Can anyone show me how solve this problem with all working out? Thanks 
This question was taken from the VCAA Specialist Maths Exam 1, Q10.
Hi Vincent
 = \frac{4\sqrt{2}}{7})
 = \frac{opposite}{adjacent})
And so by using the pythagorean theorem, we can find out the hypotenuse:
 = \frac{opposite}{hypotenuse})
 = \frac{4\sqrt{2}}{9})
Okay so we want to find sin(x), but the angle we have is sin(2x), but we do know that sin(2x) can be written as sin(x+x):
sin(x+x) = sin(x)cos(x) + sin(x)cos(x)
sin(x+x) = 2sin(x)cos(x)
But now we have a dilemma because how can we find out what sin(x) equals if we don't even know what cos(x) equals? We cant, and so we have to find out what cosine equals first, then only we can find out sin(x):
 = \frac{adjacent}{hypotenuse} = \frac{7}{9})
: \frac{7}{9} = cos^2(x) - sin^2(x))
Where
 = 1 - cos^2(x))
 - 1 +cos^2(x) )
 - 1)
 = \frac{16}{9})
 = \pm \frac{2\sqrt{2}}{3})
, but because x is in the first quadrant (given from the question), then cos(x) has to be positive!
 = \frac{2\sqrt{2}}{3})
So now back to our sin(x) equation:
We said that sin(2x) =
, and we know that sin(2x) = 2sin(x)cos(x), so:
sin(x))
)
 = \frac{1}{3})
Apologies for the late reply, and sorry for the length of this explanation, but I tried to include almost every step of working out for you so that it could help! xD
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