assignment 8

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11:30:59 `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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RESPONSE --> The hypotenuse^2 = Opposite^2 + Adjacent^2, Also, Sin(theta) is the Opposite/Radius(1) and Cos(theta) is the Adjacent/Radius(1) This is can be written as: Sin^2(theta) + Cos^2(theta) = Radius^2 since the radius is known to be 1, we can write it as Sin^2(theta) + Cos^2(theta) = 1

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11:32:54 The 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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RESPONSE --> okay, I see what we did here.

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12:03:12 `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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RESPONSE --> Sec^2(theta) = (hypotenuse/adjacent)^2 Tan^2(theta) + 1 = (sin(theta)/cos(theta))^2 + 1 =sin^2(theta) / cos^2(theta) + 1 =sin^2(theta)/cos^2(theta) + cos^2(theta)/cos^2(theta) =(sin^2(theta) + cos^2(theta)) / cos^2(theta) =1/cos^2(theta) = sec^2(theta)

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12:05:54 Starting 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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RESPONSE --> check

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12:11:12 `q003. Prove that csc^2(theta) - cot^2(theta) = 1.

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RESPONSE --> csc^2(theta) - cot^2(theta) = 1 1 = 1/sin^2(theta) - cos^2(theta)/sin^2(theta) =(1 - cos^2(theta))/sin^2(theta) since 1 - cos^2(theta) = sin^2(theta) 1 = sin^2(theta)/sin^2(theta)

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12:12:10 Rewriting 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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RESPONSE --> check

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12:26:36 `q004. Prove that sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta).

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RESPONSE --> [1/cos^2(theta) * 1/sin^2(theta)] - 1/sin^2(theta) = 1/cos^2(theta) [1/cos^2(theta) * 1/sin^2(theta)] - 1/sin^2(theta) 1/[cos^2(theta) * sin^2(theta)] - 1/sin^2(theta) 1/[cos^2(theta) * sin^2(theta)] - cos^2(theta)/[cos^2(theta) * sin^2(theta)] [1 - cos^2(theta)]/[cos^2(theta) * sin^2(theta)] [1 - cos^2(theta)] = sin^2(theta) sin^2(theta)/[cos^2(theta) * sin^2(theta)] = 1/cos^2(theta) = sec^2(theta)

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12:27:47 Rewriting 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.

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RESPONSE --> I probably used the other strategy, check

You took a different route through the algebra on that last one, but it's equally valid and actually more sophisticated than the given solution. Good work

Let me know if you have questions.