Solve the following trigonometric inequality:sin2`alpha` sin6`alpha` >=sin3`alpha` sin5`alpha`

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embizze | High School Teacher | (Level 2) Educator Emeritus

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Find all  `alpha` such that `sin2alphasin6alpha>=sin3alphasin5alpha` .

(1) This is equivalent to `sin2alphasin6alpha-sin3alphasin5alpha>=0`

(2) Using the identity `sinusinv=1/2[cos(u-v)-cos(u+v)]` we get



`1/2cos4alpha-1/2cos8alpha-1/2cos2alpha+1/2cos8alpha>=0` (Using `cos(-alpha)=cosalpha` )


`1/2[cos2(2alpha)-cos2alpha]>=0` Use `cos2u=2cos^2u-1` :



(3) Now `cos2alpha-1<=0` for all `alpha` . So we need `2cos2alpha+1<=0` so that the product is nonnegative.

(4) `2cos2alpha+1<=0`

`cos2alpha<=-1/2` Use `cos2u=2cos^2u-1` :







(This is the reference range)

(4) So the complete solution is all such `alpha` :

`npi+pi/3<=alpha<=npi+(2pi)/3` for `n in ZZ`



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