Assignment 8 QA

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course Mth164

5:50 pm3/2/15

Question: `q001. Note that there are four questions in this Assignment.

In general the sine and cosine functions and tangent function are defined for a circle of radius r centered at the origin. At angular position theta we have sin(theta) = y / r, cos(theta) = x / r and tan(theta) = y / x. Using the Pythagorean Theorem show that sin^2(theta) + cos^2(theta) = 1.

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Your solution:

sin^2(theta) + cos^2(theta) = 1.

Using this theorem I’ll use a circle with the radius 1 to prove that:

sin^2(theta) + cos^2(theta) = 1.

I’ll use the angular position pi/6 which gives us the following:

Sin^2(pi/6) + cos^2(pi/6) = 1

(½)^2 + sqrt(3)/2 = 1

¼ + ¾ = 4/4 = 1

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This proves the result for theta = pi/6. It doesn't prove it for any other angle.

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confidence rating #$&*: 3

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Given Solution:

`aThe Pythagorean Theorem applies to any point (x,y) on the unit circle, where we can construct a right triangle with horizontal and vertical legs x and y and hypotenuse equal to the radius r of the circle. Thus by the Pythagorean Theorem we have x^2 + y^2 = r^2.

Now since sin(theta) = y/r and cos(theta) = x/r, we have

sin^2(theta) + cos^2(theta) = (y/r)^2 + (x/r)^2 =

y^2/r^2 + x^2/r^2 =

(y^2 + x^2) / r^2 =

r^2 / r^2 = 1.

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Self-critique (if necessary):

I did not do this the same way but I did prove that the formula was correct. I realize in my example I do not express the radius since it is one causing it not to alter the problem but this wouldn’t be the same given a circle with a radius more than 1.

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Self-critique Rating: 3

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Question: `q002. Using the fact that sin^2(theta) + cos^2(theta) = 1, prove that tan^2(theta) + 1 = sec^2(theta).

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Your solution:

sin^2(theta) + cos^2(theta) = 1

(y/r)^2 + (x/r)^2 = 1

(Y^2 + x^2) / r^2 = 1

Which is equal to:

R^2 /r^2 = 1

tan^2(theta) + 1 = sec^2(theta).

(y/x)^2 + 1 = (r/x)^2

-1

(y/x)^2 = (r/x)^2 -1

-(r/x)^2

(y/x)^2 - (r/x)^2 = -1

(Y^2 - r^2) / x^2 = -1 Since y^2 + x^2 = r^2 we can see that y^2 - r^2 = - x^2

-X^2 / x^2 = -1

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Your solution is valid, but you do want to learn to express things in terms of sines, cosines etc..

Let's look at how your reasoning parallels the reasoning we would use with the trigonometric functions:

Let's assume that the circle is the unit circle, so that r = 1. Recall that on the unit circle model, y = sin(theta) and x = cos(theta).

Your first line could be your line

tan^2(theta) + 1 = sec^2(theta).

Your next line is

(y/x)^2 + 1 = (r/x)^2 ,

which could be written as

(sin(theta) / cos(theta) ) ^ 2 + 1 = (1 / cos(theta))^2

Your line

(y/x)^2 = (r/x)^2 -1

could be written

(sin(theta) / cos(theta))^2 = (1 / cos(theta))^2 - 1

This equation also follows from the preceding equation in sin(theta) and cos(theta).

-(r/x)^2

(y/x)^2 - (r/x)^2 = -1

This corresponds to

(sin(theta) / cos(theta) ) ^ 2 - (1 / cos(theta))^2 = -1

either by direct interpretation or just rearranging the preceding equation in sin(that) and cosine(theta).

(Y^2 - r^2) / x^2 = -1

This corresponds to

(sin^2(theta) - 1 ) / cos^2(theta))^2 = -1,

which would also follow from the preceding equation in the sine and cosine, where both terms on the left had the same denominator, allowing the numerators to be added.

Since y^2 + x^2 = r^2 we can see that y^2 - r^2 = - x^2

This says that since sin^2(theta) + cos^2(theta) = 1, we can see that sin^2(theta) - 1 = -cos^2(theta).

-X^2 / x^2 = -1

Substituting we get

-cos^2(theta) / cos^2(theta) = -1.

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confidence rating #$&*: 3

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Given Solution:

`aStarting with tan^2(theta) + 1 = sec^2(theta) we first rewrite everything in terms of sines and cosines. We know that tan(theta) = sin(theta)/cos(theta) and sec(theta) = 1 / cos(theta). So we have

sin^2(theta)/cos^2(theta) + 1 = 1 / cos^2(theta).

If we now simplify the equation, multiplying both sides by the common denominator cos^2(theta), we get

sin^2(theta)/cos^2(theta) * cos^2(theta)+ 1 * cos^2(theta)= 1 / cos^2(theta) * cos^2(theta).

We easily simplify this to get

sin^2(theta) + cos^2(theta) = 1,

which is thus seen to be equivalent to the original equation tan^2(theta) + 1 = sec^2(theta).

