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MAC 1114 February 9th &11th

by: Heya_Lanayia

MAC 1114 February 9th &11th MAC 1114

Marketplace > Florida State University > Math > MAC 1114 > MAC 1114 February 9th 11th
GPA 3.9
Analytic Trigonometry MAC1114
Dr. LeNoir

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Notes from February 9th &11th
Analytic Trigonometry MAC1114
Dr. LeNoir
Class Notes
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This 0 page Class Notes was uploaded by Heya_Lanayia on Sunday February 14, 2016. The Class Notes belongs to MAC 1114 at Florida State University taught by Dr. LeNoir in Winter 2016. Since its upload, it has received 23 views. For similar materials see Analytic Trigonometry MAC1114 in Math at Florida State University.


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Date Created: 02/14/16
MAC 1114 Analytic Trigonometry Notes for 2916 amp21116 FEBRURARY 9th 2016 Evaluate 1 Arcsin 1sqrt2 Since sin is negative it has to be in quadrants 3 or 4 If we look at the restrictions graphic above Quadrant 3 isn t an option because it isn t in the restrictions of pi2 and pi2 therefore it has to be quadrant 4 So to answer the question we need an angle in quadrant 4 that has a sin of 1sqrt2 Answer pi4 1sqrt2 sqrt22 a w 1quot min nah23 Irmmm w 2 Cos391 2 Cos1 and arccos are the same ANSWER This problem is undefined because the X values on the cosine graph only go from 1 to 1 3 Cos39112 For this one we re looking for an angle that has a cosine of 12 Since we re dealing with a negative value that means that the angle has be in either quadrant 2 or 3 Going back to the first graphic we see that the restrictions for cosine range from Opi or quadrants 1 and 2 Because 12 is negative we re gonna go with quadrant 2 because it s where the negative portion of the restriction See graphic If the cosine y value is12 then we know the reference angle is going to be pi3 However pi3 isn t in quadrant 2 So we need an angle with a reference angle of pi3 that s also in quadrant 2so we get 2pi3 ANSWER 2pi3 4 Sec sin391 12 For this one we re looking for the secant OF an angle that has a sin of 12 So let s first find the angle with a sin of 12 and then find the secant of whatever that angle is Because we re talking about sin where restrictions are located in quadrants 1 amp4 and 12 is positive we know the angle has to be in quadrant 1 where the positive portion of the restriction is If the sin y value is 12 then that means the reference angle is pi6 In this case the pi6 is in quadrant 1 so we have the first half of our answer Now we just need to find the secant of pi6 and we get 2sqrt3 ANSWER 2sqrt3 5 Tan arcsin 1 In this question we re looking for the tangent OF an angle whose sin is 1 Because we re talking about sin with the positive restriction is in quadrant 1 and 1 is positive we re going to be in quadrant 1 If the tangent is 1 and we re in the first quadrant then the angle has to be pi2 So now all we have to do is take the Tan of that angle but we have a problem Tangent is undefined at pi2 and at all odd multiples of pi2 because this is where the asymptotes are on the tangent graphs ANSWER Undefined 6 Cos arctan 12 For this one we re looking for the cosine OF an angle who s tangent is 12 Ok so lets find an angle with a tangent of 12 and then find the cosine of it problem here is that we have no idea of what this angle is But never fear the Pythagorean Theorem is here Sorry I had to But just like with the other examples because we re talking about tangent whose restrictions are in quadrants 1 and 4 and 12 is negative we re going to be in quadrant 4 where negative portion of the restriction is l Dimmt I 1 Z addmenr hypo 39 F J H we 5 39 If ANSWER 2sqrt5 7 Sin sin391 7 It s easier to just see it but if we say it out loud we re looking for the sin of an angle with a sin is 7 so you get 7 again It s a little confusing but if you have the same ratio then you use the same number given But make sure values are defined For this one if we had 12 the answer would be undefined because restrictions for sin are 1 to 1 ANSWER 7 8 Tan tan391 3 Same concept as earlier ANSWER 3 9 Cos391 cos pi5 Same concept from 7 amp8 I ANSWER pis 10 Sin391sin 5pi6 For this one we re looking for an angle s whose sine is the sine of 5pi6 So basically we re looking for an angle with the same sine as 5pi6 sqrt 32 that fits in the restrictions for sine But how do we know which angle to choose Ok so this is where things get a little confusing Just like in previous examples you re tempted to just answer with 5pi6 However this doesn t work here because 5pi6 doesn t fall within the restrictions for