Ken's specialist question thread!

[nina_rox]

Ahh thank you! Could you please explain the first one? I got a lot of fractions and still haven't got it. Thanks!!

[Lasercookie]

I actually tried this last night, but I got false on my CAS, probably because you have |a||b|cos(theta), but since theta is 90 degrees it ends up being zero which makes the whole thing zero, if that makes sense.

*facepalm*
Yeah, that makes perfect sense. Not sure why I was thinking cos(90)=1 :/

[kensan]

I actually tried this last night, but I got false on my CAS, probably because you have |a||b|cos(theta), but since theta is 90 degrees it ends up being zero which makes the whole thing zero, if that makes sense.

*facepalm*
Yeah, that makes perfect sense. Not sure why I was thinking cos(90)=1 :/

Yeah I sometimes do that too

Ahh thank you! Could you please explain the first one? I got a lot of fractions and still haven't got it. Thanks!!

Use double angle formulas

   The 1 will cancell out in the denominator



=

Hopefully I haven't made any errors, ask if you don't get any of the steps

[nina_rox]

Thank you so much!!

[nina_rox]

Another one:
What is the range of the function with equation of y=3cosec(2x)-1?
Also what is the range of just y=cosec(x)?

[pi]

Another one:
What is the range of the function with equation of y=3cosec(2x)-1?
Also what is the range of just y=cosec(x)?

[-4, 2], simply done by [3(-1) - 1, 3(1) -1] using the transformations


and [-1, 1] for normal y=csc(x)


Woops! Thanks swarley

[Greatness]

cosec(x) = 1/sin(x) so to visualise it, draw sin(x) then take the reciprocal of it. -> where there is a x intercept there will be asymptote - so at 0, TT, 2TT and so on. Then take the reciprocal of the y value of the turning points and you'll get turning points of the reciprocal graph. Now you can draw the graph. So the range of y=cosec(x) is (-inf,-1]u[1,inf) For y=3cosec(2x)-1 you can just do it by looking at the vertical translation and dilation

[Greatness]

Another one:
What is the range of the function with equation of y=3cosec(2x)-1?
Also what is the range of just y=cosec(x)?

[-4, 2], simply done by [3(-1) - 1, 3(1) -1] using the transformations


and [-1, 1] for normal y=csc(x)

xD it's the other way around cos the graph goes from infinity to the turning points cos it has been 'flipped' in a way

[pi]

LOL, my bad haha I even visualised the right graph in my head and still looked at the wrong part Just exclude (kinda) the"range" I have given lol


(good thing I'm not doing any more maths)

[nina_rox]

I'm confused now so for graph of y=cosec(x) then range is R (or -ive infinity to +ive infinity)
and for y=3cosec(2x)-1 I drew the graph and I thought the range would be something like (ive infinity, -4] then
union to [2, positive infinity]?

[pi]

The range of y=csc(x) is actually (-inf,-1]U[1,inf), not R

Your other solution is correct

[Greatness]

Take a look at this: http://mathworld.wolfram.com/Cosecant.html
Have you learnt how to graph reciprocal functions?

[nina_rox]

The range of y=csc(x) is actually (-inf,-1]U[1,inf), not R

Your other solution is correct

Thank you so much VegemitePi and swarley!

[nina_rox]

Take a look at this: http://mathworld.wolfram.com/Cosecant.html
Have you learnt how to graph reciprocal functions?

Thanks for that. Yes we have, but we haven't really discussed about the range and domain of them so I just wanted to get an approval on that.

[kensan]

Was wondering how many methods are there for working out this question.
a=-2i-2j+3k
b=3i-j-2k
Find a unit vector perpendicular to both a and b.
The bit before that i worked out the vector resolutes of a in direction b and b in direction a, can I use these answers to help me with my main question?