No, they're not equivalent. You would have to write cos(4x) as cos(2x + 2x) and go from there.
How would you do something like cos(2x +2x)? And do you think that kind of thing might come up - I thimk we've only really done it with single 2x using double angle formulae..
Well that's because they're a pain in the a**! ...and everything else is pretty straight forward once you've done a few past papers.
Yeah I know I hate them too, but I mean that it seems everyone commenting thinks that it's just a personal weakness when really they're just awful haha
I don't understand what the markscheme has done for 7c. I calculated B and then worked out the distance between AB, however they state that AB^2 = 2AP^2 Where does that come from? :/
How would you do something like cos(2x +2x)? And do you think that kind of thing might come up - I think we've only really done it with single 2x using double angle formulae..
Well, remember the double angle formulae are derivatives of the formulae given in the formula booklet regarding sin(A+B), cos(A+B) and tan(A+B). That is why you rewrite cos4x as cos(2x + 2x), because it is now in the form cos(A+B) which you know from the formula booklet equals cosAcosB - sinAsinB (in this case cos2xcos2x - sin2xsin2x).
Which simplifies to cos^2(2x) - sin^2(2x). That's interesting, lol !
No, they're not equivalent. You would have to write cos(4x) as cos(2x + 2x) and go from there.
I'm probably missing something totally obvious (which is a little worrying haha), can you explain in a little more detail how you would identify what cos(4x) is equal to?
Simply using the DA formulae I get the following: Cos(4x) == Cos(2x + 2x) == Cos(2x) * Cos(2x) - Sin(2x) * Sin(2x) == Cos^2(2x) - Sin^2(2x)
Think about cast circle and where acute angles, obtuse angles and reflex angles are. Here this might help some what. Sine you are given a positive sin value it must be in the second quadrant.
Think about cast circle and where acute angles, obtuse angles and reflex angles are. Here this might help some what. Sine you are given a positive sin value it must be in the second quadrant.
I'm probably missing something totally obvious (which is a little worrying haha), can you explain in a little more detail how you would identify what cos(4x) is equal to?
Simply using the DA formulae I get the following: Cos(4x) == Cos(2x + 2x) == Cos(2x) * Cos(2x) - Sin(2x) * Sin(2x) == Cos^2(2x) - Sin^2(2x)
Yes, I reached the same conclusion. I believe it's because you can split 4x into two equal parts. A better example would be cos3x/sin3x/tan3x, which they would be more likely to ask as it doesn't simplify so nicely.
With a question like this is it just convention to label the sides of the parallelogram clockwise starting from bottom left? I tried this question and I labelled my diagram in a different way from the markscheme and got a different answer. How are you supposed to know which corner is D?
I don't understand what the markscheme has done for 7c. I calculated B and then worked out the distance between AB, however they state that AB^2 = 2AP^2 Where does that come from? :/
Oh wow where does that come from?? I got |AP|=|BP|=6sqrt(14) What paper is this? which year?
With a question like this is it just convention to label the sides of the parallelogram clockwise starting from bottom left? I tried this question and I labelled my diagram in a different way from the markscheme and got a different answer. How are you supposed to know which corner is D?