Contoh Soal Algebra/Aljabar

Unit 1
1. Multiply and simplify by factoring: a) √( 15x2y) √( 6y2) ( √ is square root)
Step1: Multiply the two radicands together. √( 15x2y) √( 6y2) = √(90 x2y3)
Step2: Write √(90 x2y3) as the product of a perfect square = √(9x 2y2 )√(10y)
Step3: Take the square root of the radicand that is a perfect square = 3xy √(10y)
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2. State whether each number is rational, irrational, or complex.
a) -5/6 rational b) 8.93 rational (repeating) rational d) √ 6 irrational e) 5.2323 rational f) 3 + 4i complex
Note: Although not usually done so, a) b) c) d) and e) can be thought of as a complex number
a + bi where b = 0.
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3. Multiply: (2 – 3i)(2 + 3i)
Step1: 4 -6i + 6i – 9i2
Step2: Remember that i2 = -1
Step3: 4 -6i + 6i – 9i2 = 4 – 9 (-1) = 4 + 9 = 13
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4. Find the conjugate of 3 - 7i. The conjugate of a number a + bi is a – bi. Here, a = 3, b = -7 so the conjugate is 3 – (-7)I = 3 + 7i
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5. Divide: (-5 + 9i) by (1 – 2i)
Step1: Write as a fraction (-5 + 9i)/(1-2i)
Step2: Multiply numerator and denominator by (1 + 2i)
Step3: The numerator becomes (-5 + 9i)(1 + 2i) = -23 - i
Step4: The denominator becomes (1 -2i)(1 + 2i) = 5
Step5: = (-23/5) – (1/5)i
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6. Expand (3x – 2)6
(x+t)n = nC0xn+nC1xn−1t+nC2xn−2t2+• • •+nCn−2x2tn−2+nCn−1xtn−1+nCntn
Step1: (3x – 2)6 = 6C0(3x)6+6C1(3x)5(-2)+6C2(3x)4(-2)2+• • •+6C4(3x)2(-2)4+6C5(3x)(-2)5+6C6(-2)6
Step2: nCr = n!/(r!(n-r)!)
Step3: 6C0 = 1, 6C1 = 6 , 6C2 = 15 , 6C3 = 20 , 6C4 = 15 , 6C5 = 6, 6C6 = 1
Step4: (3x – 2)6 = (3x)6+(6)(3x)5(-2)+(15)(3x)4(-2)2+(20)(3x)3(-2)3+15(3x)2(-2)4+6(3x)(-2)5+(-2)6
Step5: (3x – 2)6 = 729x6+(6)(243x5)(-2)+(15)(81 x4)(4)+(20)(27x3)(-8)+15(9x2)(16)+6(3x)(-32)+64
Step6: (3x – 2)6 = 729x6 - 2816x5 + 4860 x4 - 4320 x3 + 2160x2 - 576x + 64
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7. Complete the square: Let q(x) = 3x2 - 24x + 50
Step1: a = 3, b = -24, c = 50
Step2: q[u - b/(2a)] = au2 - (b2 – 4ac)/4a
Step3: b/2a = -24/2(3) = -24/6 = -4
Step4: (b2 – 4ac)/4a = [(-24)(-24) – [4(3)(50]/4(3) = (576 – 600)/12 = -24/12 = -2
Step5: q(u – b/2a) = q[u – (-4 )] q(u +4) = 3u2 – (-2) = 3u2 +2 q(u+4) = 3u2 +2
Step6: x = u – (b/2a), therefore u = x + (b/2a) u = x – 4
Step7: q(u + 4) = 3u2 +2 q(x – 4 + 4) = 3u2 +2 q(x)= 3u2 +2 q(x) = 3 (x – 4)2 + 2





