# sin 3a )^2+(sin 2a )^2>2cos a sin 2a sin 3a ?0<a <pie /2

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the triple angle and double angle identities, such that:

`sin 3a = 3sin a - 4sin^3 a => sin^2 3a = (3sin a - 4sin^3 a)^2`

`sin 2a = 2 sin a*cos a => sin^2 2a = 4sin^2 a*cos^2 a`

Replacing `(3sin a - 4sin^3 a)^2` for `sin^2 3a` and `4sin^2 a*cos^2 a` for `sin^2 2a`  yields:

`(3sin a - 4sin^3 a)^2 + 4sin^2 a*cos^2 a > 2cos a*2 sin a*cos a*sin a(3 - 4sin^2 a)`

Factoring out `sin^2 a` to the left side yields:

`sin^2 a((3 - 4sin^2 a)^2 + 4cos^2 a) > 4sin^2 a*cos^2 a*(3 - 4sin^2 a)`

Reducing duplicate factors yboth sides yields:

`(3 - 4sin^2 a)^2 + 4cos^2 a > 4cos^2 a*(3 - 4sin^2 a)`

`(3 - 4sin^2 a)^2 + 4cos^2 a - 4cos^2 a*(3 - 4sin^2 a) > 0`

`(3 - 4sin^2 a)(3 - 4sin^2 a - 4cos^2 a)+ 4cos^2 a > 0`

`(3 - 4sin^2 a)(3 - 4(sin^2 a + cos^2 a)) + 4cos^2 a > 0`

Replacing 1 for` sin^2 a + cos^2 a` yields:

`(3 - 4sin^2 a)(3 - 4) + 4cos^2 a > 0`

`-3 + 4sin^2 a + 4cos^2 a > 0 `

Factoring out 4 yields:

`-3 + 4(sin^2 a + cos^2 a) > 0`

`-3 + 4 > 0 => 1 > 0`

Hence, testing if the inequality `sin^2 3a +sin^2 2a > 2cos a*sin 2a*sin 3a` holds yields the valid statement `1 > 0` , thus, the inequality `sin^2 3a +sin^2 2a > 2cos a*sin 2a*sin 3a` is valid.