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Rafy

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Nws m8

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ah i just subbed it into my calc, squared it then took the square root of that and simplified. not sure if its right but whatevs.
u let theta = cos inverse of whateevr

then u get sin 2 theta which equals 2 sin theta cos theta and fkin sub in values from cos theta= 2/3
 

hsc2012la

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let inverse cos (2/3) = theta
draw up the triangle and find the other side which is root(5)
therefore it'll be sin(2theta -> since theta equals inverse of cos etc). Expand it into 2sinthetacostheta and then use the dimensions of the triangle, sub it in and you should get 2(root5/3)(2/3) = 4 root5/ 9
 

SandyS1995

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let inverse cos (2/3) = theta
draw up the triangle and find the other side which is root(5)
therefore it'll be sin(2theta -> since theta equals inverse of cos etc). Expand it into 2sinthetacostheta and then use the dimensions of the triangle, sub it in and you should get 2(root5/3)(2/3) = 4 root5/ 9
do u think ill lose a mark for straight subbing into calculator??
 

Nooblet94

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Can someone, tell how they did part a?
Use the double angle formula to expand it and then draw up a right angled triangle with sides 2 and 3, find the 3rd side and use that to find sin(arccos(2/3))
 

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