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Prove that: `|sintheta"sin"(60-theta)sin(60+theta)|lt=1/4`for all values of `theta`. |
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Answer» `|sinthetasin(60-theta)sin(60+theta)|` `=1/2|sintheta(2sin(60-theta)sin(60+theta))|` Using, `cosC-cosD = -2sin((C+D)/2)((C-D)/2),` `=1/2|sintheta(cos2theta - cos120^@)|` `=1/2|sinthetacos2theta - sintheta(-sin30^@)|` `=1/2|sinthetacos2theta +1/2sintheta|` `=1/4|2sinthetacos2theta +sintheta|` Using `sinC-sinD = 2cos((C+D)/2)sin((C-D)/2)` `= 1/4|sin3theta - sintheta+sintheta|` `=1/4|sin3theta|` Now, we know, `-1 le sin3theta le 1` `=>|sin3theta| le 1` `=>1/4|sin3theta| le 1/4` `:. |sinthetasin(60-theta)sin(60+theta)| le 1/4.` |
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