Practice Problems Test 4 PDF

Title Practice Problems Test 4
Author Umme Habiba
Course Precalculus
Institution Hunter College CUNY
Pages 3
File Size 123.4 KB
File Type PDF
Total Downloads 82
Total Views 159

Summary

Practice Problems for 4th test...


Description

M125 Practice Problems for Exam #4 p L 1. Find the amplitude, period, and phase shift of y = 2 cosH2 x + ÅÅÅÅ 2 and then graph one period. p 2. Graph y = tanHx - ÅÅÅÅ L, 2

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3 1 N. b) sinHtan-1 H-1LL. c) cotHsin-1 ÅÅÅÅ L. d) cscHtan-1 2 xL. 3. Find exact values: a) cos-1 J- ÅÅÅÅÅÅÅÅÅ 2 Å 5

4. A flagpole is situated on top of a building. The angle of elevation from a point on level ground 330 feet from the building to the top of the flagpole is 63°. The angle of elevation from the same point to the bottom of the flagpole is 53°. Find the height of the flagpole to the nearest tenth of a foot. 5. A building that is 250 feet high casts a shadow 40 feet long. Find the angle of elevation, to the nearest tenth of a degree, of the sun at this time. 6. A ball on a spring is pulled 4 inches below its rest position and then released. Its amplitude is 4 inches and its period is 6 seconds. Find the equation for the ball's simple harmonic motion. ˛ 7. An object moves in simple harmonic motion according to the function d = -8 cos ÅÅÅÅ Å t , where t is 2 measured in seconds and d in inches. Find a) the maximum displacement. b) the frequency. c) the time required for one cycle (i.e. the period).

8. Verify the following identities. cos x a) tan x + ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅxÅ = sec x 1+sin

cos x 1-sin x b) ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ 1+sin x cos x

1 1 c) ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ + ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ = 2 + 2 cot2 x 1+cos x 1-cos x

9. Find the exact value of cos 75 °. 4 2 , b lies in quadrant I find 10. Given tan a = - ÅÅÅÅ3 , a lies in quadrant IV, and cos b = ÅÅÅÅ 3 a cos 2 a a) cosHa + bL. b) sinHa - bL. c) tanHa + bL. d) . e) cos ÅÅÅÅ2Å .

11. Write cos4 x as an equivalent expression that does not contain powers of trigonometric functions greater than 1. 12. a) Find all solutions in the interval @0, 2 pL : 2 sin x cos x = -cos x. b) Find all solutions of 2 sin2 2 x - 3 sin 2 x = -1 and then all solutions in the interval @0, 2 ˛L. 13. Solve the following triangles. Round sides to the nearest tenth and angles to the nearest degree. a) A = 85 °, B = 35 °, c = 30 . b) A = 20 °, a = 30, b = 40. c) a = 10, b = 3, C = 15 °. 14. The tallest trees in the world grow in the Redwood National Park in California. To measure the height of a tree, a surveyor sitting at point A to the left of the tree measures the angle of elevation to the top of the tree to be 44°, and from a point B that is 100 feet further to the left from A, the

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surveyor measures the angle of elevation to the top of the tree to be 37°. What is the height of the tree to the nearest foot? 15. To find the length AB of a small lake, a surveyor measured angle ACB to be 96°, AC to be 91 yards, and BC to be 71 yards. What is the approximate length of the lake to the nearest yard?

Trigonometric Identities Sum Identities

Difference Identities

sinHa + bL = sin a cos b + cos a sin b cosHa + bL = cos a cos b - sin a sin b tan a+tan b

sinHa - bL = sin a cos b - cos a sin b cosHa - bL = cos a cos b + sin a sin b tan a-tan b

tanHa + bL = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅ 1-tan a tan b

tanHa - bL = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅ 1+tan a tan b

Double-Angle Identities

Half-Angle Identities

sin 2 q = 2 sin q cos q cos 2 q = cos2 q - sin2 q = 1 - 2 sin2 q = 2 cos2 q - 1 2 tan q ÅÅÅÅÅÅÅÅÅÅÅ tan 2 q = ÅÅÅÅÅÅÅÅ 1-tan2 q

