Find the exact value of sin(2x) given that sin(x)=3/5 and is in quadrant II.

Please help and explain!! I know that sin(2x)= 2sin(x)cos(x) but I get lost from there.

...you know that x is in the second quadrant, so you know that cos x is negative...

the triangle is a 3, 4, 5 right triangle...

sin (2x) = 2 sin x cos x = 2 (3/5)(- 4/5) = - 24/25

qed

sin(2x) = 2 sin(x) * cos(x)
= 2(3/5)(4/5) = 24/25
sin(x) = 3/5
x = sin^-1(3/5)
= 36.86989764584377
2x = 73.73979529168754
Therefore sin(2x) is 1st quadrant .............Ans

x in quadrant II, cos(x)<0
cos(x) = -√（1-(sin(x))^2) = -√(1-(3/5)^2) = -4/5
sin(2x) = 2sin(x)cos(x) = 2*3/5*(-4/5) = -24/25

x=36.86 there fore 2x=73.73
for sin73.73=0.96
as this is in IInd quadrant subtract this value from 180 u will get the required value

cos^x+sin^x= 1............cos(x)= 4/5

sin(2x)= 2sin(x)cos(x)
=2*(3/5)*(4/5)
= 24/25
=0.96