Mathematics: Post your doubts here!

[destined007]

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_qp_41.pdf c'mon help me with number 3 and guidlines to answer this type of question? urgent

 

[ArthurBonZavi]

Well done !

But this might rather be solved quite easily by applying Lami's theorem.

 

[rz123]

^ can u please do that by applying lami's theorem? Thanks

 

[yuliana95]

I know that this must be some silly questions.. but really need your help!! Please.. It's URGENT.. :'(
Question:
Given that sin x+sin y=1/2 and cos x+cos y=1/3
Find: a. sin(x-y)
b. cos (x-y)
It would really mean a lot to me if you could help me... Thanks in advance..

 

[yuliana95]

I know that this must be some silly questions.. but really need your help!! Please.. It's URGENT.. :'(
Question:
Given that sin x+sin y=1/2 and cos x+cos y=1/3
Find: a. sin(x-y)
b. cos (x-y)
It would really mean a lot to me if you could help me... Thanks in advance..

 

[ArthurBonZavi]

^ can u please do that by applying lami's theorem? Thanks

First form a closed triangle, with the three forces as W N, 5.5 N and 7.3 N. Note that this will form a right-angled triangle which will have 7.3 N as the hypotenuse side, then you will get W by the Pythagoras' theorem and the value of the angle by the same general method of right-angled triangles.

If you don't get it, tell me.

 

[ArthurBonZavi]

I know that this must be some silly questions.. but really need your help!! Please.. It's URGENT.. :'(
Question:
Given that sin x+sin y=1/2 and cos x+cos y=1/3
Find: a. sin(x-y)
b. cos (x-y)
It would really mean a lot to me if you could help me... Thanks in advance..

Do we just have to simplify or we have to get the answers in the numerical form ?

 

[rz123]

^ can u please do that by applying lami's theorem? Thanks

First form a closed triangle, with the three forces as W N, 5.5 N and 7.3 N. Note that this will form a right-angled triangle which will have 7.3 N as the hypotenuse side, then you will get W by the Pythagoras' theorem and the value of the angle by the same general method of right-angled triangles.

If you don't get it, tell me.

sorry i didn't get it...can u post the detailed solution. like how did u draw this closed triangle..can u make it more clearer. thanks..

 

[abcde]

I know that this must be some silly questions.. but really need your help!! Please.. It's URGENT.. :'(
Question:
Given that sin x+sin y=1/2 and cos x+cos y=1/3
Find: a. sin(x-y)
b. cos (x-y)
It would really mean a lot to me if you could help me... Thanks in advance..

I'm not quite sure but here's my attempt:
(a) sin (x - y) = sin x cos y - sin y cos x
= (1/2 - sin y) cos y - sin y (1/3 - cos y)
= 1/2 cos y - 1/3 sin y

(b) cos (x - y) = cos x cos y + sin x sin y
= (1/3 - cos y) cos y + (1/2 - sin y)sin y
= 1/3 cos y - cos^2 y + 1/2 sin y - sin^2 y
= 1/2 sin y + 1/3 cos y - 1.

Please do mention the correct answers.

 

[yuliana95]

I know that this must be some silly questions.. but really need your help!! Please.. It's URGENT.. :'(
Question:
Given that sin x+sin y=1/2 and cos x+cos y=1/3
Find: a. sin(x-y)
b. cos (x-y)
It would really mean a lot to me if you could help me... Thanks in advance..

Do we just have to simplify or we have to get the answers in the numerical form ?

I'm not quite sure myself.. My teacher just gave me that problem and he didn't say anything.. I'll try to ask him..

 

[ArthurBonZavi]

(a) sin (x - y) = sin x cos y - sin y cos x
= (1/2 - sin y) cos y - sin y (1/3 - cos y)
= 1/2 cos y - 1/3 sin y

(b) cos (x - y) = cos x cos y + sin x sin y
= (1/3 - cos y) cos y + (1/2 - sin y)sin y
= 1/3 cos y - cos^2 y + 1/2 sin y - sin^2 y
= 1/2 sin y + 1/3 cos y - 1.

Please do mention the correct answers.

I am getting the same answers.

 

[ArthurBonZavi]

sorry i didn't get it...can u post the detailed solution. like how did u draw this closed triangle..can u make it more clearer. thanks..

Have a look at this :
http://en.wikipedia.org/wiki/Lami%27s_theorem

See the mark scheme for further information.

 

[EnigmaticSolitude]

Here's a question:

Thanks in advance!

 

[mrpaudel]

Here's a question:

Thanks in advance!

x=X+1 so....X=x-1 ....now... substitute this value of (x-1) as X in above equation to get y^3=X^2 .....now draw a curve.......i hope u can get this one... and..draw x=1 line..to show..the line cuts that curve in two equal halves(i.e symmetrical)

 

[yuliana95]

(a) sin (x - y) = sin x cos y - sin y cos x
= (1/2 - sin y) cos y - sin y (1/3 - cos y)
= 1/2 cos y - 1/3 sin y

(b) cos (x - y) = cos x cos y + sin x sin y
= (1/3 - cos y) cos y + (1/2 - sin y)sin y
= 1/3 cos y - cos^2 y + 1/2 sin y - sin^2 y
= 1/2 sin y + 1/3 cos y - 1.

Please do mention the correct answers.

I am getting the same answers.

I think it should be numerical.. but I am not sure either.. Thanks guys..

 

[VelaneDeBeaute]

A question from The Binomial Theorum !
:: Pure Mathematics, Chapter 9, Exercise 9B, Q.7.
Q = Find the first three terms in the expansion, in ascending powers of x , of (1+2x)^8. By substituting x = 0.01, find an approximation to (1.02)^8.
Im having problems with the bold text, can someone please explain in detail how is this gonna work out ! Ty !

 

[destined007]

A question from The Binomial Theorum !
:: Pure Mathematics, Chapter 9, Exercise 9B, Q.7.
Q = Find the first three terms in the expansion, in ascending powers of x , of (1+2x)^8. By substituting x = 0.01, find an approximation to (1.02)^8.
Im having problems with the bold text, can someone please explain in detail how is this gonna work out ! Ty !

(1.02)^8=(1+2(0.01))^8
compare this with (1+2x)^8
now put x=0.01 in the expansion of (1+2x)^8

 

[VelaneDeBeaute]

^ I didnt get it
Actually by substituting the value and solving the whole expression, we get the exact answer where as the question asks for approximate ! :sorry: :Search:

 

[destined007]

^ I didnt get it
Actually by substituting the value and solving the whole expression, we get the exact answer where as the question asks for approximate ! :sorry: :Search:

You used first three terms only, right?

 

[VelaneDeBeaute]

^ Something like that is in the book i guess !! And you see, these three term are to be found in the first part of the question! If we hadn't had to use them, why would the question asked for it?