MAXIMUM value of sin^2(120+theta)+sun^2(120-theta) is
Wajahat , 6 Years ago
Grade 12th pass
Trigonometry> MAXIMUM value of sin^2(120+theta)+sun^2(1...
2 AnswersSaurabh Koranglekar
Last Activity: 6 Years ago
Arun
Last Activity: 6 Years ago
sin(t)^2 = (1/2) * (1 - cos(2t))
sin(120 + a)^2 + sin(120 - a)^2 =>
(1/2) * (1 - cos(2 * (120 + a))) + (1/2) * (1 - cos(2 * (120 - a))) =>
(1/2) * (1 - cos(240 + 2a) + 1 - cos(240 - 2a)) =>
(1/2) * (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 - (1/2) * (2 * cos(240) * cos(2a)) =>
1 - cos(240) * cos(2a)
cos(t) = 1 when t = 360 * k
cos(t) = -1 when t = 180 + 360 * k
1 - cos(240) * (-1) =>
1 + cos(240) =>
1 - 1/2 =>
1/2
1 - cos(240) * 1 =>
1 - (-1/2) =>
3/2
3/2 is the maximum
sin(120 + a)^2 + sin(120 - a)^2 =>
(1/2) * (1 - cos(2 * (120 + a))) + (1/2) * (1 - cos(2 * (120 - a))) =>
(1/2) * (1 - cos(240 + 2a) + 1 - cos(240 - 2a)) =>
(1/2) * (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 - (1/2) * (2 * cos(240) * cos(2a)) =>
1 - cos(240) * cos(2a)
cos(t) = 1 when t = 360 * k
cos(t) = -1 when t = 180 + 360 * k
1 - cos(240) * (-1) =>
1 + cos(240) =>
1 - 1/2 =>
1/2
1 - cos(240) * 1 =>
1 - (-1/2) =>
3/2
3/2 is the maximum
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