Math 30-1 Trigonometry Prac ce Exam 4. There are two op ons for PP( 5, mm), it can be drawn in SOLUTIONS

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1 SOLUTIONS Math 0- Trigonometry Prac ce Exam Visit for more Math 0- Study Materials.. First determine quadrant terminates in. Since ssssss is nega ve in Quad III and IV, and tttttt is neg. in II and IV, is in quad IV. So, or 6 (not one of the choices) So, consider co-terminal angles of 0 by adding / subtracting 60 (and converting to radians), or use and 6 add/subtract , converts to 66 First determine quadrant terminates in. Since ssssss is nega ve in Quad III and IV, and tttttt is neg. in II and IV, is in quad IV. PP(, 4) 44 4 PP(, 4) Next, plot a point PP in quad IV and sketch angle Produced by rtdlearning.com Permission is given for any not-for-profit use by Alberta students and teachers. Then make a triangle by connec ng PP to the xx-axis. Label the sides of the triangle from the coordinates of PP use pyth. theorem to get the hyp.. This is a reasoning ques on, not a calcula on ques on! Using CAST rule, since cccccc is nega ve in Quad II and III, and tttttt is neg. in II and IV, is in quad II. Which means, is between and. (.7 rads and.4 as decimals) So, the only possible op on there is.6 (Note, all answers are presumed to be in radians as no unit is specified!) NR # Since ssssss is nega ve in Quad III and IV, and we want the smallest op on for draw an angle in st. pos. in quad III. Given: First find reference angle, let s call it αα (inside the triangle) αα ssssss NOTE: When finding reference angles (by defini on less than 90 so all trig ra os are posi ve) drop any nega ves. ANSWER:. (radian mode) Now use 4 aaaaaa hyyyy where.so hyyyy aaaaaa. ANSWER: A. hhhhhh (radius) oooooo 4. There are two op ons for PP(, mm), it can be drawn in quadrant II or III. (as the xx-coord is negative) However, it is given that tttttt is nega ve, we know we should draw PP in quad II. At this point we can pause to consider how fun it is to reason things out like that. (pause for 0 to seconds) Diagram of information given PP(, mm) We can now solve for mm by either drawing a triangle and label the hypotenuse (since unit circle), or by going straight to the unit circle formula: xx + yy ( ) + mm mm 69 mm ± Since quad II, use the + version! mm mm mm ± First find the length of : cccccc67.6 AAAA AAAA.6 cccccc67 AAAA cm (radius) Next find the length of : (half of AAAA is the opp side ) tttttt67 oooooo.6 oooooo.6tttttt67 oooooo.769 AAAA.769 AAAA 77. Finally find the arc length: (67 ) aa rrrr aa in radians So perimeter is: Rectangle top / bottom Third rectangle side, AAAA Perimeter cm aaaaaa 99. cm Final side, the arc length

2 NR # aa rr rr aa re-arrange rr 8.9 convert to radians:.9 0 rr rr (leash length + 0.m) So, leash Remember, m dog s extra reach is included here! aaaaaa m ANSWER: (yy) + (yy) (hyyyy) yy (hyyyy) hyyyy yy yy hyp yy adj aa bb Add: aaaaaaaaaa ββ 44 aaaaaaaaaa 4 bbbbbbbbbbbbbb Or first convert to degrees: 9 Converts to 7 is FALSE Principal angle 4 is TRUE Is coterminal yy-coord at xx-coord at 4 is TRUE is is tttttt y x cccccc 44 4 is TRUE x y 8. Amplitude, aa, from ff to gg decreases Graph shifts right, so cc changes 9. First factor bb for horiz. phase shi : NR #4 yy ssssss[4 xx + 4 ] yy. ssssss(0.08xx 0.64) +.8 Period is Phase shi Period is MAX is AMPL + Vert. Shi bb bb (to the left, but 4 not relevant) aa + dd ANSWER: A ANSWER: NR # Since no horiz. stretch (bb ), period is. Basic sine curve (with no horizontal stretch or shift) has a MIN at xx : So we can see distance between PP and QQ is greater than is TRUE is TRUE aa dd Median Line aa can be visualized as dist. from MAX (or MIN) to median line dd as dist. from median line to xx-intercept is FALSE 0. NR # Domain of yy tttttttt is xx + nnnn That is, where 0, since tttttttt at the top / botom of the unit circle So for yy tttttt4xx Hor. Str. of 4 Domain is: xx nnnn xx 88 + nnnn 44 For yy-int, set xx 0: But ssssss(0) 0, so yy aa ssssss(00) dd yy dd 4 is FALSE For bb <, for example. yy aa ssssss(xx) dd The horiz. str. would be the reciprocal, or here 00. For horiz str <, all points move closer to the xxaxis is TRUE ANSWER: Note: Mistake on some answer keys Period is ANSWER: 4 bb med. line dd bb bb 44

