Pre-Calculus: Solved Problems and Answers

This article contains solutions to various Pre-Calculus problems, including trigonometry, circular functions, and more. It also includes a table of contents for easy navigation.

Running Head: PRE-CALCULUS
Pre-Calculus
Name of the student:
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1PRE-CALCULUS
Table of Contents
Answer 1..........................................................................................................................................2
Answer 2..........................................................................................................................................2
Answer 3..........................................................................................................................................2
Answer 4..........................................................................................................................................3
Answer 5..........................................................................................................................................3
Answer 6..........................................................................................................................................3
Answer 7..........................................................................................................................................3
Answer 8..........................................................................................................................................3
Answer 9..........................................................................................................................................4
Answer 10........................................................................................................................................4
Answer 11........................................................................................................................................4
Answer 12........................................................................................................................................4
Answer 13........................................................................................................................................4
Answer 14........................................................................................................................................5
Answer 15........................................................................................................................................5

2PRE-CALCULUS
Answer 1.
sin θ = 0.6831
Or, θ = sin−1 ( 0.6831 )=43.08636550
Answer 2.
sec θ=−4
Or, 1
cos θ =−4
Or, cos θ=¿ −1
4 =−0.25 ¿
Answer 3.
cos θ= 2
3
Or, 2
√ ¿ ¿
Or, ¿
Or, −cos2 θ=−( 1− 4
9 )
Or, cos2 θ=(1− 4
9 )
Or, cos θ = √ 5
9 = √ 5
3
Or, θ = cos−1 √ 5
3 =41.8103149+ k∗360 degrees, here, k = -1, 0, 1, …
Or, θ ¿ 41.8103149+1∗360 degrees, here, k = 1.
Or, θ = (90 -41.8103149 ¿ + 270 degrees

3PRE-CALCULUS
Or, θ = (48.1896851 + 270) degrees = 318.1896851 degrees. = ( 4 Π
15 +3 Π
4 ).
Answer 4.
tanθ=¿ √ 6
2 ¿
Or, 1
cot θ = √ 6
2
Or, cot θ=¿ 2
√6 ¿
Or, θ = cot−1 2
√6 = √ 2
3 = 0.886 radians = 50.768 degrees = 5 Π
18 = ( 5 Π
18 +2 Π) = 2 5 Π
18 .
Answer 5.
tan1200=tan ( 90+ 30 ) 0=¿−tan300=− √ 3 ¿ = -1.73205081.
Answer 6.
sin 3150=sin ( 360−45 )0=−cos 450 =−1
√2 = -0.70710678.
Answer 7.
tan 5 Π
6 = tan ( Π
2 − Π
6 ¿=−cot ( Π
6 )=−0.57735027.
Answer 8.
In one area, the lowest angle of elevation of the Sun in winter is 270 .

4PRE-CALCULUS
The minimum distance x that a plant needing full Sun can be placed from a fence that is 6
feet high. Hence,
tan270= 6
x
Or, x = 6
tan 270 = 6
0.50952545 =11.78 feet .
Therefore, the minimum distance is found to be 11.78 feet.
Answer 9.
The radius of the circle (r) = 9.5 cm. The central angle of the circle (θ) = 5 Π
12 .
The arc length of the segment of the circle = r* θ = (9.5* 5 Π
12 ¿ cm. = (9.5*1.308996939)
cm. = 12.4354709205 cm.
Answer 10.
10 tan3 x
sec x /¿ 5 tan x
sec x = 10 tan3 x
sec x * sec x
5 tan x = 2 tan2 x
Answer 11.
cos2 x−2 cos x−8
cos x −4 = cos2 x−4 cos x+2cos x −8
cos x−4 = cos x ¿ ¿ ¿ = ¿ ¿
= (cos x + 2).
Answer 12.
sin−1( −1
2 ) = - sin−1( 1
2 ) = - Π
6 radian = −300

5PRE-CALCULUS
Answer 13.
cos θ=−3
5
For, quadrant iii, the equation will be-
-sin θ = −3
5
Or, θ = 1800 + sin−1 3
5 = 1800 +36.869897650 =216.869897650
Or, sin ( θ
2 ¿=sin ( 216.869897650
2 ) =sin ( 108.4349488750 )=¿ ¿0.9486833.
Or, tan 2θ = tan (433.73979530 ¿=3.42857143
Or, cos 2 θ = (cos 433.73979530 ¿ ¿ = 0.28.
Answer 14.
2 cos ( t ) =6 cos ( t ) − √ 12
Or, 2 cos ( t )−6 cos ( t )=− √12
Or, −4 cos ( t )=− √12
Or, cos (t) = − √ 12
−4 = 2 √3
4 = √3
2
Or, t = cos−1 √3
2
Or, t = 0.86602540378 = (30¿¿ 0+ k∗3600) ¿, k = …, -1, 0, 1, …
Therefore, the solution is 300, [θ is in the range (00 , 3600 ¿ ¿ .
Answer 15.

6PRE-CALCULUS
cos 2x – cos x - 2 = 0
Or, 2 cos2 x−1−cos x−2=0
Or, 2 cos2 x−cos x−3=0
Or, 2 cos2 x+ 2cos x −3 cos x−3=0
Or, 2 cos x ¿ ¿
Or, ¿
Therefore, either ¿
Or, cos x=−1
Or, x = cos−1 (−1 )
Or, x = 1800 +k∗3600 , k = …, -1, 0, 1, …
Putting k = 0, we get, x = 1800
Otherwise, ¿
Or, 2cos x=3
Or, cos x = 3
2 = 1.5
The value of cos (x) must always lie between the range 0 and 1. Therefore, no solution
could be found in this regard.
That is why, the only solution is = 1800, [x is in the range ( 00 , 3600 ¿ ¿ .