IGCSE MATHS Doubts

[$H00t!N& $t@r]

do you need this for math or add math since i can't seem to remember any sine/cosine curve qs in math past papers...
do you mean sine/cosine rules by any chance?

I need it for math (0580). actually there are many questions on this but i dont seem to uderstand them. there isnt much in the text book either...  I dont mean the sine and cosine rule used in triangles. Apparently, there are sine and cosine curves. do you have any idea on those?

[Dibss]

yep there are those ) alright give me an example of a qs from past papers an we'll see if i can use that to explain

[$H00t!N& $t@r]

yep there are those ) alright give me an example of a qs from past papers an we'll see if i can use that to explain

In fact one example of such questions is in this thread (page 13). It is solved but i dont grab the concept. Could you help me with that? 

[Dibss]

try this site for the basic idea of the curves. http://curvebank.calstatela.edu/unit/unit.htm
if you draw a circle and divide into quarters or four 'quadrants' it looks something like below:
 
  90 
     S  |  A
180----------0/360
     T  |  C 
       270
each quadrant is labelled to represent what is positive in that quadrant:
1) A=All +ve , S= Sin +ve , T=Tan +ve , C=Cos +ve
2) This shows the four quadrants, going anticlockwise from 0 to 90 to 180 270 to 360.
you can test this out to familiarise yourself with it. do sin or cos or tan of any number between 0-90 on your calc and you'll get a positive answer.
do sin of any no from 90-180--> positive
tan of any no form 180-270--> positive
cos of any no form 270-360-->positive

ok. the actual question:
a)
sinx=sin 50
x=50degrees
they want the angle between 90 and 180 that'll give you the same answer. The angle already given to us is in the 1st quadrant, so now we know we will get the same value in the 2nd quadrant with (180-x)
180-50=130degrees
ok i just realised this might be difficult to understand without a diagram if you haven't done this before :/

b)sinx=-sin50
sine is negative in 3rd and 4th quadrants
180+50=230
360-150=310

i'm really sorry if i haven't explained it clearly... hope it helps a little

[Fastal]

sorry for the late reply.

heres 2)d)iii

work out the inverse first.

determinant - 1.5

(1/1.5     0   ) = (2/3  0)
(  0    1.5/1.5)   (0     1)

now first, u have to learn this list of matrix transformations, dont learn them all, learn the imp. ones like shear , stretch, enlargement etc, heres it : https://studentforums.biz/index.php/topic,5575.msg167872.html#msg167872  (3rd post)

and this is what it says abt stretch

Stretch

1. x invariant  wer k=scale factor (1 0)
                                             (0 k)

2. y invariant  wer k=scale factor (k 0)
                                             (0 1)

now if u notice, out matrix is like the 2. one.

therefore, it means, that its stretch with y invariant and scale factor = k = 2/3

and abt the shear question, sry, im a little confused myself. will try to help asap/

so its a MUST to learn all the transformation matrices, theres no other way?
and about the shear question, please help

[PeaceMaker]

Paper 2 2005 May/june Q) 2, 6, 13, 14b, 17. I don't understand please explain and again, thanks!

[Fastal]

Paper 2 2005 May/june Q) 2, 6, 13, 14b, 17. I don't understand please explain and again, thanks!

i have a doubt in 14b tooh

[astarmathsandphysics]

Will look as soon as i get up

[Meticulous]

Paper 2 2005 May/june Q) 2, 6, 13, 14b, 17. I don't understand please explain and again, thanks!

Q2: sin5=h/3.17.....h=3.17*sin5

Q6: using triangle PBQ, PB=BQ= 4 cm..now use pythagoras

Q13: Easy. Count the dots first, then get the nth term.

Q14 b: % increase= (new value-old value)/old value  x 100%

Q17: it says t=no of hours after midday, so at midday t=0..substitute 0 into the equation.

[Monica]

Q2) You will just use the sin rule. It is a right triangle, you have the angle, hyp and you need the opposite(h).

Sin5= opp/hyp = h/3.17

Q6) Each side of ABCD is 8 cm and each point of PQRS is the mid point of ABCD, thus PB=4 and BQ=4.

Q14) Percentage increase= Increase/ actual *100.
                                  310 000 - 62 000 / 62 000 = 4*100 = 400%

[Fastal]

SOMEONE PLEASE

[PeaceMaker]

Thank you very much, people. If anyone could explain Q13 [May/june paper 3 2005] in detail, that would be nice.

[Fastal]

if youre talking about paper 2 2005 june q13 than here----

a) you can see in the diagram that it increases by three thus
Pattern number|1|2|3|4|
Number of dots|5|8|11|14|

b)use the forumula
a_n=a₁ +(n-1)d[d is the difference, and a₁ is the first term]
a_n=5+(n-1)3
a_n=5+3n-3
a_n=3n+2

c)subsitute formula here
3*60+2=182
d)subsitute again
89=3n+2
thus n=29

[PeaceMaker]

Thanks! Now everything is clear. Time to continue solving more papers.

[Vin]

SOMEONE PLEASE

Its clearly shear .. y axis invariant line .. im a lil confused about factor .. i think its 0.5 .. check in the ms ..