sin pi2 to pi2 which would be quadrants l amp4 Because we re talking about sin we re gonna be in quadrants l amp4 However 5pi6 is in quadrant 2 where 5pi6 is positive so we need to go to quadrant l where the positive portion of the restrictions is So if which angle from quadrant 1 will have the sin of 5pi6 sqrt32pi6 ANSWER pi6 11 Tan391tan 35pi 18 So in this one we re looking for an angle whose the tangent is the tangent of 35pi 18 So basically we re looking for an angle with that has the same tangent as 35pi 18 Because we re talking about tangent we re gonna be in quadrants l amp4 Because 35pi 18 is in quadrant 4 where tangent is negative we re gonna be in quadrant 2 where the negative portion of tangent s restriction is But how do we find that angle We use the reference angle Graphic to make things click a little better I Remember that we re simply looking for an angle that has the same tangent as 35pi 18 and because 35pi18 and pi 18 are in the same spot then they have the same tangent and falls within the tangent restriction So pi 18 works I ANSWER pi18 12 Cos391cos12pi5 Very similar to 11 In this one we re looking for an angle that has the same cosine as 12pi5 AND falls within the restrictions for cosine 0 to pi aka quadrants 1 amp2 The reason we can t just use 12pi5 similar to the concepts in 79 is because it doesn t fall within cosine s restrictions Because we re talking about cosine we re going to be in quadrants 1 amp2 and because 12pi5 is negative we re gonna be in quadrant 2 but 12pi5 is in quadrant 4 However in quadrant 4 cosine is positive so you re going use an angle from quadrant 1 positive portion of cosine s restrictions But how do we find that angle Use the reference angle and just find it in quadrant 1 The reference angle for 12pi5 is 2pi5 2pi5 has the same cosine as 12 pi5 and is in the correct restrictions so it s a correct answer I ANSWER 2pi5 13 If 0ltblt1 then cos cos39lb This one goes back to concepts from 79 this works here because you re using the same trig value and b lies withing cosine restriction ANSWER b 14 If agt1 then sec sin391 lsqrta Because agt1 then it s not within the restrictions so we have to do more work Because a is positive we re gonna be in quadrant 1 4mm g 51 IT39 4quot quot tagIra2 x Ga LJa MWHWK lmtm be 9 15 Solve sin3912 sqrt3X2 2pi3 sin Z l muHmH 1 1 ANSWER 3 sqrt3 61 62 establish each identity 1 cose tane cote cscG Basically we have to prove that this statement is true Final line bottom right corner got slightly cut off it should be cscG FEBRUARY 11th 2016 1 Coscos391 6 Because the trig ratios cos are the same and 6 lies within the cos restriction 1 to 1 then 6 is the answer ANSWER 6 2 Cos arctan1 We re looking for the cosine of an angle who s arctan is 1 So we re going to find the angle with an arctan of 1 and then find the cosine of that angle which is pi4 So now we have cospi4 which is sqrt 22 ANSWER sqrt 2 2 3 Sin391 sin 11pi13 For this one we re looking for an angle that has the same sin of 1 1pi 13 and is in the restrictions of sin pi2 to pi2 Step 1 find draw angle Step 2 Find appropriate quadrant it should be in Step 3 Use reference angles to get it there 1 1pi 13 is in quadrant 3 which isn t in the restrictions for sin pi2 to pi2 which would be quadrants 1 and 4 Because sine is negative in quadrant 3 then we re gonna go to quadrant 4 the negative portion of sine s restriction The reference angle of 1 1pi 13 is 2pi 13 so the answer is negative 2pi 13 If it helps just think of throwing the reference angle into the appropriate quadrants 4 in this case and then adding the sign that goes with the quadrant m vs I 4 Tan391 tan 7pi 10 For this one we ll looking for an angle that has the same tangent as 7pi 10 and falls within the tan restrictions pi2 to pi2 7pi 10 is in quadrant 2 but tangent s restrictions are from pi2 to pi2 so quadrants 1 and 4 Since tangent is negative in quadrant 2 we need to go to the negative portion of tangent s restriction quadrant 2 Via using the reference angle which is 3pi 10 So we re gonna throw the reference angle into quadrant 2 and because it s in the negative portion of the restriction the answer is negative ANSWER 3pi10 5 Cotcos391 0 sin391 1sqrt2 ANSWER 1 Going back to 61 62 Establish each identity aka prove the statement is true that the left side is indeed equal to the right side 2 secG1secel tanze After foiling you get 860261 look familiar 860261 Tan26 tanze 3 3sin26 4cos26 3 cos26 3sin26 4cos26 3 cos26 3 lcosze 4 cosze 33cos26 4cos26 3 cosze 3 cosze Complete the identity 5 Is this an identity N Jm m1 JdE mH hl ANSWER No this is not an identity


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