Difficult 3u questions, HELP!!! (1 Viewer)

[elizabethy]

hey , can some help me with these questions?

1. By considering coefficients of x^2n in (1+x)^n (1-x)^n(1-x)^2n,
show that
(nCr-1)(nCr+1)- (nCr-2)(nCr+2) + .........+ (-1)^r (nC1) ( nC2r-1) - (-1)^r (nC2r) = o.5 (nCr)[ (nCr)-1]

2. find the general solution to the following :

2sin^2x = 1

3. Prove :
(tanx +tan2x) ( cotx + cot3x)= 4

thanks a lot !

 

[Trev]

Just curious, it says that you go to university?
Question 2 answer:
2sin^2x=1
sinx = +- 1/(sqrt2)
basic angle x = 45 degrees, since plus/minus, it lies in every quadrant
therefore general solution is 180n +- 45 where n is a positive integer.

 

[Trev]

Question 1 is binomial theorem isn't it? We haven't done that yet. I had a go at question 3, but i'm not getting anywhere

 

[Will Hunting]

Hey elizabethy,

2. Using, cos2x = cos^2x - sin^2x = 1 - 2sin^2x
sin^2x = (1-cos2x)/2

Now, 2sin^2x = 2[(1-cos2x)/2] = 1
cos2x = 0
2x = pi/2, -pi/2 for -pi < x < pi (or, 2x = 90, 270 for -180 < x < 180)
In general, the sol'n is x = 180n +- 45 (45 is used since 2x = 90, x = 45)

3. This one's epic. My teasing it out on this forum won't do a whole lot for you. If you know your trig rules for sums of, differences of and double angles, you shouldn't have too great a problem pulling through it - eventually. There are number of routes you might take, but I'd begin by eliminating tanxcotx as 1 straight up and then working with the remainder of the LHS as best you can, removing common factors, cancelling down and using sine and cos ratios where possible. Alternatively, you might try simplifying each of the 2 factors on the LHS before expanding, cross-multiplying fractions etc. If this question's been sourced from a 3U reference, the following won't apply, but, a little later on, I might see how DMT might be employed to simplify the ratios involving 3x more directly. I'll get back to you on Q.1 - it's a form of question I haven't as yet met, but I'll look into it for ya!

 

[KFunk]

I just did 3, and as Will Hunting said it's quite an epic. Bare with me while I work up the energy to type it out .

 

[KFunk]

(tanx +tan2x) ( cotx + cot3x)= 4

I'm going to deal with it in brackets 'cause of the size of it... here it goes.

Left hand bracket (of LHS)

= sinx//cosx + (2sinxcosx)//(2cos²x -1)

=(2sinxcos²x -sinx +2sinxcos²x)//cosx(2cos²x - 1)

= (4sinxcos²x -sinx )//cosx( 2cos²x -1)

= sinx(4cos²x - 1)//cosx(2cos²x - 1)




*phew* so... onto the right bracket (on the LHS ). I've skipped a few steps here in expanding cos3x/sin3x, but I'm sure you'll be cool with that.




= cosx//sinx + (2cos³x -cosx -2sin²xcosx)//(2sinxcos²x + sinx -2sin³x)

= (4sinxcos³x - 4sin³xcosx)//sin²x(2cos²x + 1 - 2sin²x)

= 4cosx(2cos²x - 1)//sinx(4cos²x -1)




By this point I can almost assure you that there are many mistakes but take the two brackets and multiply them:

sinx(4cos²x - 1)//cosx(2cos²x - 1)
times
4cosx(2cos²x - 1)//sinx(4cos²x -1)

cos kills cos and sin kills sin =

(4cos²x - 1)//(2cos²x - 1)
times
4(2cos²x - 1)//(4cos²x -1)

=4

damn... what a biatch of an expansion.

 

[elizabethy]

Just curious, it says that you go to university?

yeah im at university....these were on behalf of my sis

 

[m_isk]

Hey KFunk there is a muchhh easier way!!
(tanx+tan2x)(cotx+cot3x)
= (sinx/cosx + sin2x/cos2x)(cosx/sinx + cos3x/sin3x)
= (sinxcos2x+sin2xcosx)/cosxcos2x (cosxsin3x + sinxcos3x)/sin3xsinx
If u look at the crap in the brackets, you will see that the Left one is equal to sin(x+2x)=sin3x
and similarly, the right one is equal to sin(3x+x) = sin4x
.'. = sin3xsin4x/sin3xsinxcos2xcosx
= sin4x/sinxcos2xcosx (crossing off sin3x)
= 2sin2xcos2x/sinxcosxcos2x (since 2sin2x=2.2sinxcosx=4sinxcosx)
=4sinxcosxcos2x/sinxcosxcos2x
=4 as required!!

 

[KFunk]

Good call, it certainly saves a lot of time... and space...

 

[elizabethy]

thanks ppl