Trigonometry Topic Test Solutions FAMAT State Convention 2012 1. E. sin x + 2sinxcosx = 0. This leads to sin x(1 + 2 cos x) = 0. So, sin x = 0 and 2π
4π
cos x = -­‐.5. The solutions are 0,
, π,
, 2π . 3
3
2. A. The cosine angle is in quadrant one and the tangent angle is in quadrant four. Build the appropriate triangles and then use the formula sin (A-­‐B)= sin A cos €
B-­‐ cos A sin B. Plugging the values in gives the following: ⎛ 24 ⎞ 5 ⎛ 7 ⎞ −12 204
. =
⎜ ⎟ − ⎜ ⎟
⎝ 25 ⎠ 13 ⎝ 25 ⎠ 13
325
3. C. All of the angles are coterminal with 30 degrees except C. The angle equivalent to 7260 is 60 degrees, not 30 degrees. €
4. D. Using Law of Cosines, plugging in the values produces the following: XZ 2 = 10 2 + 6 2 − 2(10)(6)(cos120) = 136 + 60 = 196 . Taking the square root gives the solution. 5. A. All of the angles are in radians. The values, in order, are .99, -­‐.14, .22, .71. A is the largest value. €
16π
. 7
We must find the least common multiple between the two. Since one of the periods is an integer, our answer must be an integer. Multiplying the sine period by a €ultiple of 9 as well. When you multiply by 63, you get multiple of 7, we must find a m
144, which is also a multiple of 9. So the period of the function is 144€π . 6. E. The period of the cosine graph is 9π and the period of the sine graph is ⎛ 7π 2π ⎞ ⎛ 7π 2π ⎞
+
−
⎜
7π
2π
3 ⎟ cos⎜ 6
3 ⎟ = 2cos 11π cos π + cos
= 2cos⎜ 6
7. C. cos
⎟ ⎜
⎟
€ 12
6
3
2
2
4
⎜
⎟ ⎜
⎟
⎝
⎠ ⎝
⎠
8. B. An even function occurs when f(x) = f(-­‐x). This occurs with the cosine function. €
9. D. Find the length of each side of triangle ABC. The length of each side is: a = 109, b = 53, c = 34 . Use the law of cosines to find the cosine of angle B. This produces 53 = 34 +109 − 2 34 109 cos B . Solving for cos B produces the (
solution when it is rationalized. €
)(
)
10. E. There are infinitely many solutions to the problem because a domain is not given €for the question. 2
€
€
11. C. (sin 2 x + cos2 x ) = sin 4 x + 2sin 2 x cos 2 x + cos4 x = 1. 7
7
sin 4 x + cos 4 x = 1 − 2sin 2 x cos2 x . sin2x = 2sin x cos x =
. →sin x cos x =
12
24
⎛ 49 ⎞
98 478 239
49
2
2
. sin 4 x + cos 4 x = 1 − 2⎜
. =
=
⎟ = 1 −
€ sin x cos x =
⎝ 576 ⎠
576 576 288
576
€ the complex number in polar form, it becomes 81cis120 0 . 12. D. When you express The radius of the 4th root is 3 and the four angles for the fourth roots of the complex € 30, 120, 210 and 300 in degrees. The solution is D because the angle is number are 330 degrees. The first three solutions are the correct answers for 30, 120 and 210 €
degrees. 13. A. When you build the triangle, the length of the hypotenuse is 10x 2 + 2x +1 . When you find the cosecant of the angle, it is equal to hypotenuse/opposite. In this case, that is the solution. € of the sector. So one 14. C. The perimeter of a sector is two radii and the arc length 1 2
formula is 2r + rθ = 16 and for the area of the sector, r θ = 7 . Solving for θ in the 2
14
area equation produces θ = 2 . Plugging this into the perimeter equation produces r
€ ⎛14 ⎞
€
14
2
2r + r⎜ 2 ⎟ = 16 →2r +
= 16 →2r 2 +14 = 16r
€ →2r −16r +14 = 0 . Dividing by 2 ⎝ r ⎠
r
2
produces r − 8r
€ + 7 = 0 → ( r − 7)( r −1) = 0 →r = 7,1. €
2π
1
= 4 π , C = x − 2π = 0 → x = 4 π , 1
2
€
2
D=7. The sum of the four parts is the solution. 15. B. The values are as follows: A = 3, B = 16. C. The fraction given is a popular difference €between cosine values of angles on €
the unit circle. ⎛ 2 ⎞⎛ 3 ⎞ ⎛ 2 ⎞⎛ 1 ⎞
6− 2
0
cos 75 0 = cos( 45 + 30) = cos 45 0 cos 30 0 − sin 45 0 sin 30 0 = ⎜ ⎟⎜ ⎟ − ⎜ ⎟⎜ ⎟ =
4
⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠
€
The other angle in the interval with this value of cosine is 285 0 . The angles between 75 and 285 have cosine values less than the fraction. So, 210 degrees out of the 360 210 7
