Koby Posted July 26, 2013 Share Posted July 26, 2013 Do you guys think I got these right? 1) Briefly tell when an angle is in Standard Position. ANSWER - An angle is in standard position when - 1. its vertex A is at the origin of the x-y plane, 2. its initial side AB lies along the positive x-axis, and 3. its terminal side AC has rotated away from the x-axis. The angle is POSITIVE if AC has rotated away from the x-axis in a counter-clockwise direction. The angle is NEGATIVE if AC has rotated away from the x-axis in a clockwise direction. ---------------------------------------------------------- 2) Briefly explain when an angle is in the Third Quadrant. ANSWER - The x-y plane is divided into four quadrants, defined as - Quadrant 1 - the area covered by starting at a Standard Angle of [ 0 degrees ] on the positive x-axis, and sweeping counter-clockwise by 90 degrees to [ +90 degrees ] on the positive y-axis. Quadrant 2 - the area covered by starting at a Standard Angle of [ +90 degrees ] on the positive y-axis, and sweeping counter-clockwise by 90 degrees to [ +180 degrees ] on the negative x-axis. Quadrant 3 - the area covered by starting at a Standard Angle of [ +180 degrees ] on the negative x-axis, and sweeping counter-clockwise by 90 degrees to [ +270 degrees ] on the negative y-axis. Quadrant 4 - the area covered by starting at a Standard Angle of [ +270 degrees ] on the negative y-axis, and sweeping counter-clockwise by 90 degrees to [ +360 degrees ] on the positive x-axis. So an angle is in the Third Quadrant when it is between a Standard Angle of [ +180 degrees ] and [ +270 degrees]. Alternatively, an angle is in the Third Quadrant when both the x and y coordinates for the target point are negative. ---------------------------------------------------------- 3) Briefly explain Quadrantal angles and Coterminal angles. ANSWER - Angles are "coterminal" if, in the Standard Position, they end on the same line. For example, starting from the positive x-axis, the terminal point of [ +180 degrees ] is indistinguishable from that for [ -180 degrees ], because they both end on the negative x-axis. A "quadrantal" angle is any angle which terminates on the boundary of any of the Four Quadrants, that is, on either the positive or negative x or y axis. This occur ---------------------------------------------------------- 4) Express the following in Radians - a) 135 degrees * (2*pi radians / 360 degrees) = 2.3562 radians b) -15 degrees * (2*pi radians / 360 degrees) = -0.26180 radians c) 60 degrees * (2*pi radians / 360 degrees) = 1.0472 radians d) 112.5 degrees * (2*pi radians / 360 degrees) = 1.9635 radians e) -150 degrees * (2*pi radians / 360 degrees) = -2.6180 radians f) 1025 degrees * (2*pi radians / 360 degrees) = 17.8896 radians ---------------------------------------------------------- 5) Express the following in degrees, minutes and seconds - (assuming that the figures provided are expressed in Radians) a) [ 4 radians = 229 degrees, 10 minutes, 59 seconds ] Derivation - 4 radians * (360 degrees / (2*pi radians)) =(NNN) NNN-NNNNdegrees (NNN) NNN-NNNNdegrees => 229 degrees (NNN) NNN-NNNNdegrees - 229 degrees = 0.1831181 degrees) (0.1831181 degrees) * (60 minutes / degree) = 10.9871 minutes 10.9871 minutes => 10 minutes (10.9871 minutes - 10 minutes = 0.9871 minutes) (0.9871 minutes) * (60 seconds / minute) = 59.226 seconds 59.226 seconds => 59 seconds (rounded to nearest second) Thus - 4 radians = 229 degrees, 10 minutes, 59 seconds b) [ 0.23 radians = 13 degrees, 10 minutes, 41 seconds ] Derivation - 0.23 radians * (360 degrees / (2*pi radians)) = 13.17802929 degrees 13.17802929 degrees => 13 degrees (13.17802929 degrees - 13 degrees = 0.17802929 degrees) (0.17802929 degrees) * (60 minutes / degree) = 10.6818 minutes 10.6818 minutes => 10 minutes (10.6818 minutes - 10 minutes = 0.6818 minutes) (0.6818 minutes) * (60 seconds / minute) = 40.908 seconds 40.908 seconds => 41 seconds (rounded to nearest second) Thus - 0.23 radians = 13 degrees, 10 minutes, 41 seconds c) [ pi/6 radians = 30 degrees, 0 minutes, 0 seconds ] Derivation - pi/6 radians * (360 degrees / (2*pi radians)) = 30.0000 degrees (30.0000 degrees - 30 degrees = 0) - therefore, no minutes or seconds. d) [ 3pi/2 radians =(NNN) NNN-NNNNdegrees, 0 minutes, 0 seconds ] Derivation - 3pi/2 radians * (360 degrees / (2*pi radians)) =(NNN) NNN-NNNNdegrees (270.0000 degrees - 270 degrees = 0) - therefore, no minutes or seconds. e) [ -1.4 radians = -80 degrees, -12 minutes, -51 seconds ] Derivation - -1.4 radians * (360 degrees / (2*pi radians)) = -80.21409132 degrees -80.21409132 degrees => -80 degrees (-80.21409132 degrees +80 degrees = -0.21409132 degrees) (-0.21409132 degrees) * (60 minutes / degree) = -12.8455 minutes -12.8455 minutes => -12 minutes (-12.8455 minutes - 12 minutes =-0.8455 minutes) (-0.8455 minutes) * (60 seconds / minute) = -50.73 seconds -50.73 seconds => -51 seconds (rounded to nearest second) Thus - -1.4 radians = -80 degrees, -12 minutes, -51 seconds f) [ 9.6 radians = 550 degrees, 2 minutes, 22 seconds ] Derivation - 9.6 radians * (360 degrees / (2*pi radians)) =(NNN) NNN-NNNNdegrees (NNN) NNN-NNNNdegrees => 550 degrees (NNN) NNN-NNNNdegrees - 550 degrees = 0.0394833 degrees) (0.0394833 degrees) * (60 minutes / degree) = 2.3690 minutes 2.3690 minutes => 2 minutes (2.3690 minutes - 2 minutes = 0.3690 minutes) (0.3690 minutes) * (60 seconds / minute) = 22.14 seconds 22.14 seconds => 22 seconds (rounded to nearest second) Thus - 9.6 radians = 550 degrees, 2 minutes, 22 seconds ---------------------------------------------------------- 6) What is the size, in degrees, of the angle subtended by an arc of 1 and 2/3 feet in a circle whose radius is 45 inches? ANSWER - [ Angle =25.4648 degrees ] DERIVATION - The circumference of the circle is given by - [ Circumference = C = 2*pi*R ] The angle subtended by the arc will be - [ Angle = (Arc Length / C) * 360 degrees ] The following figures are provided - [ Arc Length = (1 + (2/3)) ft * (12 in / ft) = 20 in ] [ R = radius = 45 in ] So - [ Angle = (Arc Length / C) * 360 degrees ] [ Angle = (Arc Length / (2*pi*R)) * 360 degrees ] [ Angle = (20 in / (2*pi*45 in)) * 360 degrees ] [ Angle = 0.07073553 * 360 degrees ] [ Angle =25.4648 degrees ] ---------------------------------------------------------- ---------------------------------------------------------- 8) Find the distance from the Origin to each of the points in Question 7. (If your answer is irrational, leave it in radical form). a) (3,-7) Distance = SQRT((3)^2 + (-7)^2) = SQRT(9 + 49) = 2*SQRT(29/2) b) (-4,6) Distance = SQRT((-4)^2 + (6)^2) = SQRT(16 + 36) = SQRT(52) = 2*SQRT(13) c) (0,5) Distance = SQRT((0)^2 + (5)^2) = SQRT(0 + 25) = SQRT(25) = 5 d) (6,0) Distance = SQRT((6)^2 + (0)^2) = SQRT(36 + 0) = SQRT(36) = 6 e) ( -2, -4) Distance = SQRT((-2)^2 + (-4)^2) = SQRT(4 + 16) = SQRT(20) = 2*SQRT(5) f) (0,0) Distance = SQRT((0)^2 + (0)^2) = SQRT(0 + 0) = SQRT(0) = 0 ---------------------------------------------------------- 9) Given that [ SIN X= -2 SQUARED/2 ] and that Cos(x) is negative, find the orther functions of (x) and the value of (x). [ Unable to resolve this question - need clarification if possible. ] ---------------------------------------------------------- 10) Derive the identity [ Cot^2(A) + 1 = Cosec^2(A) ] - For a given triangle of sides [ a, b, c ] with opposing angles [ A, B, C ] where [ C = 90 ] and [ c = hypotenuse ] - [ (a^2 + b^2) = c^2 ] [ Sin(A) = (a/c) ] [ Cos(A) = (b/c) ] [ Tan(A) = (a/b) ] [ Cosec(A) = 1/Sin(A) = (c/a) ] [ Sec(A) = 1/Cos(A) = (c/b) ] [ Cot(A) = 1/Tan(A) = (b/a) ] [ Cot^2(A) + 1 = (b^2 / a^2) + 1 ] [ Cot^2(A) + 1 = (a^2 + b^2) / a^2 ] But (a^2 + b^2) = c^2 [ Cot^2(A) + 1 = (c^2 / a^2) ] [ Cot^2(A) + 1 = (c/a)^2) ] [ Cot^2(A) + 1 = Cosec^2(A) ] ---------------------------------------------------------- 11) Express [ (Cot^2(x) - 1) / Cosec^2(x) ] in terms of Sin(x) - ANSWER - [ (Cot^2(x) - 1) / Cosec^2(x) = (1 - 2*Sin^2(x)) ] DERIVATION - Use identities - [ Cot(x) = Cos(x) / Sin(x) ] [ Sin^2(x) + Cos^2(x) = 1 ] Therefore - [ Cosec^2(x) = 1 / Sin^2(x) ] [ Cot^2(x) -1 = (Cos^2(x)/Sin^2(x)) - 1 ] [ Cot^2(x) -1 = (Cos^2(x) - Sin^2(x)) /Sin^2(x) ] [ Cot^2(x) -1 = (1 - Sin^2(x) - Sin^2(x)) /Sin^2(x) ] [ Cot^2(x) -1 = (1 - 2*Sin^2(x)) /Sin^2(x) ] Therefore - [ (Cot^2(x) - 1) / Cosec^2(x) = ((1 - 2*Sin^2(x)) /Sin^2(x)) / (1/ Sin^2(x)) ] [ (Cot^2(x) - 1) / Cosec^2(x) = (1 - 2*Sin^2(x))*Sin^2(x) / Sin^2(x) ] [ (Cot^2(x) - 1) / Cosec^2(x) = (1 - 2*Sin^2(x)) ] ---------------------------------------------------------- 12) Reduce (csc^2(x) - sec^2(x)) to an expression containing only tan (x). ANSWER - [ 1/Tan^2(x) - Tan^2(x) ] DERIVATION - [ Tan(x) = Sin(x) / Cos(x) ] [ Csc(x) = 1 / Sin(x) ] [ Sec(x) = 1 / Cos(x) ] [ Sin^2(x) = Tan^2(x) / (1 + Tan^2(x)) ] [ Csc^2(x) - Sec^2(x) = (1 / Sin^2(x)) - (1 / Cos^2(x)) = (Cos^2(x) - Sin^2(x)) / (Sin^2(x)*Cos^2(x)) = (1 - Sin^2(x)/Cos^2(x)) / Sin^2(x) = (1 - Tan^2(x)) / Sin^2(x) = (1 - Tan^2(x)) / (Tan^2(x) / (1 + Tan^2(x))) = (1 - Tan^2(x))*(1 + Tan^2(x)) / Tan^2(x) = (1 - Tan^4(x)) / Tan^2(x) = 1 / Tan^2(x) - Tan^2(x) ] ---------------------------------------------------------- 13) verify the following identities. a) [ (sin^2(B) -Cos^2(B)) = 2*Sin^2(B) - 1 ] ANSWER - [ Sin^2(B) - Cos^(B) = Sin^2(B) - Cos^2(B) + 0 = Sin^2(B) - Cos^2(B) + (Sin^2(B) - Sin^2(B)) = Sin^2(B) + Sin^2(B) - Cos^2(B) - Sin^2(B)) = 2*Sin^2(B) - (Cos^2(B) + Sin^2(B)) = 2*Sin^2(B) - 1 ] where - [ Cos^2(B) + Sin^2(B) = 1 ] b) (1 - Cos^2(y) + Sin^2(y))^2 + 4*Sin^2(y)*Cos^2(y) = 4*Sin^2(y) ANSWER - [ Sin^2(y) + Cos^2(y) = 1 ] [ Cos^2(y) = 1 - Sin^2(y) ] [ (1 - Cos^2(y) + Sin^2(y))^2 + 4*Sin^2(y)*Cos^2(y) = (1 - (1 - Sin^2(y)) + Sin^2(y))^2 + 4*Sin^2(y)*(1 - Sin^2(y)) = (1 - 1 + Sin^2(y)) + Sin^2(y))^2 + 4*Sin^2(y)*(1 - Sin^2(y)) = (2*Sin^2(y))^2 + 4*Sin^2(y) - 4*Sin^4(y) = 4*Sin^4(y) + 4*Sin^2(y) - 4*Sin^4(y) = 