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HONORS PRECALCULUS / BERNHARDT
Unit 4 Review Solutions
1. Two observers stand 100 feet apart on level ground. The angle of inclination
from one observe to the peak of a mountain is 10º. The angle of inclination from
the other observer is 15º. How tall is the mountain?
y
10º
100 ft
15º
x
x tan15º = y
(x + 100) tan10º = y
(x + 100) tan10º = x tan15º
x tan10º +100 tan10º = x tan15º
100 tan10º = x tan15º !x tan10º
100 tan10º = x(tan15º ! tan10º )
100 tan10º
=x
tan15º ! tan10º
" 100 tan10º %
$# tan15º ! tan10º '& tan15º = y
y ! 51.567 ft (Not much of a mountain.)
2. A bicycle is moving at 35 mph. How fast are its 20-inch diameter tires turning?
35!mi 5280!ft 12!in 1!hr 1!circumf
1!rev
!
!
!
!
!
= 588.237!rpm
1!hr
1!mi
1!ft 60!min 20" !in 1!circumf
3. If sin x = !12 / 13 and tan x < 0 , in which quadrant is x? What are the six basic
trigonometric ratios for x?
Quadrant IV, because sine is negative and tangent is negative.
opposite = –12
hypotenuse = 13
(!12)2 + x 2 = 132
x = 5 or –5
The adjacent side of the reference triangle is 5, b/c the angle is in Q4.
cos x = 5 / 13
tan x = !12 / 5
csc x = !13 / 12
sec x = 13 / 5
cot x = !5 / 12
4. Write the equation of a sinusoid that has two adjacent maxima at (1, 2) and (3, 2)
and a minimum at (6, 1).
Period = 3 – 1 = 2 (This is the horizontal distance between maxima.)
B=
2! 2!
=
=!
P
2
y=
1
3
sin (! x + c ) +
2
2
2=
1
3
sin (! (1) + c ) +
2
2
1 1
= sin (! + c )
2 2
sin !1 (1) = " + c
!
=! +c
2
!
"
=c
2
y=
1 #
!& 3
sin % ! x " ( +
2 $
2' 2
A=
2 !1 1
=
2
2
D=
2 +1 3
=
2
2
5. Find x if sec 2 x =
4
and 0 ≤ x ≤ 2π. (Hint: There are four answers.)
3
± sec 2 x = ±
sec x = ±
4
3
2
3
1
2
=±
cos x
3
cos x = ±
3
2
If cos x =
3
!
11!
, then x =
or
.
2
6
6
If cos x = !
3
5!
7!
, then x =
or
.
2
6
6
6. Find four co-terminal angles for ! =
5"
. At least one should be negative, and at
12
least one should be in degrees.
!=
5"
= 75º
12
There are an infinite number of solutions, but all solutions can be written
5!
as
+ 2k! or 75º +k ! 360º , where !k !! (i.e., k is a member of the set
12
of all integers).
7. Sketch the graphs of the six basic trig functions. Identify asymptotes, intercepts,
and extrema.
8. Practice the unit circle.
9. Prove: 1 !
sin 2 x
= cos x
1 + cos x
1 + cos x
sin 2 x
!
= cos x
1 + cos x 1 + cos x
1 + cos x ! sin 2 x
= cos x
1 + cos x
(
cos x + 1 ! sin 2 x
1 + cos x
) = cos x
cos x + cos 2 x
= cos x
1 + cos x
cos x (1 + cos x )
= cos x
1 + cos x
cos x = cos x
10. Prove:
1 ! tan x
= csc x ! sec x
sin x
1
tan x
!
= csc x ! sec x
sin x sin x
csc x !
sin x 1
"
= csc x ! sec x
cos x sin x
csc x !
1
= csc x ! sec x
cos x
csc x ! sec x = csc x ! sec x
11. Prove:
1
1
+
= !2 tan t sect
sin t ! 1 sin t + 1
sin t + 1 sin t ! 1
+
= !2 tan t sect
sin 2 t ! 1 sin 2 t ! 1
2 sin t
= !2 tan t sect
sin 2 t ! 1
2 sin t
= !2 tan t sect
1 ! cos 2 t ! 1
(
)
2 sin t
= !2 tan t sect
! cos 2 t
!2 sin t 1
"
= !2 tan t sect
cost cost
!2 tan t sect = !2 tan t sect
12. Solve on the interval [0, 2π): sin x = tan x .
sin x ! tan x = 0
sin x !
sin x
=0
cos x
1 %
"
sin x $ 1 !
' =0
#
cos x &
sin x = 0!or!1 !
1
=0
cos x
sin x = 0!or!cos x = 1
x = 0,!! !!!or!!x = 0
13. Solve on the interval [0, 2π): 4 sin 2 x = 1 .
sin 2 x =
1
4
sin 2 x =
sin x = ±
x=
1
4
1
2
! 5! 7! 11!
,! ,! ,!
6 6 6
6
14. Solve on the interval [0, 2π): cos 2x = 1 + sin x .
1 ! 2 sin 2 x = 1 + sin x
0 = sin x + 2 sin 2 x
0 = sin x (1 + 2 sin x )
0 = sin x!!or!!0 = 1 + 2 sin x
1
= sin x
2
7! 11!
x = 0,!! !!or!!x =
,!
6
6
0 = sin x!!or!!!
15. Solve the triangle(s) and the area(s): !A = 40º , a = 54, b = 62
Because 62 sin 40º < 54 < 62 , this is the ambiguous case with two triangles.
FIRST TRIANGLE
sin 40º sin B
=
54
62
!B = 47.562º
!C = 180 " ( 47.562º +40º ) = 92.438º
sin 40º sin 92.438º
=
54
c
c = 83.933
Area =
1
! 54 ! 62 !sin 92.438º = 1,672.485
2
SECOND TRIANGLE
!B = 180º "47.562º = 132.437º
!C = 180º " (132.437º +40º ) = 7.563º
sin 40º sin 7.563º
=
54
c
c = 11.057
Area =
1
! 54 ! 62 !sin 7.563º = 220.326
2
16. Solve the triangle(s) and the area(s): !A = 75º , a = 51, b = 33
Because 51 > 33, there is only one triangle.
sin 75º sin B
=
51
33
!B = 38.683º
!C = 180 " ( 38.683 + 75º ) = 66.317º
sin 75º sin 66.317º
=
51
c
c = 48.352
Area =
1
! 51! 33!sin 66.317º = 770.630
2
17. Solve the triangle(s) and the area(s): a = 6, b = 9, c = 4
6 2 = 9 2 + 4 2 ! 2 " 9 " 4 " cos A
!A = 32.089º
9 2 = 6 2 + 4 2 ! 2 " 6 " 4 " cos B
!B = 127.169º
!C = 180º " ( 32.089º +127.169º ) = 20.742º
Area =
1
! 6 ! 9 ! sin 20.742º = 9.562
2
18. Solve the triangle(s) and the area(s): !A = 60º , a = 20, b = 30
Because 30 ! sin 60º > 20 , there are no triangles.