Question

Tìm min, max

a) \(y=\sqrt{3}sinx-cosx+5\)

b) \(y=cosx+cos\left(x+\frac{\pi}{3}\right)\)

c) \(y=sinx+\sqrt{3}cosx+12\)

Answer 1

a/ \(y=2\left(\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx\right)+5=2sin\left(x-\frac{\pi}{6}\right)+5\)

Do \(-1\le sin\left(x-\frac{\pi}{6}\right)\le1\Rightarrow3\le y\le7\)

b/ \(y=2cos\left(x+\frac{\pi}{6}\right)cos\left(-\frac{\pi}{6}\right)=\sqrt{3}cos\left(x+\frac{\pi}{6}\right)\)

Do \(-1\le cos\left(x+\frac{\pi}{6}\right)\le1\Rightarrow-\sqrt{3}\le y\le\sqrt{3}\)

c/ \(y=2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)+12=2sin\left(x+\frac{\pi}{3}\right)+12\)

Do \(-1\le sin\left(x+\frac{\pi}{3}\right)\le1\Rightarrow10\le y\le14\)