1-cos q ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ sin ÅÅÅÅq2 = ± "############## 2 1+cos q ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ cos ÅÅÅÅq2 = ± "############## 2 sin q 1-cos q ÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ tan ÅÅÅÅ2q = ÅÅÅÅÅÅÅÅ sin q 1+cos q

Power-Reducing Identities 1-cos 2 q ÅÅÅÅÅ sin2 q = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 2 1+cos 2 q ÅÅÅÅÅ cos2 q = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 2 1-cos 2 q ÅÅÅÅÅ tan2 q = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 1+cos 2 q

Solutions p 1. y = 2 cos H2 x + ÅÅÅÅp2 L. A = 2, B = 2, C = - ÅÅÅÅ . 2 (You can also rewrite the function as y = 2 cos 2 Hx + ÅÅÅÅp4 L). Amplitude: 2 2p Å =˛ Period: ÅÅÅÅÅÅÅ B p C ÅÅÅÅ Phase shift: ÅÅÅÅÅ B =- 4 period p = ÅÅÅÅ Increment: ÅÅÅÅÅÅÅÅ 4ÅÅÅÅÅÅÅ 4 2

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p 4

p 4

-2

p 2

3p 4

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3p H ÅÅÅÅÅÅÅÅÅÅ , 1 L 4

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p ÅÅÅÅ 4

p ÅÅÅÅ 2

p H ÅÅÅÅÅ , -1 L 4

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7p H ÅÅÅÅÅÅÅÅÅÅ , 1 L 4 3p ÅÅÅÅÅÅÅÅ 4

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5p ÅÅÅÅÅÅÅÅ 4

3p ÅÅÅÅÅÅÅÅ 2

7p ÅÅÅÅÅÅÅÅ 4

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5p H ÅÅÅÅÅÅÅÅÅÅ , -1 L 4

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è!!!!!!!!!!!!!! è!!! è!!! 2 1+4 x2 . 5 p . b) a) ÅÅÅÅ ÅÅÅÅÅ ÅÅÅÅ - ÅÅÅÅÅ2ÅÅÅÅÅ . c) 2 6 . d) ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 6 2x 209.7 feet. 80.9°. p d = -4 cos ÅÅÅÅ t. 3 a) 8 inches. b) ÅÅÅÅ14 cycles per second. c) 4 seconds. answer omitted. "######## ###### è!!!

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-54+25 5 2 5 7 6 +4 5 -8-3 5 ÅÅÅÅÅÅÅÅÅÅÅÅ . d) - ÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅ . b) ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅ . c) - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ ÅÅ 10. a) ÅÅÅÅÅÅÅÅ 22 15 15 25 . e) - 5 . 3 1 1 11. ÅÅÅÅ cos 4 x. + ÅÅÅÅ cos 2 x + ÅÅÅÅ 8 8 2 7p 3p 11 p p , , , ÅÅÅÅÅÅÅÅ 12. a) x = ÅÅÅÅ ÅÅÅÅ Å ÅÅ Å ÅÅÅÅ Å ÅÅ Å ÅÅ . 2 6 2 6 5˛ 13 p 5p p p 5˛ ˛ p + n ˛ ; x = ÅÅÅÅ , 4˛Å , ÅÅÅÅ + n ˛, ÅÅÅÅÅÅÅÅ b) x = ÅÅÅÅ ÅÅÅÅ , ÅÅÅÅÅÅÅÅ ÅÅ , ÅÅÅÅ ÅÅÅÅ , 17 ÅÅÅÅÅÅÅÅ ÅÅ . ÅÅ + n ˛, ÅÅÅÅ 12 12ÅÅ ÅÅÅÅ 12 12 4 12 4 12 13. a) A = 85 ° B = 35 ° C = 60 ° b) A = 20 ° B = 27 ° C = 133 ° A = 20 ° B = 153 ° C = 7 ° a = 34.5 b = 19.9 c = 30 a = 30 b = 40 c = 64.2 a = 30 b = 40 c = 10.7 c) A = 159 ° B = 6 ° C = 15 ° a = 10 b = 3 c = 7.1 14. 343 feet 15. 121 yards...


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