find maximum and minimum cos x+(√3) sin x for (π/6)≤x≤π


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Question Number 71563 by gunawan last updated on 17/Oct/19

$find maximum and minimum$ $\cos x + \sqrt{3} \sin x$ $f o r$ $\frac{π}{6} ⩽ x ⩽ π$
	Answered by MJS last updated on 17/Oct/19

	$\cos x + \sqrt{3} \sin x = 2 \sin (x + \frac{π}{6})$ $\Rightarrow minimum at x = \frac{5}{6} π$
	Answered by Kunal12588 last updated on 17/Oct/19

	$a s i n θ + b c o s θ$ $= \sqrt{a^{2} + b^{2}} (\frac{a}{\sqrt{a^{2} + b^{2}}} s i n θ + \frac{b}{\sqrt{a^{2} + b^{2}}} c o s θ)$ $l e t \frac{a}{\sqrt{a^{2} + b^{2}}} = s i n ϕ o r c o s ϕ$ $\Rightarrow \frac{b}{\sqrt{a^{2} + b^{2}}} = c o s ϕ o r s i n ϕ [t h i s s t e p i s d e d u c t i o n f r o m a b o v e s t e p n o t a s u u m p t i o n]$ $[n o t e : t h i s a s s u m p t i o n i s o n l y v a l i d i f - 1 ⩽ \frac{a}{\sqrt{a^{2} + b^{2}}} ⩽ 1]$ $i f u c h o o s e b l u e o n e s$ $a s i n θ + b c o s θ = \sqrt{a^{2} + b^{2}} (s i n ϕ s i n θ + c o s ϕ c o s θ)$ $= \sqrt{a^{2} + b^{2}} s i n (θ - ϕ)$ $i f u c h o s e r e d o n e s$ $a s i n θ + b c o s θ = \sqrt{a^{2} + b^{2}} (c o s ϕ s i n θ + s i n ϕ c o s θ)$ $= \sqrt{a^{2} + b^{2}} c o s (θ + ϕ)$ $∴ m a x = \sqrt{a^{2} + b^{2}}, a l s o c h e c k a t e n d s o f d o m a i n$ $m i n = - \sqrt{a^{2} + b^{2}}, a l s o c h e c k a t e n d s o f d o m a i n$
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