Unit 2
1. Are the following systems of equations consistent, inconsistent, or dependent (infinitely many solutions)?
a. 2x – y = 4, 5x – y = 13 b. 6x – 2y = 2, 9x – 3y = 3
Solution:
a) 2x – y = 4 can be written as y = 2x – 4, 5x – y = 13 can be written as y = 5x – 13
2x – 4 = 5x – 13 9 = 3x x = 3
(Another way to solve this is to write the first equation as y = 2x – 4 and substitute in the second equation 5x – (2x – 4) =13 when you solve this you also obtain x = 3)
Substituting x = 3 in the first equation we get, 2(3) – y = 4 6 – y = 4 y = 2
But when we substitute in the second equation, 5(3) – y =13 y = 2
These equations are consistent (and independent).
Answer: consistent
b) If you divide 6x – 2y = 2 by 2 3x – y = 1
If you divide 9x – 3y = 3 by 3 3x – y = 1
Since these equations reduce to the same equation, they are dependent.
Answer: dependent
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2. Solve the set of equations for x and y:
2x – y = -1
3x – 3y = -2
You can use the substitution method.
y = 2x + 1 and substitute in 3x – 3y = -2 3x - 3(2x + 1) = -2
3x -6x -3 = -2 -3x -3 =-2 -3x = 1, x = -1/3
Substituting in y = 2x + 1 y = -2 (1/3) + 1 y = -2/3 + 1 y = 1/3
Answer: x = -1/3, y = 1/3
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3. If
write a formula for x in terms of A, B, C, D, E, and F.
Solution: G(Bx + C) = A(Ex – F) è GBx + GC = AEx – AF è GC + AF = AEx – GBx
GC + AF = (AE – GB)x è Divide each side of the equation by (GC + AF)
Answer: x = (GC + AF)/( AE – GB)
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4. A third-order determinant can be defined by:
a1 b1 c1
a2 b2 c2
a3 b3 c3
= a1 times
b2 c2
b3 c3
- a2 times
b1 c1
b3 c3
+ a3 times
b1 c1
b2 c2
Find the value of the determinant:
-1 -2 -3
3 4 2
0 1 2
=
(-1) times
4 2
1 2
-(3) times
-2 -3
1 2
0
Unit 3
1. If f(x) = (125y3 – 8) and g(x) = (5y – 2), find f(x)/g(x)
Answer: 25y2 + 10y + 4
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2. When (x3 +9x2 - 5) is divided by (x2 – 1), find the value of the quotient and the remainder.
Answer: quotient is x + 9 and remainder is x + 4
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3. When (x2 – 3x +2k) is divided by x + 2, the remainder is 7. Find the value of k.
Solution:
Quotient: x - 5
Divisor: x + 2

