final post!:trig

part 1  q:17  this going to be a cos     max-min/2= a 4-0/2=2 y= 2cos x   period=6 frequency= b

period= 2pi/|b| 6=2pi/|b| 6b=2pi 2pi/6= 1.047197551= b y= 2 cos 1.047197551x h= the distance from the origin to the eoc which is 5.5   k= max+min/2 =2 therefore y= 2cos 1.047197551 (x-5.5) +2

1. a) 5(180)/36= 25 degrees  b)  75 x pi /180 = 75pi/180

part 2

9) idk how to do that

10) a) sin^2x+ cos^2x/1-cos^2x = sin^2x+ cos^2x/sin^2x pi  cross out common factors = cos^2x

b) pi sec^2x/ csc^2x  ri  1/cos x/ 1/sin x=  1/ cos x time sinx/1 = tan x

c) ri 1/cos^2x- sin^2x/cos^2x qi /  cos^2x pi idk how to do the rest….

d) pi tan^2x/ 1/sin^2x x cos^2 x/sin^2 x/sin^2x idk how to do the rest

11) a) pi sin^2x/cos^2x = tan^2x qi  b)  sec^2x/csc^2x  pi c) idk how to do that d) 1/ sin^2x- cos^2x/sin^2x /cos^2 x idk how to do the rest

 

ps i doomed

 

conics cheat sheet

Circle

definition: the locus of a point which is equidistant from a fixed point.

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all the parameters are straight forward. the two formulas are x^2+y^2=1 and the second one is x^2/a^2 + y^2/b^2=1 but remember a and b must be the same or else their not a circle but an ellipse. remember if you are given the distance between one vertex to the origin you know that the rest are the same. example say you know that the circle is 10 units long on the x axis you know that the y axis is the same or else it’s not a circle but am ellipse.
Ellipse

definition: the locus of a point which is a constant sum of distance from two fixed points.

ImageImage these are both ways the ellipse can be placed it depends on whether a or b is bigger. if a is bigger the major axis is on the x axis but if b is bigger the major axis is on the y axis as shown by the two conics above. the first graph would look like this x^2/5^2+y^2/4^2=1and the second looks like this x^2/4^2+y^2/5^2. the major axis goes on the side which is bigger and the minor axis is goes on thee smaller side. to find a or b you must have a or b and the c which is the distance between the origin and the focus. the formula which helps us is c^2=semi major^2- semi minor^2 this formula works for both vertical and horizontal ellipses. also remember that in every ellipse they have two foci so u can use either or. now to put the formula to use say that we know that the foci are (+and -10,0) and the b is 6. we know that for starters our ellipse is horizontal because if the foci are on the x axis and the foci are always on the major axis which lets us know what kind of ellipse we’re dealing with. alright so are c is 10 and are b is 6 so we must add c^2 and b^2 to get a^2. 10^2+6^2=a^2 a^2=136 we’re not done here we want just a. sqrt(136)= a a=11.66190379. the formula for this ellipse is x^2/11.66190379^2+y^2/6^2. oh just remembered something important if u add both the focal radii it equals the same as the major axis.a focal radius is the distance from a focus to a point on the ellipse.

Hyperbola

definition: the locus of points the difference of whose distances from two fixed points is a constant.

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these are both ways the hyperbola can be it just depending on which is bigger the a or b. the first graph’s formula is y^2/b^2-x^2/a^2=1 and the second graph’s formula is x^2/a^2-y^2/b^2=1.the transverse axis is the line which is from vertex to the other one. the conjugate axis the point from the vertex to the asymptote.the transverse axis can be found by doing |L1|- |L2|.  to find the rule of the asymptote you have to pick a point  (a,b) i must put it like so y=+ or – b/a x.  for example i take a point (5,2) the formula would look like y= +or- 2/5x. to find the foci i must use Pythagoras theorem ( c^2= a^2+ b^2). for example say I’m given the formula x^2/6- y^2/10 i would do 6+10=16 then sqrt(16) to get c and c equals 4.

Parabolas

definition: the locus definition of parabola is a point that is equidistant from a fixed point and a fixed line.

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the first graph is y= a(x-h)^2 +k, the second is y= -a(x-h)^2 +k, the third is x= -a(y-k) +h and the fourth is x= a(y-k) +h. the vertical or horizontal line is know as the directrix  and the point is the focus. c is the distance between the directrix or the focus to the origin of the parabola. to calculate c u must know what a is because c=|1/4a|. for example say i know a is 2 and i know that c =|1/4a| this can help me figure out what the function is. so for here the function is y=1/8 x^2. remember on this function both focal radii will always equal the same. also, i noticed that the function x= +or-a(y-k) + h is kinda like to square root functions put together kinds like how if u put two linear together u can make an abs function.

well that’s it for conics cya =)

HISTORY

Name: Jean Claude

Education: pasted unvisity

languages: English, french, Latin, German and  Portuguese.

job: a part of the company of 100 associates

birthday: 17,3,1597

gender: male

home town:Paris,France

current home town: Montreal, Quebec

photos:Image

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merchant's+coffee+hous

que3p8c

sub posts:1 going to  the Mohawks to see if they will trade fur for weapons wise me luck =P. 2 just made some more successful trades with the natives =). 3 so far the company is doing well with fur trade thanks to the high demands from France. 4 business is going so well that we have new trading camps being built all over Hochelaga. 5 today I’m going to church so you all there! 6. I’ve decided today i’m going to just relax. 7. just got word from France that they want more fur, boy business is booming. 8. just meet up with a lovely lady from France, future wife perhaps =P 9. in one year i made 1 million from this company =P 10. got married ! to the a wonderful women =)

pythagorean identities

the first identity, i think we all know this one from like grade 6 which is when u have  a r.a.t( right angle triangle) a^2+b^2=c^2 or in our case sin^2 x + cos^2 x= h   and h=1  then next two u get by dividing sin^2 x + cos^2 x=1 by cos^2 x or sin^2 x. the outcome by dividing by cos^2 x is  tan^2 x + 1= sec^2 x  and for sin^ x is 1+ cot^2 x= csc^2 x. we get these answers by changing some of the stuff into quotients or reciprocals. ex:   sin^2 x/ sin^2 x +cos^2 x/ sin^2 x + 1/sin^2 x. sin^2 x/ sin^2 x is just simply 1  and the next part we know that the quotient of that is cot^2 x. for the last part we know the reciprocal of 1/sin^2 x is csc^2 x. well that’s it for now  have a nice day.