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16 years ago
#1

a) cos(@-10*) = cos 15*
b) tan [email protected]=.4
c) 2sin @tan=3

@=theta *=degrees

(btw this was the last question on the january 2004 paper so some of u would know the answers to this)
0
16 years ago
#2
(Original post by gemstone)

a) cos(@-10*) = cos 15*
b) tan [email protected]=.4
c) 2sin @tan=3

@=theta *=degrees
a) If cos (@-10*) = cos 15*
then @-10 = 15
=> @ = 15 + 10
=25

b) tan [email protected] = 0.4
[email protected] = tan^-1 (0.4)
@ = [tan^-1 (0.4)]/2

c) Don't quite understand what you've ritten for c....doesn't make sense.

G
0
16 years ago
#3
correction to c)

c) [email protected] [email protected]=3

(this is correct)
0
16 years ago
#4
(Original post by gemstone)
correction to c)

c) [email protected] [email protected]=3

(this is correct)
Fraid i can't figure it out....sorry..anyone else help out???

G
0
16 years ago
#5
(Original post by gzftan)
Fraid i can't figure it out....sorry..anyone else help out???

G
[email protected]@=3

[email protected]([email protected]/[email protected])=3 therefore:

Sin^[email protected]=3/[email protected]
1-Cos^[email protected]=3/[email protected]
Cos^[email protected]/[email protected]=0

Then solve as normally.
0
16 years ago
#6
2sin^[email protected] = [email protected]

Edit - just realised you meant 3/2 of [email protected] rather than 3/2 [email protected] Oh well, at least this others won't have similar confusion now 0
16 years ago
#7
(Original post by Zapsta)
2sin^[email protected] = [email protected]

Edit - just realised you meant 3/2 of [email protected] rather than 3/2 [email protected] Oh well, at least this others won't have similar confusion now Yes, you can remove the denominator by multiplying the whole equation by 2. I had thought that people here are far more capeable than that.
0
16 years ago
#8
Bhaal, sin^2(x) = 1-cos^2(x) so we actually have

2cos^[email protected][email protected] = 0

Then you factorise this a a quadratic in [email protected]; I used to find it easier to let [email protected] and then the equation becomes

2y^2 + 3y - 2 = 0

Which then looks easier to solve. But that is just a method I use.
0
16 years ago
#9
(Original post by mikesgt2)
Bhaal, sin^2(x) = 1-cos^2(x) so we actually have

2cos^[email protected][email protected] = 0

Then you factorise this a a quadratic in [email protected]; I used to find it easier to let [email protected] and then the equation becomes

2y^2 + 3y - 2 = 0

Which then looks easier to solve. But that is just a method I use.
In my working out, I missed that step and automatically put that in.

[email protected]@=3
[email protected]([email protected]/[email protected])=3 therefore:
Sin^[email protected]=3/[email protected]
1-Cos^[email protected]=3/[email protected] <--see
Cos^[email protected]/[email protected]=0
0
16 years ago
#10
(Original post by Bhaal85)
In my working out, I missed that step and automatically put that in.

[email protected]@=3
[email protected]([email protected]/[email protected])=3 therefore:
Sin^[email protected]=3/[email protected]
1-Cos^[email protected]=3/[email protected] <--see
Cos^[email protected]/[email protected]=0
What do you mean? As far as I am aware the quadratic is

cos^[email protected]+3/[email protected]=0
0
16 years ago
#11
(Original post by mikesgt2)
What do you mean? As far as I am aware the quadratic is

cos^[email protected]+3/[email protected]=0
Times by 2 to get:

2cos^[email protected][email protected] = 0 which is what you got, then factorise.

2y^2+3y-2 = 0

(2y-1)(y+2)=0

Therefore [email protected]=1/2 and [email protected]=-2. The only real answer being [email protected]=1/2, therefore @=Cos^1(1/2). For which you should get two answers.
0
16 years ago
#12
wow u ppl r amazing...seriously u r....i never expected to get 9 replies to my question i thought i might have to look somewhere else.....wow i seriously am grateful thanx sooooooooooooooooooooooo much.

cheers
0
16 years ago
#13
new question that i require help in sorry for sounding dumb but i need to make sure these r correct coz i havent got a mark scheme.

f(x)=x^2 - kx + 9, where k is a constant

(a) find the set of values of k for which the equation f(x) = 0 has no real solutions.

Given that k=4
(b) express f(x) in the form (x-p)^2 + q, where p and q are constants to b found

(c) write down the minimum value of f(x) and the value of x for which this occurs
0
16 years ago
#14
(Original post by gemstone)
new question that i require help in sorry for sounding dumb but i need to make sure these r correct coz i havent got a mark scheme.

f(x)=x^2 - kx + 9, where k is a constant

(a) find the set of values of k for which the equation f(x) = 0 has no real solutions.
Given that k=4
(b) express f(x) in the form (x-p)^2 + q, where p and q are constants to b found
(c) write down the minimum value of f(x) and the value of x for which this occurs
b^2-4ac < 0
k^2 - 36 < 0
(k+6)(k-6) < 0
-6 < k < 6

f(x) = x^2 -4x +9
= (x-2)^2 +9 - 4
= (x-2)^2 +5

5, at x=2
0
16 years ago
#15
thanx for part a and b...but how on earth do u get part (c)....can u plz explain to me how u get the 5 and x=2
0
16 years ago
#16
Ok, this is something you will need to learn. When you complete the square on a function of x (NOT when x=0, in this case you are completing the square to FIND x), you get (x+a)^2+b = f(x)

This is what you must remember: In this form, you can find the minimum point of the curve. the minimum point is the number outside the brackets (b), so if b is -5, the min. point of the curve you can possibly get is at y = -5. the corresponding x value is found bu putting the bracket to zero:

x+a = 0
x = -a

Follow this through and you'll get the same result. Hope this helps 0
16 years ago
#17
thanx a lot that really helped....this is the first time i understand the principles of completing the square coz b4 i didnt have a clue..

by the way wat yr r u in?
0
16 years ago
#18

[email protected] [email protected]=3
0
16 years ago
#19
(Original post by gemstone)

[email protected] [email protected]=3
[email protected]@/[email protected] = 3

2sin^[email protected] = [email protected]

2(1-cos^[email protected]) = [email protected]

2-2cos^[email protected] = [email protected]
2cos^[email protected] + [email protected] - 2 = 0

([email protected])([email protected]+2 )

[email protected] = 1/2 or - 2

@ = 60, 300 between 0 and 360
0
16 years ago
#20
thanx a lot that reallly sorted so many things out
0
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