Dividend: x2 – 3x +2k
x2 +2x
-5x + 2k
-5x – 10
Remainder: 2k + 10
2k + 10 = 7 2k = -3, k = -3/2
Answer: k = -3/2
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4. Show that the triangle whose vertices are A = (-1, 2), B = (4, -3), and C = (5, 3) is an isosceles triangle.
Solution:
Use the distance formula: d = √[(x2 – x1)2 + (y2 – y1)2]
AB = √50, BC = √37, AC = √37
Since 2 sides are equal, the triangle is isosceles.
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5. Quadrilateral PQRS has vertices P = (-1, 3), Q = (2, 3), R = (3, -4), and S = (-3, -3). What are the coordinates of the vertices if it is translated by the equations:
x’ = x – 1, y’ = y + 2?
Solution: This is demonstrated for 1 translation: P: x = -1, y = 3
x’ = x – 1 x’ = -1 – 1 = -2
y’ = y + 2 y’ = 3 + 2 = 5
P’: (x’, y’) = (-2, 5)
Answer: P’ = (-2, 5), Q’ = (1, 5) , R’ = (2, -2), S’ = (-4, -1)
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6. The x, y coordinate system is translated in such a way that the origin Q of the new system is (-3, -1).
a. What are the transformation equations?
Answer: x’ = x + 3, y’ = y + 1
b. Point A (3,4) in the original system is equivalent to what coordinates in the new system?
Solution:
Use the same method as in example 5.
Answer: (6, 5)
c. Point B (3, -2) in the new system is equivalent to what coordinates in the original system?
Solution:
x’ = x + 3 3 = x + 3 è x = 0
y’ = y + 1 -2 = y + 1 è y = -3
Answer: (0, -3)
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7. Town A is in a French-speaking country and town B is in an English speaking country. They are 100 miles apart and connected by a straight road. When giving directions to a point on the road between them the people in town A will say that it "x km from here "(i.e. from town A) while the people in town B will say that it is "x ' miles from here" (i.e. from town B). Thus every point on the road has an x- coordinate given it by town A and an x' coordinate given it by town B.
(For this problem 8 km = 5 miles)
a. If the border is 30 miles from town A the people in town B will say that it is "___________ miles from here".
b. Give a general formula for calculating the x ' (miles from B) coordinate in terms of the x (km from A) coordinate. x ' = ______________ * x + ___________________
Solution:
Assume that the relation between the two coordinate systems (x for the A-people and x ′ for the B-people) is x ′ = px + q.
Note the fact that the A-town itself would have coordinates x = 0 and x ′ = 100. Thus, we have 100 = p(0) + q q = 100.
Similarly, for the B-town, we have x ′ = 0 and x = 160 (100 miles = 160 km). So, 0 = p(160) + 100 where we have used the known value of q. Thus p = -100/160 = -5/8 and the formula is:
Answer: x ′ = (-5/8)x + 100
Now the given border is 30 miles from A, so it is 70 miles from B.
Thus the border has x ′ = 70. The question (a) was actually asking only this much!
We do more. The x-coordinate for the border (distance in km from A) is found by solving:
70 = ( -5/8)x + 100 x = 48 The border is 48 km from A

Unit 4
1. Write parametric equations for 2x + 4y = 11.
Solution:
Write 2x + 4y = 11 as 4y = -2x + 11 y = (-1/2)x + 11/4
Let x = t and substitute in y y = (-1/2)x + 11/4 y = (-1/2)(t) + 11/4 y = (11 -2t)/4
Answer: x = t and y = (11- 2t)/4
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2. Find the x- and y-intercepts of the equation: y = (2/5) x - 2
Solution:
The equation is already in the slope-intercept form so the y-intercept is -2
To find the x-intercept let y=0 and solve for x 0 = (2/5) x – 2 2 = (2/5) x x = 5
Answer: x-intercept is 5 and y-intercept is -2
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3. Write an equation in slope-intercept form of the line whose parametric equations are x = 3t - 4 and y = -t + 8.
Solution:
Write y = -t + 8 as t = 8 - y and substitute in x = 3t – 4 x = 3(8 – y) - 4 x = 24 – 3y -4
x = 20 – 3y 3y = -x + 20 y = (-1/3)x +20/3 in slope-intercept form
Answer: y = (-1/3)x + 20/3
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4. The midpoint of a line-segment is (-1, 2). One end of the segment is (-5,-3). Find the coordinates of the other end of the segment.
Solution:
Use the midpoint formula xM = (1/2)(x1 + x2), yM = (1/2)(y1 + y2)
The coordinates of the other endpoint are (x2,y2)
-1 = ½(-5 + x2) -2 = -5 + x2 x2 = 3
2 = ½(-3 + y2) 4 = -3 + y2 y2 = 7
Answer: (3, 7)
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5. Find the coordinates of the point which is 3/4 of the distance from the point (2, 3) to the point (6, -5).
Solution:
x = a1 + t(a2-a1), y=b1 + t(b2-b1) and use t=3/4
x = 2 + (3/4)(6-2) x = 2 + (3/4)(4) = 5
y = 3 + (3/4)(-5-3) = 3 + (3/4)(-8) = -3
The point is (5,-3)
To check, find the distance between (2,3) and (5-3) and the distance between (2,3) and (6,-5). When you simplify the radicals, the distance between (2,3) and (5-3) is 3√5 while the distance between (2,3) and (6,-5) is 4√5.
Answer: (5,-3)
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6. What is the slope-intercept equation of the line that contains the point (3, 4) and is perpendicular to the line y = 1/3x – 6?
Solution:
The slope of the line y = 1/3x – 6 is 1/3 so the slope of the perpendicular line will be -3.
Use the point slope formula, where (3,4) is on the new line.
y – y1 = m(x – x1) y – 4 = -3(x - 3) y – 4 = -3x +9 y = -3x + 15
Answer: y = -3x + 15
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7. Find the equation of the line whose x-intercept is -3 and y-intercept is 2.
Solution:
Use the intercepts formula: x/p + y/q = 1 where (p,0) and (0, q) are the intercepts.
(x/-3) + y/2 = 1 -2x + 3y = 6
Answer: -2x + 3y = 6
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8. Find the equation of a line through the points (5,3) and (1,7) in the slope-intercept form.
Solution:
Use the equation y2-y1=m(x2-x1)
7-3 = m(1-5) 4 = -4m m = -1
Substituting the point (5,3) in the formula y - y1= m(x-x1) y-3=-1(x-5) y-3=-x+5
y = -x + 8
Answer: y = -x + 8
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9. Consider the parametric Line 1: x = t+1, y = −3t+1 and another Line 2: x = −2s + 4, y = −s + 1. Find their common point.
Solution:

Write the two implied equations for the common point:
t + 1 = −2s + 4 and − 3t + 1 = −s + 1.
Solve the two equations for s and t. We get t = 3/7 and s = 9/7.
Substitute t = 3/7 in the parametric equations of Line 1: x = t + 1, y = −3t + 1
x = (3/7) + 1 x = 10/7
y = -3(3/7) + 1 = -9/7 + 1 = -2/7
Answer: The point is (10/7, -2/7)
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10. Consider the parametric line: x=3 t+3, y=t+2. Determine the slope-intercept equation for the same line.
Solution:

Make a suitable combination of the two equations to eliminate the parameter:
x = 3t + 3 x = 3t + 3
-3(y = t + 2) -3y = -3t - 6
x – 3y = -3 x + 3 = 3y y = (1/3)x + 1
Answer: y = (1/3)x + 1

Unit 5
1. If f(x) = 2x3 -3x2 + 1 find:

a. f(-1)

Solution:
f(-1) = 2(-1) 3 – 3(-1) 2 + 1 = 2(-1) – 3(1) + 1 = -2 – 3 + 1 = -4
b. f(3)

Solution:
f(3) = 2(3) 3 – 3(3) 2 + 1 = 2(27) – 3(9) + 1 = 28
c. f(0)

Solution:
f(0 = 2(0) 3 – 3(0) 2 + 1 = 0 – 0 + 1 = 1
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2. What is the value of ceil (3.7)? ceil (-3.7)?

Solution:
ceil(x) = n if n − 1 < x ≤ n for integer n
ceil(3.7) = 4
ceil(-3.7) = -3
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3. A rental car company charges $32 a day plus $0.20 a mile.
a. Write an equation that describes the function for this situation.
Solution:
f(x) = 32 + .20x
b. Find the charge when the returned car was driven 200 miles in one day.
Solution:
f(x) = 32 + .20x = 32 + .20 (200) = 32 + 40
Answer: $72
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4. Find the inverse function.
a. y = f(x) = 5x + 1
Solution:
Solve the equation y = f(x) for x and then replace y by x.
y = 5x + 1 x = (y – 1)/5
Replace y by x y = (x – 1)/5
Answer: y = (x – 1)/5

b. y = f(x) = (x-2)2
Solution:
y = (x - 2)2 ±√y = x – 2 x = ±√y + 2
Replace y by x
Answer: y = ±√x + 2 or y = 2 ± √x
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5. Find the equation of a circle where P (1,3) and Q (4,7) are the endpoints of a diameter.
Solution:
Method A:
The center must be the midpoint of PQ. Midpoint is (5/2, 5)
The radius must be half the diameter. Calculate the distance from (5/2, 5) to one of the points.

