Daily Graph 19-02-09

sir for my prospect a SHOULD be mentioned

wow!! this graph thread is looking so beautiful !! all pink pink...

btw ... nice questions :)

x+y = 2nÎ + sin^-1a
or
(2n+1)Î - sin^-1a

So, it will be infinite no. of straight lines of two types:
TYPE 1:
slope = -1
parallel to line of y-intercept = sin^-1a at regular y-intercepts of 2Î

TYPE 2:
slope = -1
parallel to line of y-intercept = Î -sin^-1a at regular y-intercepts of 2Î

cubes bhool gaye kya? or this is the question??????????

if it is 2,
then sin2x= 4 again does not exist.

general.....
sinxcosx=a
==>sin2x = 2a
So, -1/2≤a≤1/2 then only it is possible.
Then x = nÎ /2 + (-1)ⁿsin^-1a/2

[11] [11] [11] [11] [11] [11] [11] [11] [11] [11] [11] [11] [11] [11]

that means 2sinxcosx = 2;

sin(2x) = 2;

thats not possible

null graf???????

there is no solution according to me...
Bhaiyya what am i getting??
sinxcosx=1
==>2sinxcosx=2
==>sin2x=2 which is impossible so no solution...
kya yeh sahi hai?