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≥(√(1+cos^2 α))×((1/( (√(1+cos^2 α))))×sin x−((cos α)/( (√(1+cos^2 α)))) ×cos x) sin y≥(√(1+cos^2 α))(cos θ×sin x−sin θ ×cos x) with θ=cos^(−1) ((1/( (√(1+cos^2 α))))) = sin^(−1) (((cos α)/( (√(1+cos^2 α))))) sin y≥(√(1+cos^2 α))×sin (x−θ)≤(√(1+cos^2 α))≥1≤1 the only solution is when “=” sign is taken. ⇒cos α=0 and sin y=1 or ⇒sin α=±1 and sin y=1 Σ(sin α+sin y)=1+1+(−1)+1=2

$$\mathrm{sin}\:{y}\geqslant\mathrm{sin}\:{x}−\mathrm{cos}\:\alpha\:\mathrm{cos}\:{x} \\ $$$$\mathrm{sin}\:{y}\geqslant\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}×\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}}×\mathrm{sin}\:{x}−\frac{\mathrm{cos}\:\alpha}{\:\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}}\:×\mathrm{cos}\:{x}\right) \\ $$$$\mathrm{sin}\:{y}\geqslant\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}\left(\mathrm{cos}\:\theta×\mathrm{sin}\:{x}−\mathrm{sin}\:\theta\:×\mathrm{cos}\:{x}\right) \\ $$$${with}\:\theta=\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}}\right)\:=\:\mathrm{sin}^{−\mathrm{1}} \:\left(\frac{\mathrm{cos}\:\alpha}{\:\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}}\right) \\ $$$$\mathrm{sin}\:{y}\geqslant\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}×\mathrm{sin}\:\left({x}−\theta\right)\leqslant\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}\geqslant\mathrm{1}\leqslant\mathrm{1} \\ $$$${the}\:{only}\:{solution}\:{is}\:{when}\:“=''\:{sign}\:{is}\:{taken}. \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\mathrm{0}\:{and}\:\mathrm{sin}\:{y}=\mathrm{1} \\ $$$${or} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\pm\mathrm{1}\:{and}\:\mathrm{sin}\:{y}=\mathrm{1} \\ $$$$\Sigma\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:{y}\right)=\mathrm{1}+\mathrm{1}+\left(−\mathrm{1}\right)+\mathrm{1}=\mathrm{2} \\ $$

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

$$\mathrm{Thank}\:\mathrm{You}! \\ $$

Commented by Tinkutara last updated on 04/Jun/17

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Commented by Tinkutara last updated on 04/Jun/17

It can also be written in 2^(nd) line that sin y ≥ (√(1 + cos^2 α)) because maximum value of sin x − cos α cos x is (√(1 + cos^2 α)) .

$$\mathrm{It}\:\mathrm{can}\:\mathrm{also}\:\mathrm{be}\:\mathrm{written}\:\mathrm{in}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{line}\:\mathrm{that} \\ $$$$\mathrm{sin}\:{y}\:\geqslant\:\sqrt{\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \alpha}\:\mathrm{because}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:\alpha\:\mathrm{cos}\:{x}\:\mathrm{is} \\ $$$$\sqrt{\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \alpha}\:. \\ $$

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