Find-the-maximum-and-minimum-values-of-1-y-5sinx-2cosx-2-y-sin-2-x-3cos-2-x-

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

Find the maximum and minimum

values of :

1) y = 5 sinx 2 cosx

2) y = \sin^{2} x + {3 \cos}^{2} x

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

d o y o u m e a n y = 5 \sin x + 2 \cos x ?

i f y e s, t h e n

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 α \sin x + \sin α \cos x)

y = \sqrt{29} \sin (x + α) w i t h α = \tan^{- 1} \frac{2}{5}

\Rightarrow y_{m i n} = - \sqrt{29}

\Rightarrow y_{m a x} = \sqrt{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 = 5 sinx 2 cosx

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

y = 5 \sin x \times 2 \cos x

y = 5 \sin 2 x

y_{m i n} = - 5

y_{m a x} = 5

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

thank you prof

Answered by bramlexs22 last updated on 26/Mar/21

(2) y = \sin^{2} x + 3 - 3 \sin^{2} x

y = 3 - 2 \sin^{2} x

we know that 0 ⩽ \sin^{2} x ⩽ 1

so - 2 ⩽ - 2 \sin^{2} x ⩽ 0

\Leftrightarrow 3 - 2 ⩽ 3 - 2 \sin^{2} x ⩽ 3 + 0

\Rightarrow 1 ⩽ y ⩽ 3 {\begin{cases} \min = 1 \\ \max = 3 \end{cases}

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

thanks for the help sir!