Mathematics: Post your doubts here!

[daredevil]

 

[sumeru]

Oct Nov 2007, Q7, part (ii) Why are we supposed to use the radian mode instead of degree mode?
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

Well, in our P3, we mostly use the angles in radians, unless specified in the questi

But for the same question daredevil asked, why doesn't my method work?
Here's the method sumeru:

Since they ask for the shortest distance from the point p to the line that lies ON the plane, so we can use the normal vector of the plane as the directional vector for the line equation PX, where X is a point on the plane where a line from P and the line (given in the question) on the plane intersect.
So my line equation for PX becomes: <0,2,4> +b<2,-2,3>
Now, if I put into the plane, 2x-2y+3z=1 this point, <2t, 2-2t, 4+3t> to find the value of b, and then by substituting this value into the line equation of PX, am I finding the the position vector for X where the line from P intersect the line in the equation on the plane?
Why is thisi method wrong?

Are you giving me the method or asking whether it is correct or not?

 

[asd]

Well, in our P3, we mostly use the angles in radians, unless specified in the questi

Are you giving me the method or asking whether it is correct or not?

Asking you why my method is wrong

 

[daredevil]

Help me with the vector ques plz...

 

[sumeru]

View attachment 39857

Use square root(a^2+b^2) formula to get the modulus.....In this case a=(1+cos2t), b=sin2t....use trignometric rules and solve it.....and for argument.....use tanx=(b/a),
again solve them.....
for modulus tanx=sin2t/(1+cos2t)
=2sint.cost/(1+2cos^t-1)
=2sint.cost/2cos^2t
=tant
so the argument will be t...

 

[sumeru]

Asking you why my method is wrong

 

[daredevil]

Use square root(a^2+b^2) formula to get the modulus.....In this case a=(1+cos2t), b=sin2t....use trignometric rules and solve it.....and for argument.....use tanx=(b/a),
again solve them.....
for modulus tanx=sin2t/(1+cos2t)
=2sint.cost/(1+2cos^t-1)
=2sint.cost/2cos^2t
=tant
so the argument will be t...

yeah i did it like that.... but made a mess out of the modulus... lol
will try it again.... hopefully success this time :`)

 

[syed1995]

yaar i asked a friend and she said that she doesnt know the logic of this but apparently all differential questions are solved in radians.

did u know that?? A star syed1995 ??

don't you dare tag me after starting that fight with me -,-

And it's simple. In Maths P1/P3.. It's always in radians unless specified in the question. like give your answer in degrees.. or Cos(90°) or the presence of "°" or the word "degree" in the question.

 

[A star]

don't you dare tag me after starting that fight with me -,-

And it's simple. In Maths P1/P3.. It's always in radians unless specified in the question. like give your answer in degrees.. or Cos(90°) or the presence of "°" or the word "degree" in the question.

lol that fight was like days ago I thought you had gone for your hibernation

 

[asd]

Help me with the vector ques plz...

I hope its clear enough that you can read it
My webcam sucks and my cell wont connect to my laptop! :///

 

[asd]

Im guessing that's supposed to be a " . " in place of " x "? 'cause their cross product is definitely not a 0.
Well, thanks anyways!

 

[sumeru]

Im guessing that's supposed to be a " . " in place of " x "? 'cause their cross product is definitely not a 0.
Well, thanks anyways!

Ah, yes....really sorry for that silly mistake......

 

[David Hussey]

so we know
sin2A=2sinAcosA

but at times we will require sinA, so how do you find it?
is it 4sinAcosA or 0.5sinAcos or something else? :/
same thing with tan and cos
someone pls explain!

 

[syed1995]

so we know
sin2A=2sinAcosA

but at times we will require sinA, so how do you find it?
is it 4sinAcosA or 0.5sinAcos or something else? :/
same thing with tan and cos
someone pls explain!

Sin2A = 2SinACosA .. the 2A gets divided by 2 to become A.
SinA = 2Sin(A/2)Cos(A/2) .. The A will get divided by 2 as well..

The rest remains the same because that's the formula from a derivation which isn't in our course. if Theta changes that formula wouldn't change. and A and 2A are values of Theta.

simple as that.

 

[David Hussey]

and what is sinAcosA equal to? :/

 

[A star]

so we know
sin2A=2sinAcosA

but at times we will require sinA, so how do you find it?
is it 4sinAcosA or 0.5sinAcos or something else? :/
same thing with tan and cos
someone pls explain!

you just need to see what you want to prove if you want to prove something like sin (A/2) then start with sin(A) if you have to go to sin(A/4) you need to form sin(A/2) then sn(A/4) dependent on the questions which if come are mind blowing

 

[A star]

and what is sinAcosA equal to? :/

nothing unles you multiply both sides (if any) by 2

 

[David Hussey]

you just need to see what you want to prove if you want to prove something like sin (A/2) then start with sin(A) if you have to go to sin(A/4) you need to form sin(A/2) then sn(A/4) dependent on the questions which if come are mind blowing

*shoots himself*

 

[syed1995]

and what is sinAcosA equal to? :/

As Sin2A = 2SinACosA
then SinACosA = (Sin2A)/2

as you will have to divide by 2 on both sides to get rid of the 2.

 

[Faithix MFSH]

Can someone please explain the solution to the second part of this question with detailed steps.
Would be greatly helpful if you manage to explain your steps..

Info about the paper source is in the uploaded screenshot.


Would be highly appreciated.