degrees meet the requirement. = . 360 12
€
2 ⎛ − 3 ⎞
3
17. A. The values are, in order: − ⎜
+ 2 − ( −2) . When you ⎟ + 3 −
2 ⎝ 2 ⎠
3
€
2 3 3 6 3 2 3
2 7 3
simplify this, you get: . +
+
−
+2+2 = 4 +
+
2
6
6
6
2
6
€
€
−π
in degrees is -­‐20. The sin (-­‐20) is equivalent to sin 340. By the 9
cofunction formula sin θ = cos(90 − θ ) , sin 340 is equivalent to cos (-­‐250). The cos (-­‐250) is equivalent to cos 110. Cos−1 (cos110 0 ) = 110 0 . 110 degrees is equal to € the solution in radians. €
19. D. Using the law of cosines, 7 2 = 5 2 + 32 − 2(5)(3)(cosY) →49 = 34 − 30cosY . This €
−1
leads to cosY = . Therefore Y = 120 0 . So we are looking for tan60 0 = 3 . 2
18. B. €
€
€
€
€ the determinant, you get 20. B. When you evaluate sin6θ cos 4θ − cos6θ sin 4θ€= sin(6θ − 4θ ) = sin2θ = 2sin
€θ cosθ . €
21. D. Three of the four expressions are equal to tan 2 x . III is the only one that does cos2 x − (1 − sin 2 x) 2 cos2 x − cos 4 x cos2 x(1 − cos 2 x) sin 2 x
not. For I, =
=
=
= tan 2 x . cos4 x
cos 4 x
cos 4 x
cos 2 x
2 − 2cos2 x
2(1 − cos2 x)
2sin 2€x
For II, =
=
= tan 2 x . For III, 2
2
cos2x +1 2cos x −1+1 2cos x
sin2x
2sin x cos x
= 2sin x cos x . For IV, €
2 =
sin2x + (cos x − sin x)
2sin x cos x + cos2 x − 2sin x cos x + sin 2 x
tan(x + π ) = tan(x − π ) = tan x and sec 2 x − tan 2 x = 1 . The product in the numerator €
produces tan 2 x . 22. C. Bringing everything to the left produces 3tan 2 x + sec 2 x −12sec x +11 = 0 . € 2 x −1) + sec 2 x −12sec x +11 = 0 . Cleaning it up produces Substituting produces 3(sec
€
4 sec 2 x −12sec x + 8 = 0 →sec 2 x − 3sec x + 2 = 0 →(sec x − 2)(sec x −1) = 0 . Solving, 1
we get sec x = 2 and sec x = 1. So, cos x€= and cos x = 1. The solutions on the 2
interval is the €answer to the question. π
3π
23. D. The graph begins at -­‐2 and goes to negative infinity between , and
€
2
4
3π
5π
then goes from 4 to infinity between and goes from -­‐2 to negative and
4
4
5π
3π
infinity from . The range of this function is the €answer. and
4
2
€ and seconds, divide the minutes by 60 and the 24. A, To convert from minutes 42
36
7
1
71
seconds by 3600. +
= +
=
= .71. So, the angle is equal to 25.710 . €
60 3600 10 100 100
25. D. π
π
3 3− 3
€
tan − tan
1−
⎛ 3 − 3 ⎞ 12 − 6 3
€
π
3
−
3
4
6 =
3 = 3 =
tan =
= 2 − 3 . ⎜
⎟ =
12 1+ tan π tan π
6
3 3 + 3 3 + 3 ⎝ 3 − 3 ⎠
1+
4
6
3
3
€
So, a = 2, b = 1 and c = 3. Plugging into the quadratic produces 3x 2 + 2x +1 = 0 . The 2
discriminant is (2) − 4(3)(1) = 4 −12 = −8 . 26. C. The trig functions cancel out because of cofunction € relationships. The cosine and sine, tangent and cotangent and sine and cosine in the numerator and €
denominator respectively cancel out because they are equal. The sin 75 and csc 105 are reciprocals of each other because csc 105 = csc 75. 27. A. The original graph of y = tan x begins at π ⎞ π
2π
π
π ⎛
. ⎜ 4 x − ⎟ = →4 x =
→ x = . 6 ⎠ 2
3
6
2 ⎝
28. B. First find side b by the law of sines. ⎛ 6 + 2 ⎞
8
b
8
b
16 3
€
. When we =
→
=
→4
3
=
⎟b →b =
€ ⎜
0
0
sin 75
sin60
4
6+ 2
3
6+ 2
⎝
⎠
4
2
⎛
16 3
6 − 2 ⎞ 48 2 −16 6
⎜
⎟ =
rationalize b, we get = 12 2 − 4 6 . Using the 4
6 + 2 ⎜⎝ 6 − 2 ⎟⎠
€
area formula, we produce ⎛ 2 ⎞
1
(8) 12 2 − 4 6 (sin 45 0 ) →4 12 2 − 4 6 ⎜ ⎟ →2 2 12 2 − 4 6 →48 −16 3 . 2
⎝ 2 ⎠
€
(
€
)
(
)
(
)
cos x
cos x
sin x
→ sin x
→cos2 x . When 29. C. Simplifying the fraction produces cos x sin x
1
+
sin x cos x
sin x cos x
you introduce sin 2 x and − sin 2 x , you get sin 2 x + cos 2 x − sin 2 x →sin 2 x + cos2x . ⎛ sin x ⎞
30. B. When you simplify the equation, cos x⎜
⎟ = cos x →sin x = cos x . The two €
⎝ cos x ⎠
€
€
trig functions are equal at 45 0 and 225 0 . €
€