Messages posted by ramyani

hey thx. i did make a blunder

.thanks for pointing it out.sending u a salute.

i m trying to crack the problem in the right way. for the time being here is a soln of a problem by amar gupta that i got somewhere in this site.it may help

amar.gupta

cos2theta=cos(theta+phi)

sin2theta= 2sin(theta+phi)

i am taking

=A and

= B.

then 2sinA cosA =2sin(A+B)

or sinA cosA =sin(A+B)......[1]

and cos2A=cos(A+B)....[2]

square and add:

(sinA cosA)² + (cos2A)² = 1

or (sinA cosA)²= 1- (cos2A)²or (sinA cosA)²= (sin2A)²or (sinA cosA)²= (2sinAcosA)²or (sinA cosA)²= 4(sinAcosA)²
or 3(sinAcosA)² = 0

sinAcosA=0 hence either sinA =0 or cosA=0

when sinA=0 from eq.[1]:

0=sin(A+B)

or cosAsinB=0 or sinB=0 or B=n (pie) where n is an +ve integer

when cosA =0 from eq.[1]:

0=sin(A+B)

or sinAcosB=0 or cosB=0

B=[2k(pie)+(pie)/2] and B=[2k(pie) - (pie)/2] where k is an +ve integer

I dont know what exactly your question demands ,as you have not mentioned any thing that why you want to make eq. in

only.

but i think this will also work.

best of luck for 2009.

let A and B be the two roots of the equations.
equn 1=> A +B= a-1     AB=a+b

equn 2=> A+B=2/a
         => a-1=2/a
         => a^2 -- a=2
         =>(a-2)(a-1)=0
         a=2,-1

a+b=b/a
b=a^2/(1-a)
b= -4,1/2

plzzzzzz rate me

It is same.

Splitting eqn 1, we get

(cos² theta -- sin sq theta)= cos sq theta--sin sq phi
lemda cos² theta--lemda sin sq theta =cos sq theta--sin sq phi

lemda cos² theta--lemda + lemdacos sq theta--cos sq theta = --sin sq phi

cos² theta (lemda + lemda --1) = lemda --sin sq phi

cos² theta = ( lemda--sin sq phi) / ( 2 lemda--1) ------------------------(3)

squaring and adding equn (1) and (2)

lemda sq = 1+ 3 sin sq( theta+phi )lemda sq = 1+ 3 --3cos sq ( theta+phi)
lemda sq= 1+ 3--3 cos sq theta +3 sin sq phi
lemda sq = 4--[( 3 lemda+3sin sq phi)/ ( 2 lemda--1)] + 3 sin sq phi
2 lemda cube--lemda sq = 4 + 6 sin sq phi --3 lemda
6 sin sq phi + 4 = 2 lemda cube - lemda sq + 3 lemda

finish

plz rate me

Let us consider a body of mass( say water) 1 kg at 60 degree. Another body(say water) of mass 1 gm at 100 degree. Let both comes down to 10 degree.You will notice heat liberated of the 1 kg water is 1000 gm X1X50=50000 cal. For 1gm body,
heat liberated = 1 gm X1X90=90 cal.

whether it is varying or not it is zero. U can use the integral form of gauss's law and still the answer is zero.If you are not satisfied imagine no of lines of forces entering per unit are. These leaves as they enter. And so the result is zero.
Rate me plz.

range of sin is -1<=sin<=1, -1<= cos<=1

i guess it will be 0.

i think the it will remain un changed

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