3 . Use pppppppppppp bb formula to solve for ff: 0.00 ss ff 0.00 ss ff 0.00 ss ff 400 HHHH If AA is at xx 0.00 then period is double, ss. Amplitude is 0 Note: you can also use formula mmmmmm mmmmmm aa For cc determine xx-coords of max / mins Median line cc Distance (from yy-axis) of point where curve upswing intersects median line. 0 rota ons in 60 sec. ( min) rota on in 60 0 ssssss (this is the period). 66 cc Halfway between 00 (where min is) ANSWER: A and. ss (where max is). 6. Solve by factoring: cccccc + 0 ( )( ) 0 Two numbers must mult. to ( )( + ) 0 Now set each factor to zero ANSWER: A or First isolate and use log laws on LS llllll (tttttttt) + llllll () llllll (tttttttt ) Use trig iden es to simplify llllll llllll () Convert to log form Find where on unit circle yy the yy-coord is xx or First isolate cccccc 4 0 trig term: cccccc 4 Sq. root both sides Since aa bb aa bb Since is reciprocal of cccccc 4 ± 4 ± ± Find where on unit circle the yy-coord is ± (, ( ), ) (, ),, 4444, (, ). Factor ssssss xx 0 ( + )( ) 0 Set each factor to zero: or Since is reciprocal of : or Find where on unit circle the xx-coord is oooo / Note that all three solu ons are 0, or, apart + nn ; nn II diff between first solu on sols NR #6 First re-write in degrees: cccccc(0 ) Next find any two standard unit circle angles that add (or subtract) to 0 Such as: cccccc( ) Note: There are many op ons, another is cccccc( 0 ) cccc ss(αα + ββ) cccccc4 cccccc60 ssssss4 ssssss60 Finally refer to unit circle and simplify ANSWER: Sketch the angle and draw a triangle to determine trig ra os: By pyth. theorem PP(, ) Now, for ssssss( ) (0) ( ) ( ) ( From unit circle 4 From diagram cccccc aaaaaa hyyyy 4 )

4 So, NPVs tttttttt 0 8. Where on the unit circle is + Simplifies to the xx-coord 0? + 0 Where on the unit circle is the yy-coord equal to? (already covered above!) So, NPV at the top / botom of the unit circle xx then every Which we write as: xx + nnnn where nn II NR #7 Simplify: tttttttt ( + ) ( + ) + () + () Pyth. identity, this is + ssssss xx + cccccc xx ( + ) + ( + ) Simplify: + tttttttt + + ssssss xx + cccccc xx ssssss xx + cccccc xx ssssss xx ssssss xx ssssss xx Code: 44 ANSWER: 4 Code: 9. Start with appropriate addi on / subtrac on formula: ssssss(αα + ββ) + ssssss + cccccc PP( 4,) Draw using info given 4 Use unit circle for this and cccccc 4 To sketch, use: To sketch, use: adj opp In Quad II since adj side is neg and > 0 Use pyth. theorem for hyp., ( 4) + hyyyy so Step The mistake is here! Correct steps: + (cccccc AA ssssss AA) + cccccc AA + + ssssss AA + ssssss AA + + ssssss AA + NR #8 Sketch: 4 By pyth. theorem Formula sheet: ANSWER: A ssssss AA + + ( + ) ( + ) tttttttt So, tttttttt 4 ANSWER: Writen # P(, ) RR By pyth. theorem: (hyyyy) ( ) + () (hyyyy) 6 44 ββ RR ββ By pyth. theorem: To find, first find reference angle RR (inside the triangle) RR tttttt (/) or ssssss (/ 6) or cccccc (/ 6) so 80 Q(, 4) To find ββ, first find reference angle ββ RR (inside the triangle) ββ RR tttttt (4/) or ssssss (4 ) or cccccc (/) so 80 +

5 WR # Second bullet Refer to your formula sheet: We ll need all this: (leave exact) opp ssss nn 6 hyp 6 adj hyp similarly 4 So ssss nn( + ββ) Produced by rtdlearning.com Permission is given for any not-for-profit use by Alberta students and teachers. Note: There was a mistake in some answer keys! Writen # Use ssssss + cccccc, which re-arranges to ssssss cccccc, to re-write equation in terms of only cccccc 0 cccccc 0 0 cccccc + cccccc + 00 nd bullet ( )( ) 0 Factor Find two #s that mult to, and expression will expand back out to cccccc + ( )( + ) 0 0 Set each factor to zero 0 or + 0 / Where on the unit circle is the xx-coord ½? PRIMARY SOLUTIONS: Where on the unit circle is the xx-coord?,, GENERAL SOLUTION: For GENERAL SOLUTION, consider all three primary solutions together They re all (that is, 0 ) apart + nn nn II Writen # Proceed on le side: + + Write each expression in terms of ssssss and cccccc Re-write bottom expressions with a common denom. nd bullet Go back to first step, restriction at any expression in the denominator. (Can t divide by 0) where on the unit circle is the xx or yy coord equal to 0? ssssss xx + cccccc xx Pyth. identity, this is! 00, xx nn nn II Invert and multiply cccccc xx

6 Writen #4 ("aa" value) Dist. from median line to mmmmmm mmmmmm aa max or min, can also use: aa aa. 44 Median line ("dd" value) mmmmmm + mmmmmm dd dd dd Finally for cc cccccc, determine the horiz. dist of the max from the yy-axis Here the max has shifted, 7 to the right Basic cos curve starts at the MAX For bb, determine the period: This shows half of the period (dist from max to min) Half of Period 8 days (6 7) Period 66 days Actually, this was given! bb 66 bb cc HH. 44 cccccc (xx ) + nd bullet Graph yy. ssssss 0.4(xx.7) +.9 yy 6 And find the INTERSECTS For xx-max, graph over period pppppp 0.4 For yy-max, use MMMMMM aa + dd. +.9 So, total # months above 6 can be found by subtracting months Third bullet aa would be higher, as the range of Calgary temperatures (between min and max) would be greater dd would be lower, as the median temperature for Calgary (represented by dd) would be lower Also. (not needed in your answer) bb would be unchanged, as the period for each city would be the same ( months). Similarly, cc would be essentially unchanged, as the number of months after which the min / max temperature occurs would be approximately the same as both cities are in the northern hemisphere. ALL DONE!