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Self-critique (if necessary):

I’m not sure that I accomplished the overall concept correctly.

I used another method and I believe I still proved the given equation.

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Your argument is very good, and can be translated directly to an argument in sines and cosines, per my preceding note.

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Self-critique Rating: 2

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Question: `q003. Prove that csc^2(theta) - cot^2(theta) = 1.

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Your solution:

csc^2(theta) - cot^2(theta) = 1

r^2 / sin^2 - cos^2 / sin^2 = 1

(r^2 - cos^2) / sin^2 = 1

OR

r^2 / y^2 - x^2 / y^2 = 1

(r^2 - x^2)/y^2 = 1 since x^2 + y^2 = r^2 we also have y^2 = r^2 - x^2

Y^2 /y^2 = 1

confidence rating #$&*: 3

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Given Solution:

`aRewriting in terms of sines and cosines we get

1 / sin^2(theta) - cos^2(theta)/sin^2(theta) = 1.

We now multiply through by the common denominator sin^2(theta) to get

1 / sin^2(theta) * sin^2(theta) - cos^2(theta)/sin^2(theta) * sin^2(theta) = 1 * sin^2(theta), or

1 - cos^2(theta) = sin^2(theta).

This is easily rearranged to give us sin^2(theta) + cos^2(theta) = 1, which we know to be true. The original equation is thus equivalent to this true equation, and is therefore true.

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Self-critique (if necessary):

Once again I used a different method than the one in the given solution but still found the equation to be true.

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Self-critique Rating: 3

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Question: `q004. Prove that sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta).

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Your solution:

sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta).

Since this state:

csc^2(theta) - csc^2(theta)

that gives us:

sec^2(theta) = sec^2(theta).

confidence rating #$&*: 2

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Given Solution:

`aRewriting in terms of sines and cosines we get

1/cos^2(theta) * 1 / sin^2(theta) - 1/sin^2(theta) = 1/cos^2(theta).

We now multiply through by the common denominator sin^2(theta)* cos^2(theta) to get

sin^2(theta)* cos^2(theta) * 1/cos^2(theta) * 1 / sin^2(theta) - sin^2(theta)* cos^2(theta) * 1/sin^2(theta) = sin^2(theta)* cos^2(theta) * 1/cos^2(theta).

Simplifying we get

1 - cos^2(theta) = sin^2(theta), which we rearrange to get

sin^2(theta) + cos^2(theta) = 1.

Note that there are other strategies for proving identities, which you will see in your text.

Complete Assignment 8, including Class Notes, text problems and Web-based problems as specified on the Assts page.

When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_.

"

Self-critique (if necessary):

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Self-critique rating:

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Question: `q004. Prove that sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta).

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Your solution:

sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta).

Since this state:

csc^2(theta) - csc^2(theta)

that gives us:

sec^2(theta) = sec^2(theta).

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aRewriting in terms of sines and cosines we get

1/cos^2(theta) * 1 / sin^2(theta) - 1/sin^2(theta) = 1/cos^2(theta).

We now multiply through by the common denominator sin^2(theta)* cos^2(theta) to get

sin^2(theta)* cos^2(theta) * 1/cos^2(theta) * 1 / sin^2(theta) - sin^2(theta)* cos^2(theta) * 1/sin^2(theta) = sin^2(theta)* cos^2(theta) * 1/cos^2(theta).

Simplifying we get

1 - cos^2(theta) = sin^2(theta), which we rearrange to get

sin^2(theta) + cos^2(theta) = 1.

Note that there are other strategies for proving identities, which you will see in your text.

Complete Assignment 8, including Class Notes, text problems and Web-based problems as specified on the Assts page.

When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_.

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

@&

Very good.

See my note about how the x, y and r argument can be translated directly to an argument using the trigonometric functions.

*@

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Question: `q004. Prove that sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta).

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Your solution:

sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta).

Since this state:

csc^2(theta) - csc^2(theta)

that gives us:

sec^2(theta) = sec^2(theta).

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aRewriting in terms of sines and cosines we get

1/cos^2(theta) * 1 / sin^2(theta) - 1/sin^2(theta) = 1/cos^2(theta).

We now multiply through by the common denominator sin^2(theta)* cos^2(theta) to get

sin^2(theta)* cos^2(theta) * 1/cos^2(theta) * 1 / sin^2(theta) - sin^2(theta)* cos^2(theta) * 1/sin^2(theta) = sin^2(theta)* cos^2(theta) * 1/cos^2(theta).

Simplifying we get

1 - cos^2(theta) = sin^2(theta), which we rearrange to get

sin^2(theta) + cos^2(theta) = 1.

Note that there are other strategies for proving identities, which you will see in your text.

Complete Assignment 8, including Class Notes, text problems and Web-based problems as specified on the Assts page.

When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_.

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

@&

Very good.

See my note about how the x, y and r argument can be translated directly to an argument using the trigonometric functions.

*@

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