tìm min ,max

T

thaoteen21

tl

a) A=2.$sin^2x$ -cos2x=2.$\dfrac{1-cos2x}{2}$ -cos2x=1-2.cos2x
\forall x thuộc R: -1 \leq cos2x\leq1
\Leftrightarrow 3\geq 1-2.cos2x\geq-1
vậy max A=3\Leftrightarrow cos2x=-1
minA=-1\Leftrightarrow cos2x=1
b) tương tự: nhưng chú ý:
sin(x+$\dfrac{\pi}{4}$)=$\dfrac{1}{\sqrt{2}}$.(sinx+cosx)

sin(x-$\dfrac{\pi}{4}$)=$\dfrac{1}{\sqrt{2}}$.(sinx-cosx)
......:D
 
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