The distance is 5/2
Equation is (x – (5/2))2 + (y – 5)2 = 25/4
x2 – 5x + 25/4 + y2 - 10y + 25 = 25/4 x2 – 5x + y2 - 10y + 25 = 0
Answer: x2 – 5x + y2 - 10y + 25 = 0
Method B:
Use the diameter form of circle: (x − a1)(x − a2) + (y − b1)(y − b2) = 0,
where P = (a1, b1), Q= (a2, b2)
(x − a1)(x − a2) + (y − b1)(y − b2) = 0
(x − 1)(x − 4) + (y − 3)(y − 7) = 0 x2 -5x + 4 + y2 -10y +21 = 0
x2 - 5x + y2 - 10y + 25 = 0
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6. Find the center and radius of circle x2 +y2 + 4x – 6y - 3 = 0
Solution:
Complete the square.
x2 +y2 + 4x – 6y - 3 = 0 x2 + 4x + y2 – 6y = 3 ( x2 + 4x + 4) + (y2 – 6y +9) = 3 + 4 + 9 (x + 2)2 + (y - 3)2 = 16
Answer: Center is (-2, 3) with radius 4
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7. A Pythagorean triple is a sequence of three integers (a,b,c) such that a2+b2=c2. So each Pythagorean triple corresponds to a right triangle whose sides are all of integer length. For example (3,4,5) is a Pythagorean triple since 32+42=52. It is easy to check that if s and t are integers that are both odd or both even then (s2-t2)/2, s t, (s 2 +t 2)/2 is a Pythagorean triple. Thus s=3 and t=1 gives the Pythagorean triple (4, 3, 5). T he triple obtained from s=13 and t=1 is (84, 13, 85).
a. Find the Pythagorean triple that corresponds to s=37, t=1
Solution:
Substitute s=37, t=1 in the formula for the triple: (s2-t2)/2, s t, (s 2 +t 2)/2
Answer: The numbers are 37, 684, 685.
You can check that these are correct. 372 + 6842 = 6852

b. Find the values of s and t corresponding the Pythagorean triple (84, 13, 85).
Solution:
Solving (s2-t2)/2 = 84 and (s 2 +t 2)/2 = 85 2s2 = 338 s2 = 169 s = 13
Substitute in s t = 13 (the middle number) t = 1
Answer: s = 13, t = 1
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8. Find the equation of a parabola through the points (1, 0), (2,-3), and (0,5).
Solution:
The formula for a parabola is y = ax2 + bx + c
Substituting the coordinates of the 3 points, you obtain 3 simultaneous equations.
0 = a + b + c
-3 = 4a + 2b + c
5 = c
You can use Cramer’s Rule to solve these equations. a = 1, b = -6, c = 5
Answer: The equation is y = x2 – 6x + 5
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9. Is the point P (3, 4) inside, outside or on the circle with equation (x + 2)2 + (y - 3)2 = 9?
Solution:
First find the distance from the center of the circle to point P.
From the equation of the circle, the center C is at (-2, 3), and the radius r = 3.
Find the distance from C to P.
d = √([3 - (-2)] 2 + [4 - 3]2) = √(52 +12) = √(26) which is approximately equal to 5.1.
Since this distance is greater than the radius 3, the point is outside the circle.
Answer: outside the circle
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10. Find the line joining the intersection points of the two circles:
x2 + y2 -7x - 2y +7 = 0 and 3x2 +3y2 -7x + y = 0
Solution:
Method A:
Rewrite the first equation as:
3x2 + 3y2 -21x - 6y + 21 = 0
3x2 +3y2 -7x + y = 0
Subtracting, you get: -14x – 7y = -21 y = -2x + 3
Answer: y = -2x + 3
Method B:
To check, you can substitute this in one of the equations:
x2 + y2 -7x - 2y +7 = 0 x2 + (-2x + 3)2 -7x – 2(-2x + 3) +7 = 0
x2 + 4x2 -12x + 9 -7x + 4x -6 + 7 = 0 x2 – 3x + 2 = 0

