Download Plane Trig Part 2

Plane Trigonometry Part 2 (6232b)
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 counterclockwise 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 points are negative.
3) Briefly explain quadrantal angles and coterminal angles.
ANSWER
coterminal angles.
Two angles drawn in standard position and have a common terminal side are
called coterminal angles.
In the figure shown, 50° and 410° are coterminal angles.
Quadrantal Angles:
These angles have terminal side coinciding with a coordinate axis, thus a
trigonometric functional value of such an angle is determined by the coordinates
of the point where the terminal side (axis) intersects the unit circle. Recall that, by
definition, the point (x,y) on the unit circle corresponds to(cos  ,sin  ).That is,
the angles 0°, 90°, 180°, 270°, 360°, 450°, –90°, –180°, –270°, –360°, ...
The t- ratios of the quadrantal angles are:-
4) Express the following in Radians
a) 135 o
135 o * (2*pi radians / 360o) = 2.3562 radians
b) -15 o
-15 o * (2*pi radians / 360o) = -0.26180 radians
c) 60 o
60 o * (2*pi radians / 360o) = 1.0472 radians
d) 112.5 o
112.5 o * (2*pi radians / 360o) = 1.9635 radians
e) -150 o
-150 o * (2*pi radians / 360o) = -2.6180 radians
f) 1025o
1025o * (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
4 radians = 229 degrees, 10 minutes, 59 seconds
4 radians * (360 o / (2*pi radians)) = 229.1831181 o
(0.1831181 degrees) * (60 minutes / degree) = 10.9871 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
0.23 radians * (360 degrees / (2*pi radians)) = 13.17802929 degrees
(0.17802929 degrees) * (60 minutes / degree) = 10.6818 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)

radians
6

radians * (360 degrees / (2*pi radians)) = 30.0000 degrees
6
0.0000 degrees = 0)  therefore, no minutes or seconds.

Radians= 30o 0’ o”
6
d)
3
radians
2
3
radians * (360 degrees / (2*pi radians)) = 270.0000 degrees
2
(270.0000 degrees - 270 degrees = 0) - therefore, no minutes or seconds.
3
= 2700 0’ 0”
2
e) -1.4 radians
-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
9.6 radians * (360 degrees / (2*pi radians)) = 550.0394833 degrees
550.0394833 degrees => 550 degrees
(550.0394833 degrees - 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?
Solution:
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
Given:
Arc Length = (1 + (2/3)) ft * (12 in / ft) = 20 in
R = radius = 45 in
So
Angle = (Arc Length / C) * 360o
Angle = (Arc Length / (2*pi*R)) * 360 o
Angle = (20 in / (2*pi*45 in)) * 360o
Angle = 0.07073553 * 360 o
Angle =25.4648 o
7) Plot on Graph Paper the following points a) (3,-7) b) (-4,6) c) (0,5)
d) (6,0)
e) ( -2, -4)
f) (0,0)
Solution:
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 = (3) 2  ( 7) 2  9  49  2
29
2
b) (-4,6)
Distance =
( 4) 2  (6) 2  16  36  52  2 13
c) (0,5)
Distance =
(0) 2  (5) 2  25  5
d) (6,0)
Distance =
(6) 2  (0) 2  36  6
e) ( -2, -4)
Distance =
( 2) 2  ( 4) 2  4  16  20  2 5
f) (0,0)
Distance =
( 0) 2  ( 0 ) 2  0  0
2
and that Cos(x) is negative, find the other
2
functions of x and the value of x.
9) Given that sin x  
Solution:
Sin and cos are negative in 3rd quadrant and tan will be positive
sin x  
2
2
therefore
cos ec ( x )  
Cos(x) =
2
 2
2
1  sin 2 ( x )  1  ( 
2 2
2
1
)  1 
2
4
2
1
**given cos(x) is negative
2
sec(x)= -1/(1/√2)=-√2
tan(x)= sin(x)/cos(x) =1
cot(x)=1
cos(x) = 
---------------------------------------------------------10) Derive the identity Cot2(A) + 1 = Cosec2(A)
Solution:
For a given triangle of sides a, b, c with opposing angles A, B, C where C = 90
and c = hypotenuse
(a2 + b2) = c2
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)
Cot2(A) + 1 = (b2 / a2) + 1
Cot2(A) + 1 = (a2 + b2) / a2
But (a2 + b2) = c2
Cot2 (A) + 1 = (c2 / a2)
Cot2 (A) + 1 = (c/a)2)
Therefore,
Cot2(A) + 1 = Cosec2(A)
(cot 2 ( x )  1)
in terms of Sin(x)
cos ec 2 ( x )
11) Express
Solution:
Use identities Cot(x) = Cos(x) / Sin(x)
Sin2(x) + Cos2(x) = 1
Therefore
Cosec2(x) =
1
sin 2 ( x )
cot 2 ( x )  1  (
cos 2 ( x )
) 1
sin 2 ( x )
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
(1  2 sin 2 ( x ))
(cot 2 ( x )  1)
(1  2 sin 2 ( x ))  sin 2 ( x )
sin 2 ( x )


