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Question Number 136827 by otchereabdullai@gmail.com last updated on 26/Mar/21

$$\mathrm{Find}\mathrm{the}\mathrm{maximum}\mathrm{and}\mathrm{minimum}$$$$\mathrm{values}\mathrm{of}:$$$$1)\mathrm{y}=5\mathrm{sinx}2\mathrm{cosx}$$$$2)\mathrm{y}={\sin }^2\mathrm{x}+{3\cos }^2\mathrm{x}$$

Commented by mr W last updated on 26/Mar/21

$$doyoumeany=5\sin x+2\cos x?$$$$ifyes,then$$$$y=5\sin x+2\cos x$$$$y=\sqrt{5^2+2^2}(\frac{5}{\sqrt{5^2+2^2}}\sin x+\frac{2}{\sqrt{5^2+2^2}}\cos x)$$$$y=\sqrt{5^2+2^2}(\cos \alpha \sin x+\sin \alpha \cos x)$$$$y=\sqrt{29}\sin (x+\alpha )with\alpha ={\tan }^{-1}\frac{2}{5}$$$$\Rightarrow y_{min}=-\sqrt{29}$$$$\Rightarrow y_{max}=\sqrt{29}$$

Commented by otchereabdullai@gmail.com last updated on 27/Mar/21

$$\mathrm{Thanks}\mathrm{alot}\mathrm{prof}$$$$\mathrm{what}\mathrm{of}\mathrm{if}\mathrm{we}\mathrm{look}\mathrm{at}\mathrm{the}\mathrm{question}\mathrm{this}$$$$\mathrm{way}\mathrm{y}=5\mathrm{sinx}2\mathrm{cosx}$$

Commented by mr W last updated on 27/Mar/21

$$y=5\sin x\times 2\cos x$$$$y=5\sin 2x$$$$y_{min}=-5$$$$y_{max}=5$$

Commented by otchereabdullai@gmail.com last updated on 27/Mar/21

$$\mathrm{thank}\mathrm{you}\mathrm{prof}$$

Answered by bramlexs22 last updated on 26/Mar/21

$$\left(2\right)\mathrm{y}=\sin {}^2\mathrm{x}+3-3\sin {}^2\mathrm{x}$$$$\mathrm{y}=3-2\sin {}^2\mathrm{x}$$$$\mathrm{we}\mathrm{know}\mathrm{that}0⩽\sin {}^2\mathrm{x}⩽1$$$$\mathrm{so}-2⩽-2\sin {}^2\mathrm{x}⩽0$$$$\iff 3-2⩽3-2\sin {}^2\mathrm{x}⩽3+0$$$$\Rightarrow 1⩽\mathrm{y}⩽3\{\begin{array}{l}\min =1\\ \max =3\end{array}$$

Commented by otchereabdullai@gmail.com last updated on 27/Mar/21

$$\mathrm{thanks}\mathrm{for}\mathrm{the}\mathrm{help}\mathrm{sir}!$$

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