Solving this equation: x = 2, x = 1
Substitute these values and solve for y.
The coordinates of the two intersection points are (2, -1) and (1,1).
Now you can write the equation of the line through these points.
m = (y2 – y1)/(x2 – x1) = (-1 -1)/(2-1) = -2
The equation of a line through a point is y – y1 = m (x2 –x1)
Use the slope m= -2 and one of the points, i.e. (1, 1)
y – 1 = -2(x – 1) y -1 = -2x + 2 y = -2x + 3

Unit 6
1. Write the equation of circle passing through points (7, 1), (-1,-5), and (6,2).
Solution:
The equation of a circle is in the form x2 + y2 + ax + by + c =0
Substituting the values of the coordinates we get 3 simultaneous equations:
-a -5b +c = -26
7a + b + c = -50
6a + 2b +c = -40
You can subtract the second equation from the first and eliminate variable c.
You can subtract the third equation from the second and eliminate variable c.
-8a - 6b = 24
a – b = -10
-8a – 6b = 24
6a – 6b = -60
Subtracting the second equation -14a = 84 a = -6
Substituting for b: a – b = -10 (-6) – b = -10 b = 4
Substituting for a and b: -a -5b +c = -26 6 -20 + c = -26 c = -12
The equation of the circle is x2 + y2 + ax + by + c = 0 x2 + y2 - 6x + 4y - 12 = 0
Complete the square for the equation to be in circle/radius form.
(x2 - 6x) + (y2 + 4y) = 12 (x2 - 6x + 9) + (y2 + 4y + 4) = 12 + 9 + 4
(x- 3)2 + (y + 2)2 = 25
Answer: (x- 3)2 + (y + 2)2 = 25
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2. Find the equation of the circle which is centered at (3,5) and which has the line x + y = 2 as a tangent.
Solution:
Find the slope of the line L: x + y = 2 y = -x + 2 m = -1
The slope of L’ is 1 (since L and L’ are perpendicular).
The equation will be y = x + b
Since the point (3, 5) will be on this line, substitute the values 5 = 3 + b b = 2
The equation of line L’ is y = x + 2
To find where L and L’ meet on the circle, solve the simultaneous equations
x + y = 2 and y = x + 2 The point of tangency is at (0 , 2).
The distance between the center of the circle and the point of tangency is equal to the radius r of the circle and is given by the distance from (3, 5) to (0, 2).
d = √(3 – 0)2 + (5 - 2)2 = √(9 + 9) = √18
Equation of the circle is (x - 3) 2 + (y - 5) 2 = 18
Answer: (x - 3) 2 + (y - 5) 2 = 18
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3. If L is the line given by the equation y = -2 -2x, are A(-2,-5) and B(4,-3) on the same side of the line? or are they on opposite sides of the line?
Solution:
Write the equation of the line y = -2 -2x in standard form ax + by + c = 0 2x + y + 2 = 0
For any point P(x, y) we can evaluate the expression f(x,y) = 2x + y + 2. Obviously, it evaluates to 0 on the line itself. f(x,y) keeps a constant sign on one side of the line.
Evaluating the point A(-2,-5), f(x,y) = 2(-2) + (-5) + 2 = -7
Evaluating the point B(4,-3), f(x,y) = 2(4) + (-3) + 2 = 7
Answer: A and B are on opposite side of the line
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4. If L is the line given by the equation y = -2 -2x, are C(4, -17) and D(5, -19) on the same side of the line? or are they on opposite sides of the line?
