Question-14590

Question Number 14590 by Tinkutara last updated on 02/Jun/17

Answered by mrW1 last updated on 03/Jun/17

\sin y ⩾ \sin x - \cos α \cos x

\sin y ⩾ \sqrt{1 + \cos^{2} α} \times (\frac{1}{\sqrt{1 + \cos^{2} α}} \times \sin x - \frac{\cos α}{\sqrt{1 + \cos^{2} α}} \times \cos x)

\sin y ⩾ \sqrt{1 + \cos^{2} α} (\cos θ \times \sin x - \sin θ \times \cos x)

w i t h θ = \cos^{- 1} (\frac{1}{\sqrt{1 + \cos^{2} α}}) = \sin^{- 1} (\frac{\cos α}{\sqrt{1 + \cos^{2} α}})

\sin y ⩾ \sqrt{1 + \cos^{2} α} \times \sin (x - θ) ⩽ \sqrt{1 + \cos^{2} α} ⩾ 1 ⩽ 1

t h e o n l y s o l u t i o n i s w h e n “ =^{″} s i g n i s t a k e n .

\Rightarrow \cos α = 0 a n d \sin y = 1

o r

\Rightarrow \sin α = \pm 1 a n d \sin y = 1

Σ (\sin α + \sin y) = 1 + 1 + (- 1) + 1 = 2

Commented by prakash jain last updated on 03/Jun/17

Thank You!

Commented by Tinkutara last updated on 04/Jun/17

Thanks Sir!

Commented by Tinkutara last updated on 04/Jun/17

It can also be written in 2^{nd} line that

\sin y ⩾ \sqrt{1 + \cos^{2} α} because maximum

value of \sin x - \cos α \cos x is

\sqrt{1 + \cos^{2} α} .