[Toán 11] 2Sin(3x+pi/4)=1+8Sin2xCos2(2x)2Sin(3x + pi/4) = \sqrt{1+8Sin2xCos^2(2x)}

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ng.mai_96

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truongduong9083

Gợi ý:
Câu 2:

VT = cos3x+2cos23x2(cos23x+2cos23x)=2cos3x + \sqrt{2-cos^2{3x}} \leq \sqrt{2(cos^2{3x}+2-cos^2{3x})} = 2
VP = 2(1+sin22x)22(1+sin^2{2x}) \geq 2
Dấu bằng xảy ra khi {cos3x=2cos23xsin2x=0\left\{ \begin{array}{l} cos3x = \sqrt{2-cos^2{3x}} \\ sin2x = 0 \end{array} \right. nhé
 
N

newstarinsky

câu 1
ĐK...........
Bình phương 2 vế ta được
$4sin^2(3x+\dfrac{\pi}{4})=1+8sin2x.cos^22x\\
\Leftrightarrow 2(1-cos(6x+\dfrac{\pi}{2}))=1+4sin4x.cos2x\\
\Leftrightarrow 2+2sin6x=1+2sin6x+2sin2x\\
\Leftrightarrow sin2x=\dfrac{1}{2}$
 
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