plz calculate the value of sin 72 degree and cos 48 degree how to calculate these types of value  explain me in simple method for class 10  i know it  is not in my syllabus but i  m curious  for  know this 

plz explain in details u have to show how to calculate the value 

2-prove that 9999=6 u can use bodmas rule between these nos

3-

given

 

Asked by keshavbudhraja (

cos4alpha /cos2bita + sin 4 alpha /sin2bita  =1  

 

to prove 

cos4bita /cos2alpha+ sin 4 bita /sin2alpha 

 

student), 4 days, 22 hours ago

Sin 18,cos18 and sin 36 , cos 36 were earlier in 11th class which have been removed in the last revision.

Soln is like this: let x = 18 so 5x = 90, or 3x = 90 - 2x, or Cos3x= cos(90- 2x), or Cos 3x = sin2x, or

4 cos³x - 3 cos x = 2 sinx.cosx, or cos x(4 cos ² x - 3) -2 sinx.cosx = 0, or cosx(4 cos ² x - 3-2 sinx)=0, or

cosx(4 (1-sin²x) - 3- 2 sinx)=0, or cosx(-4 Sin ² x -2 sinx+1)=0 i.e. either cos x =0 or (-4 Sin ² x -2 sinx+ 1)=0. as cosx is not equal to zero as x is not equal to 90 degrees, therefore (-4 Sin ² x -2 sinx+ 1)=0 , or 4 Sin ² x +2 sinx-1 =0, solving this (by discriminant formula we get Sinx = -2 +/- sq root of ((2)² - 4.(4)(-1) upon 2*4 , or Sin x = -2 +/- (2*sq root of 5) upon 2*4 or,

Sinx = -1+/- sq root of 5 upon 4. Now as x is 18 degrees so Sin x can not be -ve,

so sinx = (-1+ sqrt of 5 ) upon 4, or Sin18 =(-1+ sqrt of 5 ) upon 4

Now we can find cos18 = sq root of (1-Sin²x) which gives Cos18= (sq root od 10+2(sq root of 5) )upon 4

and Sin 48 = Sin (30 + 18) = Sin 30 Cos 18 + Cs 30.Sin 18

Now get the answer by putting std values of Sin30 & cos 30 from table and Sin 18 and cos 18 from above values.

So Sin 72 = Sin (90 - 18) = Cos 18 = (sq root od 10+2(sq root of 5) )upon 4

SNow Cos 36 = 1 - 2 Sin²18,which gives Cos 36 = (1+sq root of 5 ) /4 and then sin 36 = sq root of (1 -Cos²36), we get

Sin 36= (sq root od 10-2(sq root of 5) )upon 4