fuck you indian, it was the easiest question !Can someone, tell how they did part a?
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.Can someone, tell how they did part a?
u let theta = cos inverse of whateevrah 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.
... 4/9 (5)^1/2fuck you indian, it was the easiest question !
I got 4/9 . (3)^1/2
yh typo, it was 5... 4/9 (5)^1/2
thats what i got!... 4/9 (5)^1/2
do u think ill lose a mark for straight subbing into calculator??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
same herethats what i got!
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))Can someone, tell how they did part a?
yeah it was 2 markerdo u think ill lose a mark for straight subbing into calculator??
i got 22hrs 46min for d)ii)
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))