Math 1151 Solutions for Fall 2004 Final Exam

1) In order to find all of the trigonometric ratios for the angle θ , we need to know all

three sides of the right triangle to which it belongs. The value

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sinθ = 1213 tells us that

the length of the side opposite angle θ is in the proportion 12:13 to the length of the hypotenuse. We can find the length of the side, which we’ll call x , adjacent to θ by

using the Pythagorean Theorem:

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x 2 + 122 = 132 ⇒ x 2 = 169 − 144 = 25 ⇒ x = 5 .

The value of cos θ is then given by

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x13 = 5

13 . We can also find this result by the

equivalent method of using the “Pythagorean Identity”:

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sin2θ + cos2θ = 1 ⇒ cos2θ = 1− sin2θ = 1− 1213( )

2= 169169 −

144169 = 25

169

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⇒ cosθ = 513 .

The rest of the values we seek can be calculated from these two:

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tanθ =sinθcosθ =

12 135/13 = 12

5 ;

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sin 2θ = 2sinθ cosθ = 2 ⋅ 1213 ⋅513 = 120

169 ;

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cos 2θ = cos2θ − sin2θ = 513

2− 1213

2

= 25169 −

144169 = − 119169 .

The other forms for cos 2θ could also be used here to obtain the same result.

2) We are starting from the Pythagorean Identity provided in the hint, sin2 θ + cos2 θ = 1 .

The tan2 θ , appearing as the first term in the expression we are asked to derive, suggests

that we will want to divide the Identity above through by cos2 θ :

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sin2θcos2θ

+cos2θcos2θ

=1

cos2θ⇒

sinθcosθ

2

+ 1 =1

cosθ

2

⇒ tan2θ + 1 = sec2θ .

By dividing the Identity through by sin2 θ , we obtain its second “alternative form”:

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sin2θsin2θ

+cos2θsin2θ

=1

sin2θ⇒ 1 +

cosθsinθ

2

=1sinθ

2

⇒ 1 + cot 2θ = sec2θ .

3) This pair of equations can be solved simultaneously straightaway by adding the two of them together and solving the resulting equation:

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x + y = 1000

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+ ( x − y = 8) ⇒ 2x = 1008 ⇒ x = 504

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⇒ y = 1000 − x = 496 .

4) The transformation equations to convert a point ( x, y ) in rectangular (Cartesian) coordinates into polar coordinates ( r, θ ) are

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r = x 2 + y 2 , θ = tan−1 yx

[adjusted to appropriate quadrant] .

For our point (√3 , 1 ) , we find the values

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r = ( 3)2 + 12 = 3+ 1 = 2 ,

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θ = tan−1 13

= tan−1 1/2

3 /2

= π

6 (the arctangent function places this value in the

first quadrant, which is acceptable since both x and y for our point are positive;

otherwise, if both coordinates were negative, we would use

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θ = π6 + π = 7π

6 ). So

our point is expressed in polar coordinates as ( 2, π/6 ) . 5) This product of two complex numbers is simple enough to deal with by treating it as a product of two binomials:

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(3+ 4i ) ⋅ (5+ 6i ) = 3 ⋅ 5 + 3 ⋅ 6i + 4i ⋅ 5 + 4i ⋅ 6i

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= 15 + 18i + 20i + 24i 2 = 15 + 18i + 20i + 24 ⋅ (−1) = − 9 + 38i .6) The statement in this problem is not a valid equation and does not represent an identity. This was apparently an oversight on the part of the test-maker(s). Any number of angles can be inserted into the expression to show that it is false [example:

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cos ( 2π3 − π3 ) = cos ( π3 ) = 12 ≠ cot ( 2π3 ) + tan ( π3 ) = − 1

3+ 3 ] .

7) A fairly straightforward way to solve this pair of equations (the first being linear, the second, non-linear) is to substitute the first equation into the second one and solve the resulting equation for x :

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x 2 + y 2 = x 2 + (3x − 5)2 = x 2 + (9x 2 − 30x + 25) = 10x 2 − 30x + 25 = 5

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⇒ 10x 2 − 30x + 20 = 0 ⇒ 10 ⋅ ( x 2 − 3x + 2) = 10 ⋅ ( x − 1) ⋅ ( x − 2) = 0 , which has the solutions x = 1 and x = 2 . We can insert these values into either of the two equations to find the corresponding values for y :

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x = 1 : y = 3 ⋅1− 5 = − 2 ; x = 2 : y = 3 ⋅2 − 5 = 1 . The two solutions to this set of equations are represented by the ordered pairs ( 1, −2 ) and ( 2, 1 ) . These can be seen as the coordinates of the intersection points between the line y = 3x − 5 and the circle x2 + y2 = 5 .

