# Find all solutions to a Trig Function

Q:  Find all solutions for X where 0 < X < 360 degrees and 9*cos(X) – 5 = 0.

Okay, the 0 < X < 360 just tells you we are looking for answers in one complete circle (no more, no less).

So, we first need to solve for X like so:

9*cos(X) – 5 = 0

9*cos(X) = 5

cos(X) = 5/9

X = cos-1(5/9)

Plug this into your calculator and you get:

X = 56.25 degrees

So, next, since, from our problem, we saw that:

cos(X) = 5/9

this means that the cosine value is positive.  In what quadrants are the cosine values positive??

So, X = 56.25 degrees is the answer for the angle in Quadrant 1.

To get the angle in Quadrant 4 that also works, you gotta do 360 – 56.25 = 303.75

Therefore, 303.75 degrees is another answer for X.

What this is saying is that:

cos(56.25)= 5/9 and cos(303.75) = 5/9 also.

So, X = 56.25 and 303.75

# Simplifying Trig Functions Using Identities

Q:  Simplify cos4t – sin4t

First thing we gotta notice is that this is a difference of squares.  cos4t is the same as cos2t squared.  So, we need to use what we know about factoring from algebra to factor this:

cos4t – sin4t can be factored like so (this just takes practice to see this and realize it):

(cos2t – sin2t)(cos2t + sin2t)

But, now we can use a trig identity because cos2t + sin2t = 1, so plug that in:

(cos2t – sin2t)(cos2t + sin2t)

(cos2t – sin2t)(1) = (cos2t – sin2t)

Now we gotta notice that even though it is simplified a bunch, we have another identity:

(cos2t – sin2t) = cos(2t)

And now we are as simplified as we get.

# Amplitude, Phase Shift and Period

Q:  Consider the function y = -5 cos (2x – .1*pi)

a)  Identify the amplitude

b)  Identify the phase shift

c)  Identify the period

The following formulas / concepts will work for sin and cos graphs:

y = a cos (b*(x – h) ) + k

Note:  a, b, h and k are just numbers that will affect the graph.

The amplitude is |a|

The phase shift if 2*pi / |b|

The phase shift is “h”

The vertical shift is “k”

A big thing to notice:  the “b” value is factored out in this formula.

So…. Let’s start with our actual example: y = -5 cos (2x – .1*pi)

Identify who is a, b, h, and k.  Notice, the b is not factored out, so let’s do that firt:

y = -5 cos (2x – .1*pi)

y = -5 cos (2*(x – .05*pi))

[if you multiply the 2 back through, you get the same as the original]

Now,

a = -5

b = 2

h = .05*pi

k = 0 (there is no + k at the end of the problem)

a)  The amplitude is |a| = |-5| = 5

b)  The phase shift is h = .05*pi (the graph is shifted .05*pi units to the right)

c)  The period is 2*pi/|b| = 2*pi/|2| = pi

# Trig Functions and Exact Values

Q:  Let X be an angle in quadrant III such that cos(X) = -12/13.  Find the exact values of csc(X) and cot(X).

This will be a little hard to explain, since I cannot currently provide a picture, but it should be manageable!  Draw a picture on your own paper and follow along with me.

Draw a right triangle and label one of the angles X (just not the right angle).  We are going to ignore the negative sign for now and then take it in to account later!  Since the cos(X) = 12/13, the side adjacent to X is 12 and the hypotenuse is 13.  Use the Pythagorean Theorem to find the missing side (the side opposite X).  You will find that the opposite side is 5.

So, Adj = 12, Opp = 5 and Hyp = 13

Now, let’s consider the negative sign:  Since X is a QIII angle, we know that only tangent and cotangent are positive values in that quadrant — all other trig functions in QIII are negative.

Now, we need to find the csc(X).  “csc” is the ratio of the hyp / opp [and, it will be negative]

So, csc(X) = -13/5

And, cot(x) is the ratio of the adj / opp [and, it will be positive].  So, cot(X) = 12/5

Q : Identify the x-values that are solutions of each multiple choice equation:

1.  8cos x +4 = 0

A. x= 2π/3 , 4π/3
B. x=4π/3, 5π/3
C. x = π/3 , 5π/3
D. x =π/3 , 4π/3

2. 10cos x – 5 √(3) = 0

A. x =5π/6 , 7π/6
B. x = π/6 , 11π/6
C. x = 7π/6 , 11π/6
D. x = π/6 , 7π/6

3. 18cos x – 9 = 0

A.  x = 4π/3 , 5π/3
B . x = 2π/3 , 4π/3
C . x = π/3 , 5π/3
D. x = π/3 , 4π/3

Here is what we know:

Divide both sides by 360:

Or, if we had divided both sides by 2π we get:

These equations give us our conversion factors…

To convert from degrees to radians:

Example: Covert 120 degrees to radians:

120 * π/180 = 120π/180 = 2π/3

So, 120 degrees = 2π/3 radians.

To convert from radians to degrees:

Example:  Covert 5π/6 radians to degrees:

5π/6 * (180/π) = 900π/(6π) = 150

So, 5π/6 radians = 150 degrees.

# More Trig Solving

Q : Find all solutions of the equation in the interval [0, 2pi )

2 cos² x = 13 sin x – 5

(Multiple Choice)

A. π/4 , 3π/4 , 5π/4 , 7π/4
B. π/6, 5π/ 6
C. 3π/4 , 7π/4
D. π/3 , 2π/3 , 4π/3 , 5π/3

# Trig problem

Q:  Find all solutions of the equation in the interval [0, 2pi )

tan²x – sec x = -1

(MULTIPLE CHOICE)

a. 0
b. pi/4 , 3pi/4 , 5pi/4 . 7pi/4
c. pi/6 , 5pi/6
d. 2pi/3 , pi , 4pi/3

# Exact Values for Trig Functions

If you are in trigonometry or pre-calculus, the below is something you want to memorize, write down, something! Here is a table with the “exact values” for the important angles we use:

 Degrees Radians sin(x) cos(x) tan(x) 0 (or 360) 0 (or 2π) 0 1 0 30 π/6 1/2 √(3)/2 √(3)/3 45 π/4 √(2)/2 √(2)/2 1 60 π/3 √(3)/2 1/2 √(3) 90 π/2 1 0 Undefined 180 π 0 -1 0 270 3π/2 -1 0 Undefined

Remember:  If you ever need to find csc, sec, or cot values, they are just reciprocals.  Csc is the reciprocal of sin, sec is the reciprocal of cos, and cot is the reciprocal of tan.

# Trig solutions

Q:  Find all solutions to the equation in the interval [0, 2pi): sin(2x) – tan(x) = 0