Pre-Cal2 8

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course M 277

1-18 11

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

008. Identities

Goals for this Assignment include but are not limited to the following:

1. Given an identity involving sine, cosine, tangent, cosecant, secant and cotangent functions prove or disprove it using the Pythagorean identities and the definitions of these functions.

Click once more on Next Question/Answer for a note on Previous Assignments.

Previous Assignments: Be sure you have completed Assignments 6 and 7 as instructed under the Assts link on the homepage and submitted the result of the Query and q_a_ from that Assignment. Note that Assignment 7 consists of a test covering Assignments 1-5.

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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 + cos^2 is the same as (y/r)^2+(x/r)^2 where r=1

Because radius = 1 and will always be hypo side of triangle formed we can say y^2 + x^2 = r^2

 y^2 + x^2 = 1^2 , so y^2 + x^2 = 1

Back to first equation

(y/r)^2+(x/r)^2 = (y+x)^2/r^2 = y^2+x^2/r^2, we just showed that y^2+x^2=1 and we know r=1 so r^2 = 1^2 = 1

 y^2+x^2/r^2 = 1/1^2 = 1/1 = 1

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 think we have approx. the same solution, let me know if I’m missing something.

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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:

1st, tan = sin/cos and sec = 1/cos so equation tan^2(theta) + 1 = sec^2(theta) = (sin/cos)^2 + 1 = (1/cos)^2

sin^2/cos^2 + 1 = 1/cos^2

sin^2 + 1*cos^2 = cos^2/cos^2

sin^2 + cos^2 = 1, which is what we just proved last problem, QED ( right?thats what you put once right and left sides are equal? I think I remember that anyway)

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):

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Self-critique Rating:

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

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Your solution:

1st, csc = 1/sin and cot = cos/sin so equation csc^2(theta) - cot^2(theta) = 1 is same as

(1/sin)^2 – (cos/sin)^2 = 1

 (1/sin)^2 – (cos/sin)^2 = 1

1/sin^2 - cos^2/sin^2 = 1, mult through by sin^2 we get

1 - cos^2 = sin^2, which is the same equation as

1 = cos^2+sin^2, which is what we just proved in 1st problem.

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):

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Self-critique Rating:

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

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Your solution:

1st, csc = 1/sin and cot = cos/sin so equation csc^2(theta) - cot^2(theta) = 1 is same as

(1/sin)^2 – (cos/sin)^2 = 1

 (1/sin)^2 – (cos/sin)^2 = 1

1/sin^2 - cos^2/sin^2 = 1, mult through by sin^2 we get

1 - cos^2 = sin^2, which is the same equation as

1 = cos^2+sin^2, which is what we just proved in 1st problem.

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):

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Self-critique Rating:

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