4*Sin^2(y) + (4*Sin^4(y) - 4*Sin^4(y)) = 4*Sin^2(y) ] c) Tan^2(A)*Sec^2(A) - Sec^2(A) + 1 = Tan^4(A) ANSWER - [ Sin^2(A) + Cos^2(A) = 1 ] [ Cos^2(A) = 1 - Sin^2(A) ] [ Sin^2(A) = 1 - Cos^2(A) ] [ Tan(A) = Sin(A) / Cos(A) ] [ Sec(A) = 1 / Cos(A) ] [ Tan^2(A)*Sec^2(A) - Sec^2(A) + 1 = (Sin^2(A) / Cos^2(A))*(1 / Cos^2(A)) - (1 / Cos^2(A)) + 1 = (Sin^2(A) / Cos^4(A)) - (1 / Cos^2(A)) + 1 = (Sin^2(A) / Cos^4(A)) - (Cos^2(A) / Cos^4(A)) + 1 = (Sin^2(A) / Cos^4(A)) - (Cos^2(A) / Cos^4(A)) + (Cos^4(A) / Cos^4(A)) = (Sin^2(A) - Cos^2(A) + Cos^4(A)) / Cos^4(A) = ((1 - Cos^2(A)) - Cos^2(A) + Cos^4(A)) / Cos^4(A) = (1 - 2*Cos^2(A) + Cos^4(A)) / Cos^4(A) = (1 - 2*Cos^2(A) + (Cos^2(A))^2) / Cos^4(A) = (1 - 2*Cos^2(A) + (1 - Sin^2(A))^2) / Cos^4(A) = (1 - 2*Cos^2(A) + 1 - 2*Sin^2(A) + Sin^4(A)) / Cos^4(A) = (2 - 2*Cos^2(A) - 2*Sin^2(A) + Sin^4(A)) / Cos^4(A) = (2*(1 - (Cos^2(A) + Sin^2(A)) + Sin^4(A)) / Cos^4(A) = (2*(1 - 1) + Sin^4(A)) / Cos^4(A) = (2*(0) + Sin^4(A)) / Cos^4(A) = Sin^4(A) / Cos^4(A) = Tan^4(A) ] d) Sin(A) / Csc(A) + Cos(A) / Sec(A) = 1 ANSWER - [ Csc(A) = 1 / Sin(A) ] [ Sec(A) = 1 / Cos(A) ] [ Sin(A) / Csc(A) + Cos(A) / Sec(A) = Sin(A)*Sin(A) + Cos(A)*Cos(A) = Sin^2(A) + Cos^2(A) = 1 ] e) Sin(x)*Tan^2(x)*Cot^3(x) = Cos(x) ANSWER - [ Tan(x) = Sin(x) / Cos(x) ] [ Cot(x) = 1 / Tan(x) = Cos(x) / Sin(x) ] [ Sin(x)*Tan^2(x)*Cot^3(x) = Sin(x)*(Sin^2(x) / Cos^2(x))*(Cos^3(x) / Sin^3(x)) = (Sin^3(x)*Cos^3(x)) / (Sin^3(x)*Cos^2(x)) = Cos^3(x) / Cos^2(x) = Cos(x) ] f) Sec^2(X) / (Sec^2(X) - 1) = Csc^2(X) ANSWER - [ Csc(X) = 1 / Sin(X) ] [ Sec(X) = 1 / Cos(X) ] [ Sec^2(X) / (Sec^2(X) - 1) = (1 / Cos^2(X)) / (1 / Cos^2(X) - 1) = (1 / Cos^2(X)) / ((1 - Cos^2(X)) / Cos^2(X)) = (1 / Cos^2(X)) / (Sin^2(X) / Cos^2(X)) = (Cos^2(X) / (Cos^2(X)*Sin^2(X)) = 1 / Sin^2(X) = Csc^2(X) ] If I got some questions wrong please tell me what I topics should I study more Link to comment Share on other sites More sharing options...
alexwang32 Posted July 26, 2013 Share Posted July 26, 2013 Umm... this is a lot of homework, don't you have a TA or someone to look at it for you? And can you use latex? it's hard to distinguish the math... Link to comment Share on other sites More sharing options...
gabrelov Posted July 28, 2013 Share Posted July 28, 2013 (edited) 9) Given that [ SIN X= -2 SQUARED/2 ] and that Cos(x) is negative, find the orther functions of (x) and the value of (x). The question number 9 seems tricky, since cosine and sine are cofunctions, maybe it looks for other trigonometric function that satisfies the answer such as csc and sec noting that it says value of x meaning only one value of x of every function but (-2)2 divided by 2 is 2 and it is not possible to get 2 for cosine and sine functions in a unit circle since answer ranges only from -1 to 1 and only functions such as tangent can get numbers above and below -1 and 1. There is something wrong with the question maybe its square root of -2 over 2? Try consulting your instructor again for that question. Even you put huge number to x at cos(x) you will never get numbers above 1 or below negative 1 as an answer. Edited July 28, 2013 by gabdecena Link to comment Share on other sites More sharing options...
Koby Posted July 29, 2013 Author Share Posted July 29, 2013 Thanks! Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now