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Mathematics: Post your doubts here!

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acosθ + bsinθ = rsin(θ+α) or rsin(θ-α) or rcos(θ+α) or rcos(θ-α)

How do I know when to use what? :O
Learn them:
acosθ + bsinθ = rcos(θ-α)
acosθ - bsinθ = rcos(θ+α)
asinθ + bcosθ = rsin(θ+α)
asinθ - bcosθ = rsin(θ-α)
When cos is first it's rcos and the signs in brackets are opposite to LHS
When sin is first it's rsin and the signs in brackets are same as LHS
 
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Learn them:
acosθ + bsinθ = rcos(θ-α)
acosθ - bsinθ = rcos(θ+α)
asinθ + bcosθ = rsin(θ+α)
asinθ - bcosθ = rsin(θ-α)
When cos is first it's rcos and the signs in brackets are opposite to LHS
When sin is first it's rsin and the signs in brackets are same as LHS
I didn't know this ... damn, tysm!!! :D
 
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Can someone please explain how to get a second vector for the question (ii) for the vector product ?

upload_2016-2-13_20-34-15.png
 
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Can someone please explain how to get a second vector for the question (ii) for the vector product ?

View attachment 59259
You need two vectors that lie in the plane. Then cross both of them to find the normal vector.
The first vector is i + 2j + k (the direction vector of l)
The second vector is (4i-2j+2k)-(2i+2j+k), since both points are in the plane the vector between them is in the plane.
Note: cross-product is NOT required for Maths P3 (only required for F. Maths). You only need to find a vector that is perpendicular to both of them and this is one of the techniques.
 
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View attachment 59269
Could anyone please help out with the second part?
ms:
View attachment 59270
So first part you have 3/3-2x + -x-2/x^2+4
For binomial expansion as mentioned in data booklet we need the bracket like (1+x)^n
So change 3/3-2x ==> 3/3(1-(2x/3)) ==> (1-(2x/3))^-1
  • 1) And expand (1-(2x/3))^-1
Similarly
-x-2/x^2+4 ==> -x-2/4((x^2/4)+1) ==> 4^-1(-x-2)((x^2/4)+1)^-1
  • 2) Expand ((x^2/4)+1)^-1 and then multiply the expansion by 4^-1(-x-2)
Now add 1 and 2 to obtain full expansion of 3/3-2x + -x-2/x^2+4
 
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So first part you have 3/3-2x + -x-2/x^2+4
For binomial expansion as mentioned in data booklet we need the bracket like (1+x)^n
So change 3/3-2x ==> 3/3(1-(2x/3)) ==> (1-(2x/3))^-1
  • 1) And expand (1-(2x/3))^-1
Similarly
-x-2/x^2+4 ==> -x-2/4((x^2/4)+1) ==> 4^-1(-x-2)((x^2/4)+1)^-1
  • 2) Expand ((x^2/4)+1)^-1 and then multiply the expansion by 4^-1(-x-2)
Now add 1 and 2 to obtain full expansion of 3/3-2x + -x-2/x^2+4
Ohh such a stupid mistake i made ... I'm soo sorry for making you do it :p Thank you! I realised my mistake ... instead of substituting value of A, i had substituted value of B ... I can only pray to God I don't make such mistakes in my exam.
Thank you. :D
 
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12728767_1678641782399234_7821530730671876873_n.jpg
 
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