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# Somolom C3 Question....... watch

1. Can someone please show me how to do parts a b and c on this question, thanks...........
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2. (Original post by melbourne)
Can someone please show me how to do parts a b and c on this question, thanks...........
If you sub in the values of x = pi/2 and y = -1 then you get an equation in a and b

Do the same for the other given values of x and y and you have a second equation in a and b.

Then solve these simultaneous equations
3. When you find the modulus, anything negative becomes positive. If you imagine the y-axis to be the f(x)-axis, then any negative y becomes it's positive counterpart for the same value of x.
4. (Original post by seankhn)
When you find the modulus, anything negative becomes positive. If you imagine the y-axis to be the f(x)-axis, then any negative y becomes it's positive counterpart for the same value of x.
1) I thought the modulus part was only for a)

2) You have the equation

y=a+bcosec x

so

-1=a+b...... and what about the cosec x??????

Sorry im being a bit dim today im tired! lol
5. (Original post by melbourne)
1) I thought the modulus part was only for a)

2) You have the equation

y=a+bcosec x

so

-1=a+b...... and what about the cosec x??????

Sorry im being a bit dim today im tired! lol
cosecx = 1/sinx
Does that help?
6. (Original post by Widowmaker)
cosecx = 1/sinx
Does that help?
no not really i knew that

I just dont get why you can get rid of the cosec x and simply have y= a+b and y= a-b (which i still dont understand is a minus)......

lol
7. (Original post by melbourne)
no not really i knew that

I just dont get why you can get rid of the cosec x and simply have y= a+b and y= a-b (which i still dont understand is a minus)......

lol
lol, i'm only trying to help.

f(x) = a + bcosecx = a + b/sinx
Points on graph are; (½π,-1) and (3π/2,-5)

-1 = a + b/sin[½π]
-1 = a + b/1
-1 = a + b (1)

-5 = a + b/sin[3π/2]
-5 = a + b/-1
a - b = -5
a = -5 + b (2)

(2) into (1)
-1 = -5 + b + b
2b = -1 + 5
b = 4/2 = 2

a - b = -5
a = -5 + b
a = -5 + 2
a = -3

Just sub the values that you know lie on the graph into the equation to form simultaneous equations.
8. (Original post by Widowmaker)
lol, i'm only trying to help.

f(x) = a + bcosecx = a + b/sinx
Points on graph are; (½π,-1) and (3π/2,-5)

-1 = a + b/sin[½π]
-1 = a + b/1
-1 = a + b (1)

-5 = a + b/sin[3π/2]
-5 = a + b/-1
a - b = -5
a = -5 + b (2)

(2) into (1)
-1 = -5 + b + b
2b = -1 + 5
b = 4/2 = 2

a - b = -5
a = -5 + b
a = -5 + 2
a = -3

Just sub the values that you know lie on the graph into the equation to form simultaneous equations.
lol sorry widowmaker i didnt mean to sound rude! hehe

I repped you for the help

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