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
2α - 36 = ( 2cos25 - 1 ) + (2 cos210 - 1 ) + .............................+ (2cos2180-1)
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
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