Assignment 8a

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course mth 164

10/24 5pm

008. IdentitiesGoals 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:

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

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

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

I am stuck here I don’t know where to go next

confidence rating #$&*: 1

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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): Where do you get r^2 / r^2 = 1 from?

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Self-critique Rating:2

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Any quantity divided by itself is 1. So r^2 / r^2 must be 1.

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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 = 1 / cos^2(theta)

Now we multiply both sides by cos^2(theta) to get

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

confidence rating #$&*: 2

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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:

Rewrite the problem first:

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

Multiply both sides by sin^2(theta):

1 - cos^2(theta) = sin^2(theta), then add cos^2(theta) to both sides to get

1 = sin^2(theta) + cos^2(theta), this proves the above equation.

confidence rating #$&*: 2

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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:

Rewrite the problem first:

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

Multiply both sides by sin^2(theta):

1 - cos^2(theta) = sin^2(theta), then add cos^2(theta) to both sides to get

1 = sin^2(theta) + cos^2(theta), this proves the above equation.

confidence rating #$&*: 2

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

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Rewrite the problem first:

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

Multiply both sides by sin^2(theta):

1 - cos^2(theta) = sin^2(theta), then add cos^2(theta) to both sides to get

1 = sin^2(theta) + cos^2(theta), this proves the above equation.

confidence rating #$&*: 2

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

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Rewrite the problem first:

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

Multiply both sides by sin^2(theta):

1 - cos^2(theta) = sin^2(theta), then add cos^2(theta) to both sides to get

1 = sin^2(theta) + cos^2(theta), this proves the above equation.

confidence rating #$&*: 2

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

.............................................

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:

#*&!#*&!#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#