trig

[Bestie]

if sin(x) = -(sqrt5)/3 and x[pi,3pi/2] find the exact values of cos(2x) and sin(x/2)
so what i did so far was
cos(2x) = 1-2sin(x)^2 = 1- 2(-sqrt(5)/3)^2 = 1-2(5/9) = -1/9 is that the ans.
cause don't i have to take into consideration the x domain.. etc... i'm not sure how would you approach this question?
are you like meant to do it first and ignore the negative/plus sign and then add the negative/or plus according to the domain at the end???

for the sin(x/2) - yea im stuck i think we have to use the sin(2x) = 2sin(x)cos(x) double angle formula though


thank you

[brightsky]

x/2 is in the second quadrant. so sin(x/2) is positive. we know that cos(2x) = 1 - 2sin^2(x). this means that cos(x) = 1 - 2sin^2(x/2). it follows that sin(x/2) = sqrt((1-cos(x))/2). we know that sin(x/2) > 0, so we only take the positive square root. x is in the third quadrant, so if sin(x) = -sqrt(5)/3, then cos(x) = - sqrt(1 - (sqrt(5)/3)^2). substitute this value into sin(x/2) = sqrt((1-cos(x))/2) to find the value of sin(x/2).

[Bestie]

oh ok, thank you
is the cos(2x) one being -1/9 correct?