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## Graphing Conic Sections

Q:  Identify the center, foci, vertices, and co vertices then graph the following conic section:
x2 + 9y2 = 36

A:  Look at our given equation:  x2 + 9y2 = 36 and recognize by its form that it is an ellipse.

To match the standard form of an ellipse, we need it equal, so divide both sides by 36 to get:

x2/36 + 9y2/36 = 36/36

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## Solve the system of equations (conics)

Q:  Solve the system of equations (both of which are conic sections)
1:   x2 + y2 – 20x + 8y + 7 = 0
2:  9x2 + y2 + 4x + 8y + 7 = 0

A:  I am going to solve this by using the elimination method (since I see that I can cancel out all of the y’s)

I am going to multiply equation 1 by (-1) and then add it to equation 2:

2:  9x2 + y2 + 4x + 8y + 7 = 0

1: -x2 – y2 + 20x – 8y – 7 = 0   [after it has been multiplied by -1]

Now, add them together to get equation 3:

3:  8x2 +24x = 0

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## Find 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

This is part of your answer, but not all of it.  The calculator will only give one answer, and will not consider answers in multiple quadrants.  We will use this answer to find any other solutions for X.

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

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## Simplifying Trig Expressions Using Identities

Q:  Simplify cos4t – sin4t

A:

First thing we gotta notice is that this is a difference of squares.  cos4t is the same as (cos2t)2.  So, we need to use what we know about factoring from algebra to factor this:

cos4t – sin4t can be factored using the difference of squares formula/concept (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.

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## Amplitude, Phase Shift and Period

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

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 2π / |b|

The phase shift is “h”

The vertical shift is “k”

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

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

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 – .1π)

y = -5 cos (2*(x – .05π))

[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 = .05π (the graph is shifted .05π units to the right)

c)  The period is 2π/|b| = 2π/|2| = π

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## 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).

A:

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

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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

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Here is what we know:

Divide both sides by 360:

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