# If sin(Θ)=, at π/2 less then Θ greater than π, find the exact value of sin(2Θ) and cos(Θ/2)

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The problem does not provide the value of `sin theta` , but it tells that `theta` is in quadrant 2, hence, `sin theta > 0` .

You need to evaluate `sin 2 theta` , hence, you need to determine to what quadrant `theta` belongs, such that:

`pi/2 < theta < pi => 2*(pi/2) < 2 theta < 2pi => 2 theta in (pi, 2pi) => sin( 2 theta) < 0`

You need to use the double angle identity, such that:

`sin (2 theta) = 2 sin theta*cos theta`

Supposing that `sin theta = 1/2` , for `theta in (pi/2,pi),` then, `cos theta < 0` and you may use fundamental formula of trigonometry to find it, such that:

`cos^2 theta = 1 - sin^2 theta`

`cos theta = -sqrt(1 - 1/4) => cos theta = - sqrt 3/2`

`sin (2 theta) = 2 sin theta*cos theta => sin (2 theta) = -2*(1/2)*(sqrt3/2) = -sqrt3/2`

You need to evaluate `cos(theta/2)` , hence, you need to find to what quadrant `(theta/2)` belongs, such that:

`pi/2 < theta < pi => pi/4 < (theta)/2 < pi/2`

Since (theta)/2 in (pi/4, pi/2) that is in quadrant 1 yields that `cos((theta)/2) > 0` .

`cos(theta)/2 = sqrt((1 + cos theta)/2)`

`cos(theta)/2 = (sqrt (2 - sqrt3))/2`

**Hence, evaluating `sin(2 theta)` and `cos((theta)/2)` , using `sin theta = 1/2` , yields `sin (2 theta) = -sqrt3/2` and **`cos(theta)/2 = (sqrt (2 - sqrt3))/2.`