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Show inequality 0 < (sin(3.14/8+a))^2-(sin(3.14/8-a))^2<1/squareroot2

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xalal | Honors

Posted May 22, 2013 at 12:21 PM via web

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Show inequality 0 < (sin(3.14/8+a))^2-(sin(3.14/8-a))^2<1/squareroot2

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 22, 2013 at 12:42 PM (Answer #1)

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You need to convert the difference of squares `sin^2(pi/8+a) - sin^2(pi/8-a) ` into a product, such that:

`sin^2(pi/8+a) - sin^2(pi/8-a) = (sin(pi/8+a) - sin(pi/8-a))(sin(pi/8+a) + sin(pi/8-a))`

Converting the factor `sin(pi/8+a) - sin(pi/8-a)` into a product yields:

`sin(pi/8+a) - sin(pi/8-a) = 2cos((pi/8+a+pi/8-a)/2)sin((pi/8+a-pi/8+a)/2)`

`sin(pi/8+a) - sin(pi/8-a) = 2cos(pi/8)sin(a)`

Converting the factor `sin(pi/8+a) + sin(pi/8-a)` into a product yields:

`sin(pi/8+a) + sin(pi/8-a) = 2sin((pi/8+a+pi/8-a)/2)cos((pi/8+a-pi/8+a)/2)`

`sin(pi/8+a) + sin(pi/8-a) = 2sin(pi/8)cos(a)`

Replacing `2sin(pi/8)cos(a)` for `sin(pi/8+a) + sin(pi/8-a)`  and `2cos(pi/8)sin(a)` for `sin(pi/8+a) - sin(pi/8-a)` yields:

`sin^2(pi/8+a) - sin^2(pi/8-a) = 2cos(pi/8)sin(a)*2sin(pi/8)cos(a)`

Using the double angle formula yields:

`sin^2(pi/8+a) - sin^2(pi/8-a) = sin(2*pi/8)sin(2a)`

`sin^2(pi/8+a) - sin^2(pi/8-a) = sin(pi/4)sin(2a)`

`sin^2(pi/8+a) - sin^2(pi/8-a) = sqrt2/2*sin(2a)`

You should use the definition of sine function such that:

`-1 <= sin 2a <= 1 => -sqrt2/2 <= sqrt2/2sin 2a <= sqrt2/2`

Since `-sqrt2/2 <= 0` , yields:

`0 <= sqrt2/2sin 2a <= sqrt2/2`

Replacing` sin^2(pi/8+a) - sin^2(pi/8-a) ` for `sqrt2/2*sin(2a)` yields:

`0 <=sin^2(pi/8+a) - sin^2(pi/8-a) <= sqrt2/2`

Hence, using trigonometric identities and sine function definition yields that `0 <=sin^2(pi/8+a) - sin^2(pi/8-a) <= sqrt2/2` holds.

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