Find the maximum value of sin² (120+a) +sin² (120-a)....give ans with explanation
kvnmurty:
max is 3/2
yah..ur ans is ri8...but plz expln
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sin² (120+a) + sin² (120-a)
1/2 { 1 - cos (240+2a) } + 1/2 { 1 - cos (240-2a) }
1 - 1/2 [ cos (240 + 2a) + cos ( 240 - 2a) ]
1 - 1/2 [ cos 240 cos 2a - sin 240 sin 2a + cos 240 cos 2a + sin 240 sin 2a ]
1 - cos 240 sin 2a
1 - 1/2 sin 2a as cos 240 = cos (360 - 240)
maximum value is obtained when sin 2a is minimun and is 0 when a = 0.
so 1
1/2 { 1 - cos (240+2a) } + 1/2 { 1 - cos (240-2a) }
1 - 1/2 [ cos (240 + 2a) + cos ( 240 - 2a) ]
1 - 1/2 [ cos 240 cos 2a - sin 240 sin 2a + cos 240 cos 2a + sin 240 sin 2a ]
1 - cos 240 sin 2a
1 - 1/2 sin 2a as cos 240 = cos (360 - 240)
maximum value is obtained when sin 2a is minimun and is 0 when a = 0.
so 1
As cos 240 is -1/2, Given value is 1 + 1/2 sin 2a. So Maximum value is 3/2
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sin² (120+a) + sin² (120-a)
1/2 { 1 - cos (240+2a) } + 1/2 { 1 - cos (240-2a) }
1 - 1/2 [ cos (240 + 2a) + cos ( 240 - 2a) ]
1 - 1/2 [ cos 240 cos 2a - sin 240 sin 2a + cos 240 cos 2a + sin 240 sin 2a ]
1 - cos 240 sin 2a
1 - 1/2 sin 2a as cos 240 = cos (360 - 240)
maximum value is obtained when sin 2a is minimun and is 0 when a = 0.
so 1
1/2 { 1 - cos (240+2a) } + 1/2 { 1 - cos (240-2a) }
1 - 1/2 [ cos (240 + 2a) + cos ( 240 - 2a) ]
1 - 1/2 [ cos 240 cos 2a - sin 240 sin 2a + cos 240 cos 2a + sin 240 sin 2a ]
1 - cos 240 sin 2a
1 - 1/2 sin 2a as cos 240 = cos (360 - 240)
maximum value is obtained when sin 2a is minimun and is 0 when a = 0.
so 1
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