# Given that a belongs to (pi/2,pi) and cos 2a = -1/2, calculate sin a

hala718 | Certified Educator

cos2a= -1/2

But we know that cos(2a) = cos^ a -sin^2 a= -1/2

==> sin^2 a= cos^2 a + 1/2.....(1)

But sin^2 a + cos^2 a= 1

==> sin^2 a= 1-cos^2 a......(2)

2sin^2 a= 3/2

==> sin^2 a= 3/4

==> sina= +-sqrt(3)/4

But, since (a) is in the second quadrant, then it is positive

==>  a= pi -pi/3= 2pi/3

==> sina= sqrt(3)/2

giorgiana1976 | Student

If the angle a is in the interval (pi/2,pi), that means that a belongs to the second quadrant.

In the second quadrant, the function sine is positive.

cos 2a = cos (a+a) = cosa*cosa - sina*sina =

= (cos a)^2 - (sin a)^2

From the fundamental formula of trigonometry, we'll have:

(cos a)^2 + (sin a)^2 = 1

So, (cos a)^2 = 1 - (sin a)^2

cos 2a = 1 - (sin a)^2 - (sin a)^2

cos 2a = 1 -2(sin a)^2

We'll subtract 1 both sides:

cos 2a - 1 = -2(sin a)^2

We'll divide by -2 both sides:

(1-cos 2a)/2 = (sin a)^2

(1+ 1/2)/2 = (sin a)^2

3/4 = (sin a)^2

sin a = +/-sqrt3/2

We'll choose the positive value because a is in the second quadrant, so:

sin a = sqrt3/2