find-maximum-and-minimum-cos-x-3-sin-x-for-pi-6-x-pi-

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