1996 hsc maths ext1 (1 Viewer)

1zephyr · Oct 24, 2014

question 4a
prove that sin3theta/sintheta-cos3theta/costheta=2 (for sintheta and costheta=/=0
cheers bruhssssssssssssss

ilikecake · Oct 24, 2014

Common denominate the fractions, then notice the numerator simplifies to a sine expansion identity in the form of sin(A - B), which can then be simplified and expanded again to get the answer.

Drongoski · Oct 24, 2014

1zephyr said:
question 4a
prove that sin3theta/sintheta-cos3theta/costheta=2 (for sintheta and costheta=/=0
cheers bruhssssssssssssss

LHS =

$\frac {sin 3 \theta}{sin \theta} - \frac {cos 3\theta}{cos \theta} = \frac {sin 3\theta cos \theta - cos 3 \theta sin \theta}{sin \theta cos \theta} = \frac {sin(3 \theta - \theta)}{sin \theta cos \theta} = \frac {sin 2 \theta}{sin \theta cos \theta} = \frac {2sin \theta cos\theta}{sin \theta cos \theta} = \cdots$

1zephyr · Oct 24, 2014

cheers lads

mreditor16 · Oct 24, 2014

Drongoski said:
LHS =

$\frac {sin 3 \theta}{sin \theta} - \frac {cos 3\theta}{cos \theta} = \frac {sin 3\theta cos \theta - cos 3 \theta sin \theta}{sin \theta cos \theta} = \frac {sin(3 \theta - \theta)}{sin \theta cos \theta} = \frac {sin 2 \theta}{sin \theta cos \theta} = \frac {2sin \theta cos\theta}{sin \theta cos \theta} = \cdots$

that's a much neater way than mine haha

1zephyr · Oct 24, 2014

mreditor16 said:
that's a much neater way than mine haha

ayyy brah i had a gander at what they put in the book and i couldnt see what they were doing then drongonski the beast comes up with a much simpler way. wot a god

Drongoski · Oct 24, 2014

1zephyr said:
ayyy brah i had a gander at what they put in the book and i couldnt see what they were doing then drongonski the beast comes up with a much simpler way. wot a god

In case you are not aware I was responsible for the Gonski Report.

1996 hsc maths ext1 (1 Viewer)

1zephyr

Member

ilikecake

New Member

Drongoski

Well-Known Member

1zephyr

Member

mreditor16

Well-Known Member

1zephyr

Member

Drongoski

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)