8) We start the demonstration of this identity by writing cot (2θ) in terms of its definition:

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cot (2θ ) =cos (2θ )sin (2θ )

=cos2θ − sin2θ2 sinθ cosθ

.

If we now divide the numerator and denominator of this ratio by sin2 θ , which is suggested by the “1” as the second term of the numerator, we obtain

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cos2θ − sin2θ2 sinθ cosθ

⋅1 sin2θ

1 sin2θ=

cos2θ

sin2 θ− sin2 θ

sin2 θ

2 sinθ cosθsin2 θ

=cot 2θ − 1

2 s i n θ cosθs i n θ ⋅ sinθ

=cot 2θ − 1

2 cotθ.

9)

The information in the problem gives us the lengths of two sides of the triangle and the angle between them (called the “included angle”) [see diagram on left]. We are thus able to find the side opposite the included angle by using the Law of Cosines:

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c 2 = a2 + b2 − 2ab cosγ = 32 + 42 − 2 ⋅3 ⋅ 4 ⋅ cos 30o = 9 + 16 − 24 ⋅ 32 = 25− 12 3

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⇒ c = 25− 12 3 ≈ 4.2154 ≈ 2.053 . In order to find the sines of the other two angles in the triangle, we can use the Law of Sines:

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sinαa

=sinβb

=sinγc

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⇒sinα3

=sin 30o

25− 12 3=

1 2

25− 12 3⇒ sinα =

3

2 25− 12 3≈ 0.7306 and

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⇒sinβ4

=1 2

25− 12 3⇒ sinβ =

2

25− 12 3≈ 0.9741 .

The area of this triangle can be found most directly [see diagram on right above] by using

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A = 12 (base) (height) = 1

2 ⋅b ⋅ a sinγ = 12 ⋅ 4 ⋅ 3 ⋅

12 = 3 .

We could also use Heron’s formula for the area of a triangle, but it is more complicated and makes it more difficult to obtain the precise answer of 3 .

10) A polynomial of degree four has four complex roots. We also know that a polynomial which has only real coefficients must have its complex roots appear in conjugate pairs a + bi and a − bi . We are given three of the roots as 1, −1 , and −i = 0 − 1 · i , so there must be a fourth root which is a complex root and the conjugate of the given complex root, making it i = 0 + 1 · i . If a polynomial has a root (or zero) r , then it has a factor ( x − r ) . Our polynomial must then have the full factorization (over the complex numbers)

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( x − 1) ⋅ ( x − [−1]) ⋅ ( x − [−i ]) ⋅ ( x − i ) = ( x − 1) ⋅ ( x +1) ⋅ ( x + i ) ⋅ ( x − i ) . We can then multiply this out, taking advantage here of the fact that we have terms of the difference of two squares:

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( x − 1) ⋅ ( x +1) ⋅ ( x + i ) ⋅ ( x − i ) = ( x 2 − 12) ⋅ ( x 2 − i 2 )

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= ( x 2 − 1) ⋅ ( x 2 − [−1]) = ( x 2 − 1) ⋅ ( x 2 + 1) = (x 2)2 − 12 = x 4 − 1 . We can see now that we have the four roots of the equation x4 − 1 = 0 , which are the four complex fourth-roots of 1 : 1 , i , −1 , and −i . 11) We note that this is an arithmetic series with a first term of a = 1 and a constant difference between terms of d = 2 . We may use the formula for the terms of an arithmetic sequence, a

n = a + ( n − 1 ) · d , to work out how many terms are in the

series:

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an = 1001 = 1+ (n − 1) ⋅ 2 ⇒ 2 (n − 1) = 1000 ⇒ n − 1 = 500 ⇒ n = 501 . From there, we can now find the sum either from the arithmetic series formula:

