Qualification > Math
Additional Math Help HERE ONLY...!
jellybeans:
Helloooo, :D
ummm i need help with Q2 please? :) i can find the first two values but... how do you find the rest? :o and how do you know how many 'values' can fit in within the range? :-\
THANKS IN ADVANCE ;D
BlackBunny103:
--- Quote from: astarmathsandphysics on June 07, 2010, 12:24:07 pm ---here
--- End quote ---
Thanxx a lot, Mr Astar :D
slvri:
okay
3sin(x/2-1)=1
sin(x/2-1)=1/3
convert the given range for x into a range for (x/2-1) which is the basic angle here
0<x<6pi
0<x/2<3pi
-1<x/2-1<3pi-1
-1<x/2-1<8.425
now denote x/2-1 by a(alpha but i cant write that here so a it is)
sin a is positive in 1st and 2nd quadrants
a=sin-1(1/3)=0.3398
in the 1st quadrant a=0.3398
in 2nd a=pi-0.3398=2.802
but we can go till 8.425 as indicated in the range
so another value of a=2pi+0.3398=6.623
but no other value for 2nd quadrant cause it goes outside range(2.802+2pi=9.085>8.425
so a=0.3398,2.802,6.623
x/2-1=0.3398,2.802,6.623
x/2=1.3398,3.802,7.623
x=2.6796,7.604,15.246
x=2.68,7.60,15.2 correct to 3sf
is this correct?
jellybeans:
--- Quote from: slvri on June 07, 2010, 03:25:40 pm ---okay
3sin(x/2-1)=1
sin(x/2-1)=1/3
convert the given range for x into a range for (x/2-1) which is the basic angle here
0<x<6pi
0<x/2<3pi
-1<x/2-1<3pi-1
-1<x/2-1<8.425
now denote x/2-1 by a(alpha but i cant write that here so a it is)
sin a is positive in 1st and 2nd quadrants
a=sin-1(1/3)=0.3398
in the 1st quadrant a=0.3398
in 2nd a=pi-0.3398=2.802
but we can go till 8.425 as indicated in the range
so another value of a=2pi+0.3398=6.623
but no other value for 2nd quadrant cause it goes outside range(2.802+2pi=9.085>8.425
so a=0.3398,2.802,6.623
x/2-1=0.3398,2.802,6.623
x/2=1.3398,3.802,7.623
x=2.6796,7.604,15.246
x=2.68,7.60,15.2 correct to 3sf
is this correct?
--- End quote ---
YEP its correct ;) Thank Youuuu :D !
btways if the range is from 0<x< 6pi radians..
isnt the max supposed to be = 18.85? :( why is it ...8.425? :o :o :o :o
youre right ;D ..but why isnt it 6pi = 18.85? :-\ thanks!!!!!
slvri:
thats because theyve given you the range for x but the angle whos sine is being taken in the question is x/2-1, not x so whatever angle you find after taking inverse sine will be for x/2-1, not x itself. thats why you have to convert the given range for x to a range for x/2-1
that means for the limits given (0 and 6pi) you divide by 2 to get x/2 and then subtract 1 to get x/2-1 which is the required range
and by the way i gave add math in june 2009 and i gave a level math in june 2010
so yeah just felt like helping others out for the add math exam tomorrow :)
oh and HI ASTAR....remember me?
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