sign up I login
 advanced
refer a friend - earn nickels!!
iit009   iit009 is offline iit009's messages in the community
Message
Catalogs Discussion Forums -> Trignometry -> Pls.help guys .no one is replying me. please reply -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   
is the answer 446.1538 mm
Catalogs Discussion Forums -> About IITs and JEE -> download paper and soln. -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
0 replies   
u can download soln. from www.prernaclasses.com
u can download question paper from www.locuseducation.org
Catalogs Discussion Forums -> Trignometry -> Please calculate-rate assured -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
7 replies   
please rate me

Catalogs Discussion Forums -> Trignometry -> Please calculate-rate assured -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
7 replies   
=Sin( /14) Sin(3 /14) Sin(5 /14) Sin(7 /14) Sin(9 /14) Sin(11 /14) Sin(13 /14) 
=Sin( /14) Sin(3 /14) Sin(5 /14) Sin( /2)sin( - 5 /14)
       sin( - 3 /14)sin(?- ?/14)
=[ Sin( /14) Sin(3 /14) Sin(5 /14)]^2
=[cos( /2- /14)  cos( /2-3 /14)  cos( /2-5 /14)]^2
=[cos(3 /7) cos(2 /7) cos( /7)]^2
= [cos( /7) cos(2 /7) cos(3 /7)]^2
Multiplying and dividing by 2sin( /7) we get
=[1/2sin( /7)* (2sin( /7) cos( /7))* cos(2 /7) cos(3 /7)]^2
=[1/4sin( /7)* (2sin(2 /7) cos(2 /7))* cos(4 /7)]^2
We can write cos(3 /7)=cos( - 4?/7)
                                         =cos(4 /7)
=[1/8sin( /7)* (2sin(4 /7) cos(4 /7))]^2
=[ 1/8sin( /7)* sin(8 /7)]^2
We can write sin(8 /7)=sin( + /7)
                                          = -sin ( /7)
=[1/-8]^2
=1/64            ( answer)

Catalogs Discussion Forums -> Trignometry -> Please calculate-rate assured -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
7 replies   
i dont know whats the problem but when i post it is showing pie symbol as ?

Catalogs Discussion Forums -> Trignometry -> Please calculate-rate assured -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
7 replies   
=Sin(?/14) Sin(3?/14) Sin(5?/14) Sin(7?/14) Sin(9?/14) Sin(11?/14) Sin(13?/14)
=Sin(?/14) Sin(3?/14) Sin(5?/14) Sin(?/2)sin(?-5?/14) sin(?- 3?/14)sin(?- ?/14)
=[ Sin(?/14) Sin(3?/14) Sin(5?/14)]^2
=[cos(?/2- ?/14)  cos(?/2-3?/14)  cos(?/2-5?/14)]^2
=[cos(3?/7) cos(2?/7) cos(?/7)]^2
= [cos(?/7) cos(2?/7) cos(3?/7)]^2
Multiplying and dividing by 2sin(?/7) we get
=[1/2sin(?/7)* (2sin(?/7) cos(?/7))* cos(2?/7) cos(3?/7)]^2
=[1/4sin(?/7)* (2sin(2?/7) cos(2?/7))* cos(4?/7)]^2
We can write cos(3?/7)=cos(?-4?/7)
                                         =cos(4?/7)
=[1/8sin(?/7)* (2sin(4?/7) cos(4?/7))]^2
=[ 1/8sin(?/7)* sin(8?/7)]^2
We can write sin(8?/7)=sin(?+?/7)
                                          = -sin (?/7)
=[1/-8]^2
=1/64            ( answer)
Catalogs Discussion Forums -> Trignometry -> Please calculate-rate assured -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
7 replies   
=Sin(?/14) Sin(3?/14) Sin(5?/14) Sin(7?/14) Sin(9?/14) Sin(11?/14) Sin(13?/14)
=Sin(?/14) Sin(3?/14) Sin(5?/14) Sin(?/2)sin(?-5?/14) sin(?- 3?/14)sin(?- ?/14)
=[ Sin(?/14) Sin(3?/14) Sin(5?/14)]^2
=[cos(?/2- ?/14)  cos(?/2-3?/14)  cos(?/2-5?/14)]^2
=[cos(3?/7) cos(2?/7) cos(?/7)]^2
= [cos(?/7) cos(2?/7) cos(3?/7)]^2
Multiplying and dividing by 2sin(?/7) we get
=[1/2sin(?/7)* (2sin(?/7) cos(?/7))* cos(2?/7) cos(3?/7)]^2
=[1/4sin(?/7)* (2sin(2?/7) cos(2?/7))* cos(4?/7)]^2
We can write cos(3?/7)=cos(?-4?/7)
                                         =cos(4?/7)
=[1/8sin(?/7)* (2sin(4?/7) cos(4?/7))]^2
=[ 1/8sin(?/7)* sin(8?/7)]^2
We can write sin(8?/7)=sin(?+?/7)
                                          = -sin (?/7)
=[1/-8]^2
=1/64            ( answer)
Catalogs Discussion Forums -> Mechanics -> Rotation -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
28 replies   
i think a) option should be correct.am i correct?
Catalogs Discussion Forums -> Mechanics -> Please answer me -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
9 replies   
is answer is square root 10/7gLsin(theta)
Catalogs Discussion Forums -> Thermal Physics -> thermo -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
8 replies   
i also got this answer only


Catalogs Discussion Forums -> Thermal Physics -> thermo -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
8 replies   
For an ideal gas if the molar heat capacity varies as C=Cv +3aT^2 find the equation of the process in the variables(V,T), where a is constant.
Catalogs Discussion Forums -> Thermal Physics -> please answer -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
5 replies   
but ans is 4/23 and 24/115
Catalogs Discussion Forums -> Thermal Physics -> Equilibrium -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
11 replies   
it is impossible that to things at diff. temp. be in ther. equilibrium as if we keep 2 bodies one at 0 degree and other at 100 degree then final temp becomes 50 degree but if we keep 2 bodies in contact both at 50 degree it never happens that one body reaches temp of 100 degree and the4 other at 0 degree.2nd law of th
Catalogs Discussion Forums -> Thermal Physics -> Equilibrium -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
11 replies   
it is impossible that to things at diff. temp. be in ther. equilibrium as if we keep 2 bodies one at 0 degree and other at 100 degree then final temp becomes 50 degree but if we keep 2 bodies in contact both at 50 degree it never happens that one body reaches temp of 100 degree and the4 other at 0 degree.2nd law of th
Catalogs Discussion Forums -> Thermal Physics -> please answer -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
5 replies   
but answer is 4/23 and 24/115
 
 
Go to:   

Top Offers for goIITians
Correspondence Courses
Brilliant Tutorials
Narayana Institute
Aakash Institute
Classroom/Crash Courses
Narayana - Kota , Delhi , Others
Brilliant Tutorials - Class , Crash
Aakash Institute - Medical , Engg
Online Test Series
Brilliant Tutorials
Narayana Institute
Aakash Institute
Mahesh Tutorials
AMITY      Sri Chaitanya