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Question Number 135855 by liberty last updated on 16/Mar/21

What are the possible value of cos α×sin β if sin α×cos β=−(1/2)

$${What}\:{are}\:{the}\:{possible}\:{value}\:{of} \\ $$$$\mathrm{cos}\:\alpha×\mathrm{sin}\:\beta\:\:{if}\:\mathrm{sin}\:\alpha×\mathrm{cos}\:\beta=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by EDWIN88 last updated on 16/Mar/21

We have −1≤sin (α+β)≤1 and −1≤sin (α−β)≤1 now from −1≤sin (α+β)≤1 we get −1≤sin αcos β+cos αsin β≤1 −1≤−(1/2)+cos αsin β≤1 ⇒−(1/2)≤cos αsin β≤(3/2)...(i) from −1≤sin (α−β)≤1 we get −1≤sin αcos β−cos αsin β≤1 −1≤−(1/2)−cos αsin β≤1 −(3/2)≤cos αsin β≤(1/2)...(ii) combine equation (i) and (ii) we get ⇒ −(1/2)≤cos αsin β≤(1/2)

$$\mathrm{We}\:\mathrm{have}\:−\mathrm{1}\leqslant\mathrm{sin}\:\left(\alpha+\beta\right)\leqslant\mathrm{1}\:\mathrm{and}\:−\mathrm{1}\leqslant\mathrm{sin}\:\left(\alpha−\beta\right)\leqslant\mathrm{1} \\ $$$$\mathrm{now}\:\mathrm{from}\:−\mathrm{1}\leqslant\mathrm{sin}\:\left(\alpha+\beta\right)\leqslant\mathrm{1}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$−\mathrm{1}\leqslant\mathrm{sin}\:\alpha\mathrm{cos}\:\beta+\mathrm{cos}\:\alpha\mathrm{sin}\:\beta\leqslant\mathrm{1} \\ $$$$−\mathrm{1}\leqslant−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\alpha\mathrm{sin}\:\beta\leqslant\mathrm{1}\:\Rightarrow−\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{cos}\:\alpha\mathrm{sin}\:\beta\leqslant\frac{\mathrm{3}}{\mathrm{2}}...\left(\mathrm{i}\right) \\ $$$$\mathrm{from}\:−\mathrm{1}\leqslant\mathrm{sin}\:\left(\alpha−\beta\right)\leqslant\mathrm{1}\:\mathrm{we}\:\mathrm{get} \\ $$$$−\mathrm{1}\leqslant\mathrm{sin}\:\alpha\mathrm{cos}\:\beta−\mathrm{cos}\:\alpha\mathrm{sin}\:\beta\leqslant\mathrm{1} \\ $$$$−\mathrm{1}\leqslant−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos}\:\alpha\mathrm{sin}\:\beta\leqslant\mathrm{1} \\ $$$$−\frac{\mathrm{3}}{\mathrm{2}}\leqslant\mathrm{cos}\:\alpha\mathrm{sin}\:\beta\leqslant\frac{\mathrm{1}}{\mathrm{2}}...\left(\mathrm{ii}\right) \\ $$$$\mathrm{combine}\:\mathrm{equation}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\Rightarrow\:−\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{cos}\:\alpha\mathrm{sin}\:\beta\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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