Exercise 4.6.5
We want to find the tangent of the principle angle ( between 0 and π for secant ) whose secant is – 2. When labeling the triangle, the ratio of the hypotenuse to the adjacent side must be – 2. However, we cannot put – 2 on the hypotenuse, since we cannot put a negative number on the hypotenuse. We can put negative numbers on the adjacent and opposite sides, however, so we put + 2 on the hypotenuse and – 1 on the adjacent side. The opposite side will have length equal to the square root of 3. The tangent of the angle will equal the opposite side divided by the adjacent side, so the answer is the negative of the square root of three.
Exercise 4.6.4
We want to find the principle angle whose cosine is the same as the cosine of negative pi over three.
Principle angles for cosine lie between 0 and pi inclusive. The angle negative pi over three lies in quadrant IV, so it has a positive cosine. The principle angle having the same cosine as negative pi over three lies in quadrant I. Sketch the x and y axes and draw a unit circle. Draw the terminal side of the angle – π / 3. Draw a vertical line through the point where the terminal side of of the angle – π / 3 intersects the unit circle. That vertical line will intersect the circle at a point P in the second quadrant. Draw a line through P and the origin. That line is the terminal side of the angle we are looking for. It is the angle pi over 3.
The principle angle whose cosine is the same as the cosine of negative pi over three is positive pi over 3.
Exercise 4.6.3
We are asked to find the cosecant of an angle whose cosecant is – 11.6.
This is like asking ‘What is John’s name?’
The cosecant of an angle whose cosecant is – 11.6 is – 11.6.
Exercise 4.6.2
Invert the principle trigonometric functions whose graphs were drawn in problem 4.6.1:
arcsin ( x )
arcos ( x )
arctan ( x )
arccsc ( x )
arcsec ( x )
arccot ( x )
Exercise 4.6.1
Principal sine function:
Principal cosine function:
Principal tangent function
Principal cosecant function
Principal secant function
Principal cotangent function
Exercise 4.5.8
Draw a right triangle and label the adjacent side t
and the opposite side 1. Then the base
angle of the triangle will represent A,
the angle whose cotangent is t.
The only effect of the fact that A is in quadrant II is that t
will be a negative number. The length
of the hypotenuse will be 
. The other five trigonometric
functions of A may be found, now that the three sides of the triangle are
known:
Exercise 4.5.7
Since the secant equals the hypotenuse divided by the
adjacent side of the triangle, and the secant is 4, label the hypotenuse 4 and
the adjacent side 1. The opposite side
represents the y-coordinate, which is negative in quadrant IV, so label
the opposite side equal to 
. Using this
triangle, we can find the other five trigonometric functions of A.
Exercise 4.5.6
If a is the height of the tree and b is the width of the river, then the situation may be diagramed as follows using right triangles:
Then one may establish the following ratios:
So 
. Substituting this
value into the second equation yields
Exercise 4.5.5
Let a represent the height of the building and h the height of the antenna. Then the following relationships hold:
a = 50 tan 40°, a + h = 50 tan 52°. Therefore,
50 tan 40° + h = 50 tan 52°. So
h = 50 ( tan 52° - tan 40° ) » 22 ft.
Exercise 4.5.4
Let b represent the width of the river. Then 100 / b = tan 31°.
Thus, b = 100 / tan 31° » 166.4 ft.
Exercise 4.5.3
Let a denote the height of the flagpole. Then a / 50 = tan 48°, so a = 50 tan 48° » 55.5 ft.
Exercise 4.5.2
b / 3.56 = cot 39.7°, so b = 3.56 cot 39.7° = 3.56 / tan 39.7° » 4.29.
c / 3.56 = sec 39.7°, so c = 3.56 sec 39.7° = 3.56 / cos 39.7° » 4.63
Exercise 4.5.1
a / 12 = sin 45° = 
, so 
.
b / 12 = cos 45° = 
, so 
.
Exercise 4.4.5
The amplitude is | - 3 | = 3.
The vertical shift is 3.
The period is 2 π / 2 = π.
The phase shift is ( π / 2 ) / 2 = π / 4.
The base cycle of the graph may be found by solving the inequality 0 ≤ 2 x – π / 2 ≤ 2 π for x:
π / 4 ≤ x ≤ π - π / 4
or
π / 4 ≤ x ≤ 3 π / 4.
Two cycles could extend from -5 π / 4 to 3 π / 4, for example.
Exercise 4.4.4
This is the sine graph shifted one unit upward.
Exercise 4.4.3
Exercise 4.4.2
Exercise 4.4.1
Exercise 4.3.7
Exercise 4.3.6
Exercise 4.3.5
Exercise 4.3.4
Exercise 4.3.3
Exercise 4.3.2
Exercise 4.3.1
Exercise 4.2.13
(a) 68° (b) 15° (c) 65° (d) 65°
Exercise 4.2.12
Exercise 4.2.11
Exercise 4.2.10
Exercise 4.2.9
Exercise 4.2.8
cos( - A ) = cos A sin( - A ) = - sin A
tan ( - A ) = sin ( - A ) / cos ( - A ) = - sin A / cos A = - tan A
Exercise 4.2.7
Exercise 4.2.6
Exercise 4.2.5
Exercise 4.2.4
sin 30º = opp/hyp = 1 / 2
csc 30º = hyp/opp = 2
cos 30º = adj/hyp = √3 / 2
sec 30º = hyp/adj = 2 / √3
tan 30º = opp/adj = 1 / √3
cot 30º = adj/opp = √3
Exercise 4.2.3
sin 60º = opp/hyp = √3 / 2
csc 60º = hyp/opp = 2 / √3
cos 60º = adj/hyp = 1 / 2
sec 60º = hyp/adj = 2
tan 60º = opp/adj = √3
cot 60º = adj/opp = 1 / √3
Exercise 4.2.2
sin 45º = opp/hyp = √2 / 2
csc 45º = hyp/opp = √2
cos 45º = adj/hyp = √2 / 2
sec 45º = hyp/adj = √2
tan 45º = opp/adj = 1
cot 45º = adj/opp = 1
Exercise 4.2.1
sin A = cos B
tan A = cot B
sec A = csc B
Exercise 4.1.5
Exercise 4.1.4
Exercise 4.1.3
42.57º = 42º + 0.57º
Exercise 4.1.2
Exercise 4.1.1
