cos^2 5@+ cos^2 10@+cos^2 15@...........+cos^2 180@

Asked by vermahiteshi243 | 30th Aug, 2021, 09:53: AM

Expert Answer:

Let α = cos25 + cos210 + cos215 + cos225 + .............................+ cos2180
 
2α - 36  = ( 2cos25 - 1 )  + (2 cos210 - 1 )  + .............................+ (2cos2180-1)
 
2α - 36  = cos10 + cos20 + cos30 +....................................+ cos360

2α - 36  = ( cos10 + cos360 ) + ( cos20 +.cos350 ) ..................+ (cos180 + cos190)
 
2α - 36  =  ( 2 cos185 cos 175 ) + ( 2 cos185 cos165 ) + ..................+ ( 2 cos185 cos5 )  
 
2α - 36  = 2 cos185 [ cos175 + cos165 + cos 155 + ........................+ cos15 + cos 5 ]
 
2α - 36  = 2 cos185 [ ( cos175 + cos 5 ) + ( cos165 + cos15) ..................... + ( cos95 + cos85) ]
 
2α - 36  = 2 cos185 [ ( 2 cos90 cos85 ) + ( 2 cos90 cos75 ) + .................... + ( 2 cos90 cos 5) ]
 
2α - 36  = 4 cos185 cos90 [ cos85  +  cos75  + .....................+ cos 5 ]
 
In above expression , in RHS , we get cos90 . Hence we have , 2α - 36  = 0  or  α = 18 
 
Hence we get ,  α = cos25 + cos210 + cos215 + cos225 + .............................+ cos2180 = 18
 
 
 

Answered by Thiyagarajan K | 17th Sep, 2021, 12:27: PM