1.

Prove that: `|sintheta"sin"(60-theta)sin(60+theta)|lt=1/4`for all values of `theta`.

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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