#$&* course Mth 164 Question: `q001. Note that there are four questions in this Assignment.
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Given Solution: `aThe Pythagorean Theorem applies to any point (x,y) on the unit circle, where we can construct a right triangle with horizontal and vertical legs x and y and hypotenuse equal to the radius r of the circle. Thus by the Pythagorean Theorem we have x^2 + y^2 = r^2. Now since sin(theta) = y/r and cos(theta) = x/r, we have sin^2(theta) + cos^2(theta) = (y/r)^2 + (x/r)^2 = y^2/r^2 + x^2/r^2 = (y^2 + x^2) / r^2 = r^2 / r^2 = 1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Using the fact that sin^2(theta) + cos^2(theta) = 1, prove that tan^2(theta) + 1 = sec^2(theta). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: tan^2(theta) + 1 = sec^2(theta) sin^2(theta)/cos^2(theta) + 1 = 1 / cos^2(theta) sin^2(theta)/cos^2(theta) * cos^2(theta)+ 1 * cos^2(theta)= 1 / cos^2(theta) * cos^2(theta) sin^2(theta) + cos^2(theta) = 1 confidence rating #$&*: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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Prove that csc^2(theta) - cot^2(theta) = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: csc^2(theta) - cot^2(theta) = 1 1/sin^2(theta) - cos^2/sin^2(theta) = 1 - cos^2(theta) = sin^2(theta) sin^2(theta) + cos^2(theta) = 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aRewriting in terms of sines and cosines we get 1 / sin^2(theta) - cos^2(theta)/sin^2(theta) = 1. We now multiply through by the common denominator sin^2(theta) to get 1 / sin^2(theta) * sin^2(theta) - cos^2(theta)/sin^2(theta) * sin^2(theta) = 1 * sin^2(theta), or 1 - cos^2(theta) = sin^2(theta). This is easily rearranged to give us sin^2(theta) + cos^2(theta) = 1, which we know to be true. The original equation is thus equivalent to this true equation, and is therefore true. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q004. Prove that sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sec^2(theta)csc^2(theta) - csc^2(theta) = sec^2(theta) 1/cos^2(theta)1 / sin^2(theta) - 1/sin^2(theta) = 1/cos^2(theta) sin^2(theta)cos^2(theta)1/cos^2(theta)1 / sin^2(theta) - sin^2(theta)cos^2(theta)1/sin^2(theta) = sin^2(theta)* cos^2(theta)1/cos^2(theta) 1 - cos^2(theta) = sin^2(theta) sin^2(theta) + cos^2(theta) = 1 confidence rating #$&*: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): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. Prove that sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sec^2(theta)csc^2(theta) - csc^2(theta) = sec^2(theta) 1/cos^2(theta)1 / sin^2(theta) - 1/sin^2(theta) = 1/cos^2(theta) 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) 1 - cos^2(theta) = sin^2(theta) sin^2(theta) + cos^2(theta) = 1 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aRewriting in terms of sines and cosines we get 1/cos^2(theta) * 1 / sin^2(theta) - 1/sin^2(theta) = 1/cos^2(theta). We now multiply through by the common denominator sin^2(theta)* cos^2(theta) to get sin^2(theta)* cos^2(theta) * 1/cos^2(theta) * 1 / sin^2(theta) - sin^2(theta)* cos^2(theta) * 1/sin^2(theta) = sin^2(theta)* cos^2(theta) * 1/cos^2(theta). Simplifying we get 1 - cos^2(theta) = sin^2(theta), which we rearrange to get sin^2(theta) + cos^2(theta) = 1. Note that there are other strategies for proving identities, which you will see in your text. Complete Assignment 8, including Class Notes, text problems and Web-based problems as specified on the Assts page. When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `q004. Prove that sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sec^2(theta)csc^2(theta) - csc^2(theta) = sec^2(theta) 1/cos^2(theta)1 / sin^2(theta) - 1/sin^2(theta) = 1/cos^2(theta) sin^2(theta)cos^2(theta)1/cos^2(theta)1 / sin^2(theta) - sin^2(theta)cos^2(theta)1/sin^2(theta) = sin^2(theta)* cos^2(theta)1/cos^2(theta) 1 - cos^2(theta) = sin^2(theta) sin^2(theta) + cos^2(theta) = 1 confidence rating #$&*: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): ------------------------------------------------ Self-critique rating: #*&!#*&!