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Question Number 136827 by otchereabdullai@gmail.com last updated on 26/Mar/21
Find the maximum and minimum values of: 1) y=5sinx 2cosx 2) y=sin^2 x+3cos^2 x
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum} \\ $$$$\mathrm{values}\:\mathrm{of}:\: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{y}=\mathrm{5sinx}\:\mathrm{2cosx} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{y}=\mathrm{sin}^{\mathrm{2}} \mathrm{x}+\mathrm{3cos}^{\mathrm{2}} \mathrm{x} \\ $$
Commented by mr W last updated on 26/Mar/21
do you mean y=5 sin x+2 cos x ? if yes, then y=5 sin x+2 cos x y=(√(5^2 +2^2 ))((5/( (√(5^2 +2^2 )))) sin x+(2/( (√(5^2 +2^2 )))) cos x) y=(√(5^2 +2^2 ))(cos α sin x+sin α cos x) y=(√(29)) sin(x+α) with α=tan^(−1) (2/5) ⇒y_(min) =−(√(29)) ⇒y_(max) =(√(29))
$${do}\:{you}\:{mean}\:{y}=\mathrm{5}\:\mathrm{sin}\:{x}+\mathrm{2}\:\mathrm{cos}\:{x}\:? \\ $$$${if}\:{yes},\:{then} \\ $$$${y}=\mathrm{5}\:\mathrm{sin}\:{x}+\mathrm{2}\:\mathrm{cos}\:{x} \\ $$$${y}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}\:\mathrm{sin}\:{x}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}\:\mathrm{cos}\:{x}\right) \\ $$$${y}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:{x}+\mathrm{sin}\:\alpha\:\mathrm{cos}\:{x}\right) \\ $$$${y}=\sqrt{\mathrm{29}}\:\mathrm{sin}\left({x}+\alpha\right)\:{with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\Rightarrow{y}_{{min}} =−\sqrt{\mathrm{29}} \\ $$$$\Rightarrow{y}_{{max}} =\sqrt{\mathrm{29}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 27/Mar/21
Thanks alot prof what of if we look at the question this way y=5sinx2cosx
$$\mathrm{Thanks}\:\mathrm{alot}\:\mathrm{prof} \\ $$$$\mathrm{what}\:\mathrm{of}\:\mathrm{if}\:\mathrm{we}\:\mathrm{look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{question}\:\mathrm{this} \\ $$$$\mathrm{way}\:\:\:\:\mathrm{y}=\mathrm{5sinx2cosx} \\ $$
Commented by mr W last updated on 27/Mar/21
y=5 sin x ×2 cos x y=5 sin 2x y_(min) =−5 y_(max) =5
$${y}=\mathrm{5}\:\mathrm{sin}\:{x}\:×\mathrm{2}\:\mathrm{cos}\:{x} \\ $$$${y}=\mathrm{5}\:\mathrm{sin}\:\mathrm{2}{x} \\ $$$${y}_{{min}} =−\mathrm{5} \\ $$$${y}_{{max}} =\mathrm{5} \\ $$
Commented by otchereabdullai@gmail.com last updated on 27/Mar/21
thank you prof
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{prof} \\ $$
Answered by bramlexs22 last updated on 26/Mar/21
(2) y=sin^2 x+3−3sin^2 x y =3−2sin^2 x we know that 0 ≤sin^2 x≤1 so −2≤−2sin^2 x ≤ 0 ⇔ 3−2≤3−2sin^2 x ≤ 3+0 ⇒1 ≤ y ≤ 3 { ((min=1)),((max=3)) :}
$$\left(\mathrm{2}\right)\:\mathrm{y}=\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{3}−\mathrm{3sin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{y}\:=\mathrm{3}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}\: \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\mathrm{0}\:\leqslant\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\leqslant\mathrm{1} \\ $$$$\mathrm{so}\:−\mathrm{2}\leqslant−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}\:\leqslant\:\mathrm{0} \\ $$$$\Leftrightarrow\:\mathrm{3}−\mathrm{2}\leqslant\mathrm{3}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}\:\leqslant\:\mathrm{3}+\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}\:\leqslant\:\mathrm{y}\:\leqslant\:\mathrm{3}\: \begin{cases}{\mathrm{min}=\mathrm{1}}\\{\mathrm{max}=\mathrm{3}}\end{cases} \\ $$
Commented by otchereabdullai@gmail.com last updated on 27/Mar/21
thanks for the help sir!
$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{help}\:\mathrm{sir}! \\ $$

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