Advanced Math (2) Semester 1 Review Name SOLUTIONS

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dvanced Math (2) Semester Review Name SOLUTIONS IMPORTNT!!!!! PLESE EMIL ME IF YOU FIND TYPO!. Find the value of x to the nearest yard. 53 YRDS x 5 yards 32 cos32 = 5 x 5 x = cos32 x = 53 yards 2. In D, =, = 65, and a =. Find side b to the nearest tenth. b=5.2 sin = sin 65 b b = sin65 sin = 5.2 65 3. In D, = 57, a = 8, and b = 7. Find the number of solutions to the triangle. ONE sin57 8 = sin 7 = sin - 7sin 57 = 7.2 or 32.8 8 38.2 57 80 so only solution. 7 57 8. The angle of depression from the top of a cliff 750 m high to the base of a streetlight is 68. To the nearest meter how far is the street light from the foot of the cliff? 303 feet 5. Does csc x = csc (-x) for all values of x? NO (Q and Q don t have same sign for csc (or sin).)

6. Find the equation of the graph below. y = 2sinx ϖ 2ϖ 7. Find the equation of the graph below. y=2cos[(2ϖ/3)(x )] y= 2cos[(2ϖ/3)(x 2.5)] y= 2sin[(2ϖ/3)(x 0.25)] y= 2sin[(2ϖ/3)(x.75)] 3ϖ 8. Graph the equation y = -2sin p (x - 3). 2 9. Simplify. a) csc coscot b) - coscos - cos 2 sin 2 sec csc tan csc Á Ë c) sin 2 cot tan csc 2 sin2 cos 2 sin 2 tan sin2 sin2 ˆ 2

0. What quadrant is x in if cos x < 0 and sin x < 0? Q3. Evaluate csc p 6. since sin p 2 = 2, csc p 6 =2 2. Graph y = 2 - cos x p ˆ Á. Ë 3 - p 3 5p 3. onvert p 5 to degrees. 5. Find the length of side to the nearest tenth. 0 60 x 5 sin 60 = sin 5 0 = 35.26 or.7 = 35.26 = 8.7 x 2 =0 2 5 2-2 0 5 cos8.7 x 2 = 297.7 x =7.2 6. What is the horizontal shift of the graph of y = -3cos x - p ˆ Á? nswer: p to the right. Ë 2 2 7. Find the equation that matches the graph below: y = 3 cos 2x ϖ 2ϖ 3ϖ 3

8. Find cosq if q is an angle in standard position with its terminal ray passing through the point ( 2, ). cos x = -2 2 5 = - 5 5 ( 2, ) q 9. cot = cot = adj opp = 2 2 = 2 2 0 0 2-2 = 8 = 2 2 20. If sec x = 2 find all values of x to the nearest degree for 0 x < 360. x = 60, 300 2. If sec x is positive, in what quadrants could x be? Quadrants I or IV. 22. If tanx = 3 find all values of x to the nearest degree for 0 x < 360. x = 60, 20 23. Graph a full period of the following function. y = 2sinÁ p Ë 2 x ˆ One period goes from x=0 to x=.

2. 50 ft building casts a shadow of 250 feet at 3:00 pm. What is the angle of elevation from the end of the shadow to the top of the building? q = 3 25. cosq = - 5 and sin q > 0. Find tan q. tan q = - 3 26. Find the exact value of cos 75. cos 75 = 6-2 27. Find the exact value of sin 2 cos 2 - cos 2 sin 2. sin (2 2 ) = sin 30 = 0.5 28. Derive the formula for sin 2a from the sin (a a) formula. sin 2a = sin (a a) = sin a cos a cos a sin a = 2 sin a cos a 29. Derive the formula for cos 2a from the cos (aa) formula. cos 2a = cos (aa) = cos a cos a sin a sin a = cos 2 a sin 2 a 30. Solve the following equations for LL values of x in radians in the interval [0, 2ϖ). Ï a tanx sinx sin x = 0 x = Ì 0, 3p, p, 7p Ó SET = 0, FTOR sinx ( tanx ) = 0 sinx = 0 or tanx = 0 b) sin x cos 2x - cos x sin 2x = - Ï 7p x = Ì 2 2, ϖ Ó 2, 9ϖ 2, 23ϖ 2 This simplifies using the sin(a b) formula: ( ) = - 2 sin x - 2x spin TWIE sin2x = - 2 2x = 7p 6,p 6,9p 6, 23p 6 x = 7p 2,p 2,9p 2, 23p 2 5

