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Question Number 143450 by bramlexs22 last updated on 14/Jun/21

w h e n x + y = \frac{2 π}{3}; x ⩾ 0; y ⩾ 0

t h e m a x i m u m a n d t h e m i n i m u m

o f \sin x + \sin y i s___

Answered by EDWIN88 last updated on 16/Jun/21

y = \frac{2 π}{3} - x \Rightarrow \sin y = \sin (\frac{2 π}{3} - x)

\sin y = \frac{1}{2} \sqrt{3} \cos x + \frac{1}{2} \sin x

let \sin x + \sin y = f (x)

f (x) = \sin x + \frac{1}{2} \sqrt{3} \cos x + \frac{1}{2} \sin x

f (x) = \frac{3}{2} \sin x + \frac{1}{2} \sqrt{3} \cos x

f (x) = k \cos (x - \emptyset) \to {\begin{cases} k = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} \\ \emptyset = \tan^{- 1} (\sqrt{3}) \to \emptyset = \frac{π}{3} \end{cases}

(1) f (x) = \sqrt{3} \cos (x - \frac{π}{3}) \to \max = \sqrt{3}

when x = y = \frac{π}{3}

(2) f {(x)}_{\min} = \frac{\sqrt{3}}{2} when x = 0 \land y = \frac{2 π}{3}

or x = \frac{2 π}{3} \land y = 0

Answered by mnjuly1970 last updated on 14/Jun/21

y := \frac{2 π}{3} - x

M := s i n (x) + \frac{\sqrt{3}}{2} c o s x - \frac{1}{2} s i n (x)

:= \frac{1}{2} s i n (x) + \frac{\sqrt{3}}{2} c o s (x)

m a x (M) := \sqrt{\frac{1}{4} + \frac{3}{4}} := 1

m i n (M) := - M = - 1

Commented by bramlexs22 last updated on 15/Jun/21

y = 120 ° - x

\sin y = \sin (120 ° - x) = \frac{1}{2} \sqrt{3} \cos x - (- \frac{1}{2}) \sin x

= \frac{1}{2} \sqrt{3} \cos x + \frac{1}{2} \sin x