course mth 164 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_.
......!!!!!!!!...................................
20:31:20 Previous Assignments: Be sure you have completed Assignments 6 and 7 as instructed under the Assts link on the homepage at 164.106.222.236 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:44:12 `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.
......!!!!!!!!...................................
RESPONSE --> sin=y/1+cos=x/1=1 sqrt(tan)=1 y/x=tan a^2+b^2=c^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:44:44 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:47:22 `q002. Using the fact that sin^2(theta) + cos^2(theta) = 1, prove that tan^2(theta) + 1 = sec^2(theta).
......!!!!!!!!...................................
RESPONSE --> y/r+x/r=1 y/x+1=r/x confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:48:22 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).
......!!!!!!!!...................................
RESPONSE --> ok r = 1 because of the radius of the unit circle self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:52:51 `q003. Prove that csc^2(theta) - cot^2(theta) = 1.
......!!!!!!!!...................................
RESPONSE --> r/x-x/y=1 Therefore 1+x/y=r/x r=1 on unit circle confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:53:30 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.
......!!!!!!!!...................................
RESPONSE --> ok used x and ys instead of sin and cos self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:55:28 `q004. Prove that sec^2(theta) * csc^2(theta) - csc^2(theta) = sec^2(theta).
......!!!!!!!!...................................
RESPONSE --> sec*csc-csc=sec csc subtracts to leave result confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:56:28 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.
......!!!!!!!!...................................
RESPONSE --> ok misread problem as csc -csc self critique assessment: 2
................................................."