Evaluating the point C(4, -17), f(,y) = 2(4) + (-17) +2 = -7
Evaluating the point D(5, -19), f(x,y) = 2(5) + (-19) + 2 = -7
Answer: C and D are on opposite sides of the line
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5. The line L is given parametrically by the equations x(t) = 3t + 4, y(t) = 4t + 7 and C is the circle of radius 10 with center at (4,7).
a. What are the values of the parameter t which correspond to the points on L at which L meets C?
Solution:
The equation of the circle is (x – 4)2 + (y – 7)2 = 100 x2 -8x + 16 + y2 – 14y + 49 = 100
x2 - 8x + y2 – 14y = 35 Since L meets the circle, you can substitute x(t) = 3t + 4, y(t) = 4t + 7
(3t + 4)2 – 8(3t + 4) + (4t + 7)2 – 14(4t + 7) = 35 25t2 = 100 t2 = 4 t = 2, t = -2
Answer: t = 2, t = -2
b. What are the points of intersection of the line L and the circle C? Be sure to give both coordinates for each.
Solution:
When t = 2 x(t) = 3(2) + 4, y(t) = 4(2) + 7 x = 10, y = 15
When t = -2 x(t) = 3(-2) + 4, y(t) = 4(-2) + 7 x = -2, y = -1
Answer: (10, 15) and (-2, -1)
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6. Find the distance from the point (1, 1) to the line 3x – 3y + 1 = 0 .
Solution:
The distance from (p, q) to ax + by + c = 0 is (|ap + bq + c|)/√(a2 + b2) for point (p, q)
For the line 3x – 3y + 1, a = 3, b = -3, c = 1, p =1, q=1
Substituting in the formula (|(3)(1) + (-3)(1) + 1|)/√(32 + (-32)
1/√18 √18/18 = 3√2/18 = √2/6
Answer: √2/6
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7. Simplify the following expression. [1 - sin 4x] / [1 + sin 2x]
Solution:
[1 - sin 4x] / [1 + sin 2x] [(1 - sin2x) (1 + sin2x)]/ (1 + sin2x) 1 - sin2x = cos2x
Answer: cos2x
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8. A bicycle traveled a distance of 100 meters. The diameter of the wheel of this bicycle is 40 cm. Find the number of rotations of the wheel (to the nearest whole number).
Solution:
For every one rotation of the wheel, the bicycle moves a distance equal to the circumference of the wheel. The circumference C of the wheel is given by the formula C = πd C = 40π cm
The number of rotations N of the wheel is obtained by dividing the total distance traveled, 100 meters = 10000 cm, by the circumference.
N = 10000 cm/40 π cm = 80 rotations (rounded to the nearest unit)
Answer: 80 rotations
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9. Simplify the following trigonometric expression: [sec(x) sin 2x] / [1 + sec(x)]
Solution:
Substitute sec (x) that is in the numerator by 1/cos (x) and simplify.
[sec(x) sin 2x] / [1 + sec(x)] = sin 2x / [cos x (1 + sec (x)] = sin 2x / [cos x + 1]
Substitute sin 2x by 1 - cos 2x, factor and simplify.
= [1 - cos 2x] / [cos x + 1] = [(1 - cos x)(1 + cos x)] / [cos x + 1] = 1 - cos x
Answer: 1 - cos
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10. From a distance of 1753 feet the top of a tower forms an angle of 40 degrees with the ground. What is the height of the tower (to the nearest foot)?
(sin 40 degrees = .6428, cos 40 degrees = .7660, tan 40 degrees = .8391 ).
Solution:


tan 40° = x/1753 .8391 = x/1753 x = (.8391)( 1753) = 1471 feet
Answer: 1471 feet

Unit 7

Unit 8