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Oh sorry.

1111.111111..... recurring? I haven't done this topic yet.

Don't worry, you were correct.

Simplify cotxsecx.

Wait: you're right AkaiHanabi.

Band 3-4 (60-79%)? I don't know. All I know is raw marks usually align upwards.

68/84 is 81% raw - that's a pretty high percentage for MX1.

i) x/sin36 = 1/sin72 x= sin36/sin72 --- make x the subject = sin144/sin72 --- change sin36 to supplemantary value i.e. second quadrant (180-36) = sin(2*72)/sin72 --- recognise relationship between 72 and 144 i.e. 72*2 = 144 (denominator has 72) = 2sin72cos72/sin72 --- recognise double angle...

No. But make sure you express your ideas clearly enough, especially in the 5+ mark questions. Dot-points work well for the extended response type questions (i.e. over half a page).

We're not all as good at maths as you Namu.

Yeah.

But the question says 'show'. Okay, maybe I don't know what I'm talking about. :S

without the gaps between do [ s u p ] 2 [ / s u p ]

cos135 = (1-tan267.5) / (1+tan267.5) Substitute in tan67.5 as 1 + rt2. RHS = [1 - (1+rt2)2] / [1 + (1+rt2)2] = [1-(1+2rt2+2)] / [1+(1+2rt2+2)] = (-2-2rt2) / (4+2rt2) = (-1-rt2) / (2+rt2) rationalise denominator = -rt2/2 irrationalise denominator i.e. multiply by...

cos2@ = (1-tan2@) / (1+tan2@) RHS = [1-(sin2@/cos2@)] / [1+(sin2@/cos2)] ----- express tan as sin/cos = [(cos2@-sin2@)/cos2@] / [(cos2@+sin2@)/cos2@] ----- treat numerator fraction as one and create a common denominator of cos2@, do this for the denominator fraction too = cos2@ - sin2@...

Find the sum of all the integers between 1 and 200 that are not multiples of 7.

:) Yay!

This is extension 1.

If it was a decimal you could still do a substitution though. If if was 9.3 instead of 9*3 then you could change it to 3*3.1. I don't know what I'm talking about.

Mate, type it up in the mathematics forum. You'll get heaps of responses there. http://community.boredofstudies.org/newthread.php?do=newthread&f=12