating trig functions with negative angles, we follow the same steps: draw the terminal

side, find the reference angle, and then use the special triangles. The only matter is that

negative angles travel “clockwise”

Positive angle θ travels “counterclockwise”

from the positive x-axis

Negative angle –θ travels “clockwise” from

the positive x-axis

(Ex12.) ! = " #4

! = " #4 = "45! , reference angle 45! , so we have

opp = y = !1 , adj = x = 1, and hyp = r = 2

sin !"4 = opp

hyp = !12

csc !"4 = hyp

opp =2

!1 = ! 2

cos !"4 = adj

hyp = 12

sec !"4 = hyp

adj =21 = 2

tan !"4 = opp

adj = !11 = !1 cot !"4 = adj

opp = 1!1 = !1

!!"

1

!12

9

(Ex13.) ! = " 11#6

! = " 11#6 = "11#

6 $ 180!# = "330! , reference angle 30!

opp = y = 1, adj = x = 3 , and hyp = r = 2

sin !11"6 = opp

hyp = 12 csc !11"

6 = hypopp = 2

1 = 2

cos !11"6 = adj

hyp =32 sec !11"

6 = hypadj = 2

3

tan !11"6 = opp

adj = 13

cot !11"6 = adjopp =

31 = 3

[Evaluating Trigonometric Functions With Trigonometric Identities]

(Ex14.) sin(15! )

“Warning: Although we found that sin 30! = 1

2 , this does NOT imply that sin(15! ) = 1

4 ”

We will calculate sin(15! ) by using two trigonometric identities

Method 1: Use sin2 !2

"#$

%&'=1( cos!

2 and let ! = 30!

sin(15! ) = ± sin2 30!

2!"#

$%&

= ±1' cos 30!

2

= ±1' 3

2

2

= ±2 ' 34

= ±2 ' 32

Since sin! " 0 when 0! !" ! 90! , so we have

sin(15! ) = 2 ! 3

2" 0.2588

1

3

2

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Method 2: Use sin(! " #) = sin! cos# " cos! sin# and let ! = 45! and ! = 30!

sin(15! ) = sin(45! ! 30! )= sin 45! cos 30! ! cos 45! sin 30!

= 12" 32 ! 1

2" 12

= 3!12 2

# 0.2588

Remark: It is a good problem to show that 2 ! 32

=3 !12 2

algebraically

(Ex15.) cos(105! )

Method 1: Use cos(! + ") = cos! cos" # sin! sin" and let ! = 60! and ! = 45!

cos(105! ) = cos(60! + 45! )= cos(60! )cos(45! )! sin(60! )sin(45! )

= 12 "

12! 3

2 " 12

= 1! 32 2

# !0.2588

Method 2: Use cos(90! !" ) = sin" and let ! = "15!

cos(105! ) = cos 90! ! (!15! )( )= sin(!15! )

= ! 3!12 2

" !0.2588

notes- trig, evaluating trigonometric functions · terminal side and the negative x-axis and it has...

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