c) sin 2 x - = 0 sin 2 x = / sin x = ±/2 PLUS OR MINUS! Ï p x = 6, 5ϖ 6, 7ϖ 6, ϖ Ì Ó 6 d) 3 sin x = sin x = /3 NO SOLUTION! 3. Find LL possible angles and sides. = 86.5, a = 8.6, = 53.5 2 = 3.5, a 2 =., 2 = 26.5 2 REMEMER sin - x has solutions in QI and QII. 5 0 32. Find the missing angles. = 23.2, = 0.3, = 6.5 7 3 9 33. Solve each equation on the interval 0 x < 360. a) 3 csc x - 5 = 0 csc x = 5/3 sin x = 3/5 x = {36.9, 3. } b) 2 sin 2 x - sin x - = 0 (2 sin x )( sin x ) = 0 sin x =.5, sin x = x = {90, 20, 330 } c) tan 2 x - = 0 (tan x ) ( tan x ) = 0 tan x = or tan x = x= {5, 35, 225, 35 } 6

3. Verify the identity: 35. Verify the identity: cosq sinq sinq cosq = 2secq cosq( - sinq) sinq ( sinq)(- sinq) cosq = cosq(- sinq) - sin 2 q cosq(- sinq) cos 2 q - sinq sinq = cosq 2 cosq = 2secq = sinq cosq = sinq cosq = cot 2 q sinq cscq = cscq cscq cot 2 q cscq sinq sinq = csc 2 q - cscq = (cscq -)(cscq ) = cscq cscq = 36. boat sails from port on a course N30 W for 75 miles and then turns to a SW course for 250 miles. a) How far must the boat sail to return to port? 2.7 miles 30 5 5 30 x port x 2 = 250 2 75 2-2(250)(75)(cos75 ) x = 2.7 b) What course must the boat sail to return to port? N 62. E sin 75 = 75 2.7 =7. course is 5 7. 7

37. Label the points whose coordinates are: a) (, -20 ) b) (-5, 35 ) c) (3, 300 ) d) D(-, -50 ) D 38. onvert to rectangular coordinates. a) (5, -225 ) b) (-, 50 ) c) (3, 270 ) - 5 2 2,5 2 ˆ Á 2 3,-2 Ë 2 ( ) (0,-3) 39. onvert to polar coordinates. Round to the nearest degree. 7 3 a) (-, -) b) ( 2, - 7 2 ) c) (-7, 0) ( 7,9 ) ( 7,330 ) ( 7,80 ) 0. Find four polar coordinates for the point (2, 0 ). a) (2, ) b) (2, ) c) ( 2, ) d) ( 2, ) (2, 00) b) (2, 320 ) c) ( 2, 220 ) d) ( 2, 0 ). Find the equation of the following graph. y = sec x 8

2. Read the following carefully, then fill in the blank. a) If tan q = k, then tan ( q) = k. b) If sec q = k, then sec ( q) = k. c) If cos q = r, then cos (q 2ϖ)= _r. d) If tan q= r, then tan (qϖ) = r. 3. Find the smaller angle made between the two hands of a clock at :57. 20 6 3 57 60 30 =66.5. The minute hand of a clock is 9 inches long. How far (in inches) has the tip of the minute hand moved (around in a circle) in 25 minutes? 25 2p 9=7.5ϖ 60 5. Planet Moog is 2.8 x 0 7 km from Earth and has an apparent size of about 0.00. What is the approximate diameter of planet Moog? You don't need to evaluate your expression for this diameter..00 360 2p 2.8 07 = 88.7 km 6. onvert the following. a) 2.2 to degrees, minutes, and seconds. 2 0.2 60 = 2 2.6 = 2 2 0.6 60 =2 2 36 b) 76 6 36 decimal degrees. (I changed this problem!) 76 6 60 36 3600 = 76. 9