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Koby

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Koby

Koby

Posted
Posted

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.

 

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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.

 

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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

 

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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

 

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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

 

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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 ]

 

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----------------------------------------------------------

 

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

 

 

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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. ]

 

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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

 

 

 

 

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gabrelov

gabrelov

Posted
Posted (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 by gabdecena
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