Post Your Problem Here (2 Viewers)

[velox]

1. Prove that (1/(sinXcosX)) tan X = cot X
I tried it but got nowhere

2. Show that (6,-4), (5,-1) and (2,8) are collinear

3. Solve the following equations, leaving your answers in surd form if necessary

(i) x + x^-1 = 2
(ii) x(x+3) = 9
For (ii) can this be made into a simple quadratic? x^2 + 3x 9 = 0, and then solved? Can
The same be done to (i) aswell?

4. Solve the simultaneous equations

2a 7b + 3c = 7
a +3b+2c = -4
4a + 5b c = 9

 

[queenie]

question 1:
L.H.S.
1/sinxcos - sinx/cosx
= (1-sin^2x)/sinxcosx
1-sin^2x= cos^x

:. cos^2x/sinxcosx = cosx/sinx
= cot x
= R. H. S.

 

[queenie]

for 2.. to prove that three points are collinear, u must proove that their gradients are equal. To do this u must find the gradient of the first and the second, and the first and the third, and see if they are equal.

For 3 ii) i would do wat u said.. expand and then simplfify using the quad formula.

for 4):
let 2a-7b + 3c = 7 be equation 1
and so on..

eliminate a from 1 and 2.
you should get somethng like -13b-c=15 (unless i did something wrong)

then eliminate a from 2 and 3:
you should get something like
-19b + 7= 5

Then work on these two equations like normal siml equations to find a and b. Once you have found both a and b, sub these into one of the original equations to find c.

Remember to check in all three equations to see if you have got all three values right..

 

[Heinz]

1. Prove that (1/(sinXcosX)) tan X = cot X

(1/(sinXcosX)) sin^2x/sinxcosx
= (1 -sin^2x) /sinxcosx
= cos^2x/sinxcosx
= cosx/sinx
= cotx

2. Show that (6,-4), (5,-1) and (2,8) are collinear

The equation of the line passing through (6,-4), (5,-1) is

(y+1)/(x-5) = (-4+1)/(6-5)
y+1 = -3(x-5)
y = -3x + 14

test (2, 8)
y = -3x + 14
y = -3(2) + 14
= 8

:. (6,-4), (5,-1) and (2,8) lie on the same line (collinear)


3. Solve the following equations, leaving your answers in surd form if necessary

(i) x + x^-1 = 2
(ii) x(x+3) = 9
For (ii) can this be made into a simple quadratic? x^2 + 3x 9 = 0, and then solved? Can
The same be done to (i) aswell? yes

i) (x-1)^2
ii) x^2 + 3x - 9 = 0 quadratic equation

4. Solve the simultaneous equations

2a 7b + 3c = 7
a +3b+2c = -4
4a + 5b c = 9

4a 14b + 6c = 14 (first line)
4a +12b+8c = -16 (second line)
4a + 5b c = 9

second line minus first line
26b + 2c = -30
:. 13b + c = -15
:. 91b + 7c = -105
third line minus first line
19b - 7c = -5

solve these two silmutaneously to get b and c (b = -1 and c = -2)

Sub these two values into any of the original equations to get a (=3)

Edit: My mistake, your right queenie

 

[velox]

thanks

 

[queenie]

Originally posted by Heinz

ii) x^2 + 3x - 9 = 0 (discriminate <0 :. no real solutions)

um no discriminate = b^2-4ac
= 3^2 - (4 x 1 x -9)
which on my calculator is positive, which means that there are two real solutions..

Originally posted by Heinz
Proving that the gradients are equal simply means that theyre parallel not collinear
[=

um if u show that the gradient of ab = the gradient of ac
then they are colliener, they are not 'simply' parallel.... wat u did was not needed, by proving that ab=ac, u are saying that they are on the same line anyway, because they have a common point.

 

[Heinz]

Originally posted by queenie
um no discriminate = b^2-4ac
= 3^2 - (4 x 1 x -9)
which on my calculator is positive, which means that there are two real solutions..



um if u show that the gradient of ab = the gradient of ac
then they are colliener, they are not 'simply' parallel..

fixed the typo for that question (tends to happen when your doing it in your head)

And sorry, I didnt read your entire method.

 

[CM_Tutor]

Originally posted by queenie
um if u show that the gradient of ab = the gradient of ac
then they are colliener, they are not 'simply' parallel.... wat u did was not needed, by proving that ab=ac, u are saying that they are on the same line anyway, because they have a common point.

This is correct, but anyone using it should remember to state the argument. ie:

Since m_AB = m_AC, AB || AC and they share a common point (A), and so A, B and C are collinear.

 

[velox]

hmm i still see it as heinz saw it in the first place ???^^^

 

[CM_Tutor]

Wrx - take a sheet of paper. Draw a line on it, and mark two points - A and B - on the line. Now can you draw another line through A, that is parallel to the existing line AB that is not the same line?

 

[velox]

thanks, some times baby explanations really help thanks

 

[ToO LaZy ^*]

could someone plz check if the answer to the following question is right..


the diagram shows the graphs of the functions y=cosx and y=sin2x between x=0 and x=pi/2
the two graphs intersect at x=pi/6 and x=pi/2. calculate the area of the shaded region.


the answer that i got was 6/4...?
but not sure

 

[CM_Tutor]

Originally posted by ToO LaZy ^*
the diagram shows the graphs of the functions y=cosx and y=sin2x between x=0 and x=pi/2
the two graphs intersect at x=pi/6 and x=pi/2. calculate the area of the shaded region.


the answer that i got was 6/4...?
but not sure

I got area = 3 /4 sq unit.

 

[ToO LaZy ^*]

ah crap..!..
could you show me the working for it..?...

 

[ToO LaZy ^*]

btw..happy birthday

 

[CM_Tutor]

Area = int (from 0 to pi/6) sin 2x dx + int (from pi/6 to pi/2) cos x dx
= [(-1/2)cos 2x] (from 0 to pi/6) + [sin x] (from pi/6 to pi/2)
= (-1/2)cos(pi/3) - (-1/2)cos 0 + sin(pi/2) - sin(pi/6)
= (-1/2) * (1/2) - (-1/2) * 1 + 1 - (1/2)
= -1/4 + 1/2 + 1 - 1/2
= 3/4 sq units.

It is worth remembering the area under sin x or cos x for one quarter of a period is 1. The shaded area is clearly less than 1, so it's easy to see your answer was too big, even if it is not necessarily clear what the answer should have been.

and Thanks for the birthday wishes - I'm going to post in the Non-School thread wishing me a happy birthday later tonight.

 

[CrashOveride]

Well you get integral of cosx from pi/2 to pi/6

Then get integral of sin2x from pi/6 to 0.

Add these two to get shaded area, 3/4 units²

 

[CrashOveride]

This is fairly simple question just that i get more equations than they say are possible:

Q:

A line has gradient 3/4 and its perp. distance from (1, -3) is 2 units. Find the two possible equations of the line.

Define this line as: ax + by + c = 0
-a/b = 3/4

So a=3, b=-4 OR a=-3, b = 4

Put each into the perp. dist. formula to find your value for C. However, coz its absolute value, for each case there will be two diff values of C. Thus two answers for each set of a & b, thus 4 answers in total. Anyone?

 

[ToO LaZy ^*]

ok..got it..
thx CM

EDIT: OMG...i just realised how stupid i was..!!
i subtracted them instead of adding them!..derrr!!..me is dumb!

 

[CM_Tutor]

CrashOveride - there are only two lines. If you actually find the 'c' values, you'll find that two of the answers are the other two, multiplied by -1.