How to find maximum and minimum values of 4sin^2x+4cos^4x?

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Arvind A

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Given, 4sin2x+ 4cos4x Let, y = 4sin2x + 4cos4x = 4(sin2x + cos4x) = 4(sin2x + cos2x (cos2x)) = 4(sin2x + cos2x (1-sin2x)) = 4(sin2x + cos2x - cos2x sin2x) = 4(1 - (sinx cosx)2) = 4(1...
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Given, 4sin2x+ 4cos4x Let, y = 4sin2x + 4cos4x = 4(sin2x + cos4x) = 4(sin2x + cos2x (cos2x)) = 4(sin2x + cos2x (1-sin2x)) = 4(sin2x + cos2x - cos2x sin2x) = 4(1 - (sinx cosx)2) = 4(1 - ¼sin22x) = 4 - sin22x but the range for sin22x is 0 ? sin22x ? 1 = 4-0 ? 4-sin22x ? 4-1 = 4 ? 4-sin22x ? 3 = 3 ? 4-sin22x ? 4 = 3 ? y ? 4 Min value: 3 Max value: 4 Min-Max table Min value Max value Range sin?, sin2?, sin9? …. sinn? -1 +1 -1 ? sinn? ? 1 cos?, cos4? , cos7? … cosn? -1 +1 -1 ? cosn? ? 1 sin2?, sin24?, sin29? … sin2n? 0 +1 0 ? sin2n? ? 1 cos2?, cos23?, cos28? … cos2n? 0 +1 0 ? cos2n? ? 1 sin?cos? -½ +½ -½? sin?cos? ? ½ read less
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