€
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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 the length of the side opposite angle sin
=
12
13
tells us that
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:
The value of cos θ x
2
is then given by
+ 12
2 x
13
=
= 13
2
5
13
⇒ x
€
2
= 169 − 144 = 25 ⇒ x = 5 .
. We can also find this result by the equivalent method of using the “Pythagorean Identity”: sin
2
θ + cos
2
€
θ = 1 ⇒ cos
2
θ = 1 − sin
2
θ = 1 −
( ) 2
=
169
169
cos
=
5
13
.
The rest of the values we seek can be calculated from these two:
−
144
169
=
25
169 tan
= sin cos
=
12 13
5/13
=
12
5
; sin 2
= 2sin
cos
= 2
12
13
5
13
=
120
169
;
cos 2 θ = cos
2
θ − sin
2
θ =


5
13


2
− 
12 
 13 
2
=
25
169
θ could also be used here to obtain the same result.
−
144
169
= −
119
169
.
€
2) We are starting from the Pythagorean Identity provided in the hint, sin 2
θ + cos 2
θ = 1 .
The tan 2
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 cos 2
θ : sin 2 cos 2
θ
θ
+ cos 2 cos 2
θ
θ
=
1 cos
2
θ
⇒
 sin θ

 cos θ



2
+ 1 =


1
 cos θ



2
⇒ tan
2
θ + 1 = sec
2
θ .
By dividing the Identity through by sin 2
, we obtain its second “alternative form”: sin 2 θ sin 2 θ
+ cos 2 θ sin 2 θ
=
1 sin
2
θ
⇒ 1 +
 cos θ

 sin θ



2
=


1
 sin θ



2
⇒ 1 + cot
2
θ = sec
2
θ .
3) This pair of equations can be solved simultaneously straightaway by adding the two of them together and solving the resulting equation: x + y = 1000
+ ( x
y = 8 )
2 x = 1008
x = 504
y = 1000
x = 496 
 .

€
4) The transformation equations to convert a point ( x , y ) in rectangular (Cartesian) coordinates into polar coordinates ( r , θ ) are r = x
2
+ y
2
, θ = tan
− 1


 y x

 [adjusted to appropriate quadrant] .

For our point ( √ 3 , 1 ) , we find the values r = ( 3)
2
+ 1
2
= 3 + 1 = 2 ,
€
θ = tan − 1



1
3

 = tan

− 1



1/2
3 /2



=
π
6
(the arctangent function places this value in the x and y for our point are positive; otherwise, if both coordinates were negative , we would use
=
6
+
=
7
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:
€
( 3 + 4 i ) ⋅ ( 5 + 6 i ) = 3 ⋅ 5 + 3 ⋅ 6 i + 4 i ⋅ 5 + 4 i ⋅ 6 i
= 15 + 18 i + 20 i + 24 i
2
= 15 + 18 i + 20 i + 24 ⋅ ( − 1) = − 9 + 38 i .
€
€
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 : cos (
2
3
π
−
π
3
) = cos (
π
3
) =
1
2
≠ cot (
2 π
3
) + tan (
π
3
) = −
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 : x
2
+ y
2
= x
2
+ ( 3 x
5)
2
= x
2
+ ( 9 x
2
30 x + 25) = 10 x
2
30 x + 25 = 5
⇒ 10 x
2
− 30 x + 20 = 0 ⇒ 10 ⋅ ( x
2
− 3 x + 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 : 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 = 3 x
−
5 and the circle x 2 + y 2 = 5 .