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Sn = n2 ⋅ 2a + (n −1) ⋅ d[ ] = 501

2 ⋅ 2 ⋅1+ (501−1) ⋅ 2[ ] = 5012 ⋅ 2 + 500 ⋅ 2[ ]

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= 5012 ⋅ 2 ⋅ 501 = 5012 or 251,001 ,

or from the formula using just the first and last terms of the series:

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Sn = n2 ⋅ a + an( ) = 501

2 ⋅ 1+ 1001( ) =501 ⋅1002

2 = 501 ⋅ 501 , as above. 12) We notice that each successive term of this series is larger than the preceding one by the same factor of 2 , making this a finite geometric series. The formula for the sum

of this series is

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sn = a ⋅ 1− rn

1− r , where a is the first term, r is the ratio between

successive terms, and n is the number of terms in the series. We have already noted that r = 2 ; the first term is a = 1 , and we see that, since the exponents of the terms start with 1 and end at 64 , the number of terms is n = 64 . The sum of this series is thus

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s64 = 2 ⋅ 1− 264

1− 2= 2 ⋅ 1− 2

64

−1= 265 − 2 .

13) In this geometric series, the ratio between successive terms is r = 1/3 and the first term is a = 1/3 . When the absolute value of the ratio is | r | < 1 , the sum for an

infinite geometric series is

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s = a1− r , so the sum for our series is

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s =13

1− 13

=1323

= 12 .

14) While the given equation is in a standard form for a conic section, it is not very convenient to extract information from it. Instead, we will “complete the squares” for the x- and y-terms to put them into “binomial square” form; from there, it is relatively simple to re-write this equation in the form for the equation of an ellipse:

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4x 2 + y 2 − 8x + 4y + 4 = 0 ⇒ 4 x 2 − 8x + y 2 + 4y = − 4

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⇒ 4 ⋅ ( x 2 − 2x + 1) + ( y 2 + 4y + 4 ) = − 4 + 4 ⋅1 + 4

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⇒ 4 ⋅ ( x − 1)2 + ( y + 2)2 = 4 ⇒( x − 1)2

1+( y + 2)2

4= 1 .

We can read information about the ellipse from this last equation fairly easily: we find that the center of the ellipse lies at ( 1, −2 ) . 15) To find the three complex roots of the equation x3 − 1 = 0 is equivalent to finding

the solutions of x3 = 1 , which are the three complex cube-roots of 1 . To accomplish this, we need to apply DeMoivre’s Theorem for roots of complex numbers. We will treat 1 as the complex number 1 + 0 · i , which we will then express in polar form as 1 · ( cos 0 + i sin 0 ) [we know the angle is zero radians, since 1 lies on the positive real number line, which is the positive x-axis in the complex plane]. DeMoivre’s Theorem then gives the three cube-roots as

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wk = zn = z1/ n = r1/ n ⋅ cos(θ + 2kπn ) + i sin(θ + 2kπ

n )

, k = 0, 1, K , (n −1) ⇒

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w0 = 11/ 3 ⋅ cos( 0 + 2 ⋅0 ⋅π3 ) + i sin( 0 + 2 ⋅0 ⋅π

3 )

= 1 ⋅ cos(0) + i sin(0)[ ] = 1 (+ 0 i ) ;

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w1 = 11/ 3 ⋅ cos( 0 + 2 ⋅1⋅π3 ) + i sin( 0 + 2 ⋅1⋅π

3 )

= 1 ⋅ cos( 2π3 ) + i sin( 2π3 )[ ] = − 12 + 32 i ;

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w2 = 11/ 3 ⋅ cos( 0 + 2 ⋅2 ⋅π3 ) + i sin( 0 + 2 ⋅2 ⋅π

3 )

= 1 ⋅ cos( 4π3 ) + i sin( 4π3 )[ ] = − 12 −32 i .

