The Student Room Group
Reply 1
I find that a library card is generally good for proving your identity with.
i have a fake id if you want it. and i know someone who can get you a fake apssport for £300 if yo want...


soz. i dont know tbh
Reply 3
djwarfield
Got a couple problems if you could help me out:

problem 1:

csc x/cot x - cot x/csc x = tan x/csc x

problem 2:

cos2y = 1-tan(squared)y/1+tan(squared)y

got a couple others if yer up to it


problem 1:

csc x = 1/sinx. cot x = cos x / sin x. So cscx/cotx - cotx/cscx = 1/cosx - cosx = 1-cos^2x/cosx = sin^2x/cosx = tanx/cscx.

2: 1-tan^2y/1+tan^2y = 1-tan^2y/sec^2y = cos^2y - sin^2y = cos2y.
Reply 4
Heres another that kinda gets me....

Find all angles theta, for 0<theta<360degrees, that satisfy the equation:
2csctheta+1=-2


thanks again
Reply 5
djwarfield
Heres another that kinda gets me....

Find all angles theta, for 0<theta<360degrees, that satisfy the equation:
2csctheta+1=-2


thanks again


2csc x = -3 so csc x = -3/2 so sinx = -2/3 so x = arcsin (-2/3) and work out any other solutions...
Reply 6
thanks....I had some notes written down so I just remebered how to do it...

ok one more then ill leave you alone

find the angle between vectors u = (2,5) and v = (-2,4)

thanks again
Reply 7
djwarfield
thanks....I had some notes written down so I just remebered how to do it...

ok one more then ill leave you alone

find the angle between vectors u = (2,5) and v = (-2,4)

thanks again


u.v = modulus(u)modulus(v)cos(theta), where theta is the angle between them.

modulus u = sqrt(2^2+5^2)=sqrt29

modulus v = sqrt((-2)^2+4^2)=sqrt20=2sqrt5

u.v = 2 x -2 + (5 x 4) = 16

so 16=2sqrt5sqrt29cos(theta)

cos(theta)=8/(sqrt29sqrt5)

theta = cos^(-1) of (8/sqrt29sqrt5)

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