€
8) We start the demonstration of this identity by writing cot (2 θ ) in terms of its definition: cot ( 2 θ ) = cos ( 2 θ )
= cos
2
θ
.
sin ( 2 θ )
θ − sin
2
2 sin θ cos θ
If we now divide the numerator and denominator of this ratio by sin 2
θ , which is suggested by the “1” as the second term of the numerator, we obtain
€ cos
2
θ
2 sin
− sin
2
θ cos θ
θ
⋅
1 sin
2
1 sin
2
θ
θ
= cos
2
θ sin
2
θ
2
− sin
2 sin
2 sin
2 sin θ cos θ
θ
θ
θ
=
2 cot
2
θ − 1 s i n θ cos θ s i n θ ⋅ sin θ
= cot
2
θ − 1
2 cot θ
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: c
2
= a
2
+ b
2
− 2 ab cos γ = 3
2
+ 4
2
− 2 ⋅ 3 ⋅ 4 ⋅ cos 30 o
= 9 + 16 − 24 ⋅
2
3
= 25 − 12 3
⇒ 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: sin α
= sin β sin γ a b
= c sin 30 o
⇒ sin α
= =
1 2
⇒ sin α =
3
≈ 0.7306
and
€
3
25 − 12 3 25 − 12 3 2 25 − 12 3 sin β 1 2 2
⇒ = ⇒ sin β = ≈ 0.9741 .
4
25 − 12 3 25 − 12 3
The area of this triangle can be found most directly [see diagram on right above] by using
A =
1
2
( base ) ( height ) =
1
2
⋅ b ⋅ a sin
=
1
2
⋅ 4 ⋅ 3 ⋅
1
2
= 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)
( 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:
( x − 1) ⋅ ( x + 1) ⋅ ( x + i ) ⋅ ( x − i ) = ( x
2
− 1
2
) ⋅ ( x
2
− i
2
)
= ( x
2
− 1) ⋅ ( x
2
− [ − 1]) = ( x
2
− 1) ⋅ ( x
2
+ 1) = ( x
2
)
2
− 1
2
= x
4
− 1 .
€ the four complex fourth-roots of 1 : 1 , i ,
−
1 , and
−
€
We can see now that we have the four roots of the equation x 4
−
1 = 0 , which are 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 series:
= a + ( n
−
1 ) · d , to work out how many terms are in the a n
= 1001 = 1 + ( n − 1) ⋅ 2 ⇒ 2 ( n − 1) = 1000 ⇒ n − 1 = 500 ⇒ n =
From there, we can now find the sum either from the arithmetic series formula:
501 .
S n
= n
2
⋅ [ 2 a + ( n − 1) ⋅ d ] =
501
2
⋅ [ 2 ⋅ 1 + (501 − 1) ⋅ 2 ] =
501
2
€
=
501
2
⋅ 2 ⋅ 501 = 501
2 or 251,001 , or from the formula using just the first and last terms of the series:
⋅ [ 2 + 500 ⋅ 2 ]
S n
=
€ n
2
⋅ ( a + a n
) =
501
2
⋅ ( 1 + 1001 ) =
501 ⋅
2
1002
= 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 s n
= a ⋅
1 − r
1 − r n
, 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
€ s
64
= 2 ⋅
1 − 2
1 −
64
2
= 2 ⋅
1 − 2
− 1
64
= 2
65
− 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 s =
1 a
− r
, so the sum for our series is
€ s =
1
1
3
1
3
=
1
3
2
3
=
1
2
.
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
4 x
2
+ y
2
8 x + 4 y + 4 = 0
4 x
2
8 x + y
2
+ 4 y =
4
⇒ 4 ⋅ ( x
2
− 2 x + 1 ) + ( y
2
+ 4 y + 4 ) = − 4 + 4 ⋅ 1 + 4
⇒ 4 ⋅ ( x − 1)
2
+ ( y + 2)
2
= 4 ⇒
( x − 1)
2
+
( y + 2)
2
= 1 .
1 4
€ that the center of the ellipse lies at ( 1,
−
2 ) .
€
15) To find the three complex roots of the equation x 3
−
1 = 0 is equivalent to finding the solutions of x 3 = 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 w k
= n z = z
1/ n
= r
1/ n
⋅

 cos(
+ 2 k
n
) + i sin(
+ 2 k
n
)


, k = 0, 1, K , ( n − 1) ⇒ w
0 w
1
= 1
1/ 3
= 1
1/ 3
⋅

 cos(
0 + 2 ⋅ 0 ⋅ π
3
⋅

 cos(
0 + 2 ⋅ 1 ⋅ π
3
) + i sin(
0 + 2 ⋅ 0 ⋅ π
3
) + i sin(
0 + 2 ⋅ 1 ⋅
3
π
)


=
)