16) There are a couple of ways to show that the term x + 1 is a factor of the polynomial x49 + 1 . The less convenient of these would be to carry out the polynomial

division of ( x49 + 1 )/( x + 1 ) to find that there is no remainder (the quotient is the

alternating sum x48 − x47 + x46 − … + x2 − x + 1 ). A more efficient method is to recall that if a number x = r is a zero of a polynomial, then the term ( x − r ) is a factor of said

polynomial. So, in order to test the term ( x + 1 ) = ( x − [−1] ) , we use the value x = −1

in our polynomial to find that ( −1 )49 + 1 = ( −1 ) + 1 = 0 . We see then that ( x + 1 ) is indeed a factor of this polynomial. 17) The hint we are offered in this problem is to use the “angle-addition” formula

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sin (α + β ) = sinα cosβ + cosα sinβ . If we write 3θ as 2θ + θ , we then find that

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sin 3θ = sin (2θ +θ ) = sin 2θ cosθ + cos 2θ sinθ . We can then use the “double-angle” identities,

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sin 2θ = 2 sinθ cosθ , cos 2θ = cos2θ − sin2θ , to re-write our expression as

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sin 3θ = (2 sinθ cosθ ) ⋅ cosθ + ( cos2θ − sin2θ ) ⋅ sinθ

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= 2 sinθ cos2θ + cos2θ ⋅ sinθ − sin3θ . Finally, we can use the Pythagorean Identity, sin2 θ + cos2 θ = 1 ⇒ cos2 θ = 1 − sin2 θ ,

to eliminate the appearances of cos2 θ , leaving only terms involving powers of sin θ :

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sin 3θ = 2 sinθ ⋅ (1 − sin2θ ) + (1 − sin2θ ) ⋅ sinθ − sin3θ

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= 2 sinθ − 2 sin3θ + sinθ − sin3θ − sin3θ = 3 sinθ − 4 sin3θ . 18) Equations which has linear terms of both sin θ and cos θ cannot be solved by algebra alone. Instead, we use a “trick” based on the “angle-addition” formula

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sin (α + β ) = sinα cos β + cosα sin β . Starting from the equation

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a sinθ + b cosθ = c , we divide the entire

equation through by

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a2 + b2 to obtain

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aa2 + b2

sinθ + b

a2 + b2

cosθ = c

a2 + b2 .

cos α sin α By identifying the two factors in parentheses as indicated, we can re-write our original equation as

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cosα sinθ + sinα cosθ = sin (θ +α ) = ca2 + b2

.

(continued)

We can find the angle α from the values we find for its sine and cosine, and the remaining equation can be solved for θ + α using the inverse sine operation. (Note that

if the absolute value of the number on the right-hand side is

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c

a2 + b2> 1 , the

equation will have no solutions at all.)

For our equation, we have

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1 sinθ − 3 cosθ = 2 , which we divide through by

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(1)2 + ( 3)2 = 1+ 3 = 2 , giving us

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12 sinθ − 3

2 cosθ = 1 . We now identify the

factor ½ with cos α and the factor −(√3)/2 with sin α ; this would put the angle in the fourth quadrant, so α = −π/3 . This leaves us to solve the equation

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sin (θ + [− π3 ]) = sin (θ − π3 ) = 1 ⇒ θ − π

3 = sin−1(1) = π2 (only value)

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⇒ θ = π2 + π

3 = 5π6 .

19) We solve this problem by again applying DeMoivre’s Theorem for roots of complex numbers, as we did in Problem 15. We first need to express the complex number in polar form. Its modulus (which corresponds to the absolute value of a real number) is

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z = r = a2 + b2 = ( 12 )2 + ( 3

2 )2 = 1

4 + 34 = 1

and its angle is given by

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θ = tan−1 ba [appropriate quadrant] = tan−1 3 2

1 2

= tan−1 3( ) = π

3 .

So the polar form of our complex number is

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z = 1 ⋅ ( cos π3 + i sin π3 ) . The two

complex square roots are then

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w0 = 11/ 2 ⋅ cos(π3 + 2 ⋅0 ⋅π

2 ) + i sin(π3 + 2 ⋅0 ⋅π

2 )

= 1 ⋅ cos( π6) + i sin( π

6)

= 3

2 + 12 i ;

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w1 = 11/ 2 ⋅ cos(π3 + 2 ⋅1⋅π

2 ) + i sin(π3 + 2 ⋅1⋅π

2 )

= 1 ⋅ cos( 7π6) + i sin( 7π

6)

= − 3

2 − 12 i .

Because these numbers are both square roots of z , it is reasonable that w

1 = −w

0 .

G. Ruffa – 6/09