1
cos ec 2 ( x )
sin 2 ( x )
2
sin ( x )
(cot 2 ( x )  1)
 1  2 sin 2 ( x )
2
cos ec ( x )
12) Reduce (csc2(x) - sec2(x)) to an expression containing only tan (x).
Solution:
sin( x )
1
1
tan( x ) 
, csc( x ) 
, sec( x ) 
cos( x )
sin( x )
cos( x )
Csc2(x) - Sec2(x)
1
1


2
sin ( x ) cos 2 ( x )

cos 2 ( x )  sin 2 ( x )
sin 2 ( x )  cos 2 ( x )
sin 2 ( x )
1
cos 2 ( x )

sin 2 ( x )

1  tan 2 ( x )
sin 2 ( x )
Using identity: Sin2(x) = Tan2(x) / (1 + Tan2(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 )
1

 tan 2 ( x )
2
2
tan ( x )
tan ( x )
Csc2(x) - Sec2(x) =
1
 tan 2 ( x )
2
tan ( x )
13) Verify the following identities.
a) (sin(B)+Cos(B))(sin(B)-Cos(B)) = 2Sin2(B) - 1
Sin2(B) – Cos2(B)
= Sin2(B) - Cos2(B) + 0
= Sin2(B) - Cos2(B) + (Sin2(B) - Sin2(B))
= Sin2(B) + Sin2(B) - Cos2(B) - Sin2(B))
= 2Sin2(B) - (Cos2(B) + Sin2(B))
= 2Sin2(B) - 1
b)(1 - Cos2(y) + Sin2(y))2 + 4 Sin2(y) Cos2(y) = 4Sin2(y)
Solution:
using identity:
Cos2(y) = 1 - Sin2(y)
(1 - Cos2 y + Sin2 y)2 + 4Sin2 y Cos2 y )
= (1 - (1 - Sin2y) + Sin2 y)2 + 4Sin2 y*(1 - Sin2 y)
= (1 - 1 + Sin2 y) + Sin2 y)2 + 4Sin2 y*(1 - Sin2 y)
= (2Sin2 y)2 + 4Sin2 y - 4Sin4 y
= 4Sin4 y + 4Sin2 y - 4Sin4 y
= 4Sin2 y + (4Sin4 y - 4Sin4 y)
= 4Sin2(y)
c) Tan2(  ) Sec2(  ) - Sec2(  ) + 1 = Tan4(  )
Answer:
Sin2(  ) + Cos2(  ) = 1
Tan(  ) = Sin(  ) / Cos(  )
Sec(  ) = 1 / Cos(  )
Tan2(  ) Sec2(  ) - Sec2(  ) + 1
d)
sin  cos 

1
csc  sec 
Answer:
Csc(  ) = 1 / Sin(  )
Sec(  ) = 1 / Cos(  )
sin  cos 

csc  sec 
= Sin(  )*Sin(  ) + Cos(  )*Cos(  )
= Sin2(  ) + Cos2(  )
=1
e)Sin (x) Tan2(x) Cot3(x) = Cos(x)
Answer:
Tan(x) = Sin(x) / Cos(x)
Cot(x) = 1 / Tan(x) = Cos(x) / Sin(x)
Sin(x) Tan2(x) Cot3(x)
 sin( x ) 

sin 2 ( x ) cos 3 ( x )

cos 2 ( x ) sin 3 ( x )
sin 3 ( x )  cos 3 ( x )
 cos( x )
(sin 3 ( x )  cos 2 ( x ))
sec 2 ( x )
 csc 2 ( x )
2
(sec ( x )  1)
Answer: f)
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



 csc 2 ( x )
2
2
2
2
(1  cos ( x ))
sin ( x )
cos ( x )  sin ( x ) sin 2 ( x )
( 2 )
cos 2 ( x )
cos ( x )