=
1 ⋅ [ cos( 0) + i sin( 0) ] = 1 ( + 0 i ) ;
1 ⋅
[ cos(
2 π
3
) + i sin(
2 π
3
)
]
= −
1
2
+
2
3 i ; w
2
= 1
1/ 3
⋅

 cos(
0 + 2 ⋅ 2 ⋅
3
π
) + i sin(
0 + 2 ⋅ 2 ⋅
3
π
)


=
1 ⋅
[ cos(
4 π
3
) + i sin(
4 π
3
)
]
= −
1
2
−
2
3
16) There are a couple of ways to show that the term x + 1 is a factor of the polynomial x 49 + 1 . The less convenient of these would be to carry out the polynomial division of ( x 49 + 1 )/( x + 1 ) to find that there is no remainder (the quotient is the alternating sum x 48
− x 47 + x 46
−
… + x 2
− 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 i .

€
€ 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 ) indeed a factor of this polynomial.
49 + 1 = (
−
1 ) + 1 = 0 . We see then that ( x + 1 ) is
17) The hint we are offered in this problem is to use the “angle-addition” formula sin ( α + β ) = sin α cos β + cos α sin β . If we write 3 θ as 2 θ + θ , we then find that
θ =
θ + θ
=
θ
θ +
θ
θ .
We can then use the “double-angle” identities,
€
€
θ =
θ
θ
θ =
2 to re-write our expression as
θ =
θ
θ
⋅
θ +
2
θ
θ
−
−
2
2
θ
θ
⋅
θ
=
θ
2
θ +
2
θ ⋅
θ −
3
θ
€
Finally, we can use the Pythagorean Identity, sin 2
θ + cos 2
θ = 1
⇒
cos 2
θ = 1
−
sin 2
θ , to eliminate the appearances of cos 2
€
θ , leaving only terms involving powers of sin θ :
θ =
θ ⋅
−
2
θ
+
−
2
θ
⋅
θ −
3
θ
€
=
θ −
3
θ +
θ −
3
θ −
3
θ =
θ −
3
θ
18)
€
θ and cos θ cannot be solved by algebra alone. Instead, we use a “trick” based on the “angle-addition” formula
α + β
=
α
β +
α
β .
Starting from the equation a sin
+ b cos
= c , we divide the entire equation through by a
2
+ b
2 to obtain
€


 a 2 a
+ cos b
2

 sin
+


 a 2 b
+ b sin
2

 cos

= a 2 c
+ b 2
.
By identifying the two factors in parentheses as indicated, we can re-write our original
€ cos α sin θ + sin α cos θ = sin ( θ + α ) = c
.
a 2 + b 2
(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 c a
2
+ b
2
> 1
, the equation will have no solutions at all.)
(1)
2
For our equation, we have
+ ( 3)
2 factor ½ with cos
θ −
θ =
, which we divide through by
= 1 + 3 = 2 , giving us
1
2
θ −
2
3
θ =
. We now identify the
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 sin (
+ [
π
3
]) = sin (
€ π
3
) = 1
π
3
= sin − 1
(1) =
π
2
(only value)
=
π
2
+
π
3
=
5 π
6
.
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
z = r = a
2
+ b
2
= (
1
2
)
2
+ (
2
3
)
2
= and its angle is given by
= tan
− 1

 b a


[ appropriate quadrant ] = tan
− 1



So the polar form of our complex number is z = 1 ⋅ ( cos
€ complex square roots are then
1
4
3 2
1 2

 = tan

− 1 ( )
=
π
3
+
3
4
= 1
+ i sin
π
3
3
) . The two
.
w
0
= 1
1/ 2

⋅
 cos(


π
3
+ 2 ⋅ 0 ⋅
2
π
π
) + i sin( € 3
+ 2 ⋅ 0 ⋅
2
π
)




= 1 ⋅

 cos(
π
6
) + i sin(
π
6
)


=
2
3
+
1
2 i ; w
1
= 1
1/ 2

⋅
 cos(


π
3
+ 2 ⋅ 1 ⋅
2
π
) + i sin(
π
3
+ 2 ⋅ 1 ⋅
2
π
)




= 1 ⋅

 cos(
7
6
π
) + i sin(
7
6
π
)


=
−
2
3
−
Because these numbers are both square roots of z , it is reasonable that w
1
=
− w
0
.
G. Ruffa – 6/09
1
2 i .