Converting between Degrees and Radians

●To convert a degree measure to radians, multiply by

●To convert a radian measure to degrees, multiply by

A.

B. -

C.

D.

ANSWERS

A. ()

=radians or

B. ()

=radians or

C. = radians () Or

D. -radians () Or -

Inverse (arc) Function

❏Functions only have inverse’s if the function is one -to - one.A

❏One-to-One function means that each y-value of the function can be matched with no more than one x- value.

❏Arcsine - inverse of sine

❏Arccosine- inverse of cosine

❏Arctangent- inverse of tangent

Sine / Cosine / Tangent
Where inverse exist / 1&4 / 1&2 / 1&4

The unit circle is necessary for most inverse problems, and remember what negative and positive in each quadrant.

Tips and tricks

❏Inverse sine and cosine cannot equal more than 1

❏COT- [ cosine/sine]

●S- opp/hypo

●C- adj/hypo

●T - opp/adj

Problems

1)arccos( - )

2)arctan(-)

3)arcsin( sin 1 )

4)arctan (4/3) and cos ( 3/hypo) *Use Pythagorean identity

Answer

1)-½

2)-

3)1

4)+ =

+ =

=

4 continued) 5= C

hypo= 5, so cos = ⅗

LAW OF SINES

-IF ABC HAS SIDE LENGTHS a, b and c REPRESENTING THE LENGTHS OF THE SIDE OPPOSITES THE ANGLES WITH MEASURES

-A, B AND C, THEN SIN A/a = SIN B/b = SIN C/c

-CAN WORK WITH; AAS, ASA AND SSA

EXAMPLE 1:

FIND C TO THE NEAREST TENTH

ANSWER:

13.1

EXAMPLE 2:

FIND c TO THE NEAREST DEGREE

ANSWER:

20

HINT:

C=SIN^-1(.342)

LAW OF COSINES

-IN ABC, IF SIDES WITH a, b and c ARE OPPOSITE ANGLES WITH MEASURES A, B AND C, RESPECTIVELY, THEN THE FOLLOWING ARE TRUE.

-a^2=b^2+c^2-2bcCosA

-b^2=a^2+c^2-2acCosB

-c^2=a^2+b^2-2abCosC

-USE FOR SAS

EXAMPLE 1:

FIND b. ROUND TO THE NEAREST TENTH

ANSWER:

9.7

EXAMPLE 2:

FIND c. ROUND TO THE NEAREST TENTH

ANSWER:

18.9

HERON’S FORMULA

EXAMPLE 1:

FIND AREA OF TRIANGLE

ANSWER:

2.9

AREA OF A TRIANGLE GIVEN SAS

-THE AREA OF A TRIANGLE IS ONE HALF THE PRODUCT OF THE LENGTHS OF TWO SIDES AND THE SIN OF THEIR INCLUDED ANGLE

EXAMPLE 1:

FIND THE AREA OF THE TRIANGLE

ANSWER:

35.67

EXAMPLE 2

FIND THE AREA OF THE TRIANGLE

ANSWER

21,217.9

THE AMBIGUOUS CASE

-BEFORE SOLVING THE PROBLEM, DETERMINE HOW MANY OF POSSIBLE SOLUTIONS THE TRIANGLE MIGHT HAVE.

EXAMPLE 1

FIND ALL SOLUTIONS TO THE GIVEN TRIANGLE, IF POSSIBLE.

ANSWER:

NOT POSSIBLE

EXAMPLE 2

FIND ALL SOLUTIONS FOR THE GIVEN TRIANGLE, IF POSSIBLE

ANSWER

ONLY ONE TRIANGLE IS POSSIBLE, 126

Find One Trig Value When Given Another

To evaluate inverse sine functions, you’ll have to use the Unit Circle or the Special Right Triangles

If you use the Unit circle, you’ll need to find the number or fraction on the Y values of the unit circle. Like for the first example, the question is sin^-1 ½. To find the inverse, you’ll have to find ½ on the Y values of the coordinates.

If you want to use the special right triangle, you’ll basically have to memorize it. Afterwards you can convert from degrees to radians. There’s two type, the 30-60-90 Right Triangle and the 45-45-90 Right Triangle

To Evaluate Inverse Cosine Functions and Inverse Tangent Functions , you’ll have to use the Unit Circle or the Special Right Triangles also

Exercise Problems

  1. arcsin(-½)
  2. sin-1√3/2
  3. sin-1 2

Solutions

  1. Because sin(-π/6) = -½ for -π/2 ≤ y ≤ π/2,

arcsin(-½) = -π/6

2. Because sin π/3 = √3/2 for -π/2 ≤ π/2,

sin-1 √3/2 = π/3

3. It is not possible to evaluate y = sin-1x when x = 2 because there is no angle whose sine is 2.

4.

5.

6.

7.


2 / ).

Amplitude and Period

What is amplitude of a trig. function?

Amplitude of a periodic function is the absolute value of half the difference of the min. & max. value of a function. * |a|

What is the period of a trig. function?

To find the period of a function, you would take *b* and put it under (denominator) 2Π as shown below

~ A general ‘sine’ formula looks like this: y = a sin(bx - c) + d ~

How Period and Amplitude look on a graph:

Problems:

Find the Amplitude and Period of these functions

1)y = 3+2 sin (3x-2)

Amplitude =

Period =

2) y = 2 + ½ sin (⅓x + 2)

Amplitude =

Period =

3) y = 4 cos 2Πx

Amplitude =

Period =

4) y = cos(x)

Amplitude =

Period =

5) y = - 3 sin(2x + 1)

Amplitude =

Period =

6)

-cos (2x-Π)+1-Π

Amplitude =

Period =

Answers!

1)Amplitude = 2

Period = 2Π/3

2) Amplitude = ½

Period = 6Π

3) Amplitude = 4

Period = 1

4) Amplitude = ½

Period = 4Π

5) Amplitude = 3 **The reason for the Amplitude being a positive 3 is due to using the Absolute Value of ‘a’, even though it is a negative number. The absolute value makes it positive thus the answer is 3 and not -3**

Period = Π

6) Amplitude = 1

Period = Π