Search results

Find the relationship between a, b, c and d such that \frac{ab\left ( c+d \right )+cd\left ( a+b \right )+10\left ( ab+bc+cd+da \right )}{\left ( a+b \right )\left ( c+d \right )+10\left ( a+b+c+d \right )}=\frac{ab\left ( c+d \right )+cd\left ( a+b \right )+\left ( ab+bc+cd+da \right )}{\left...

You are very polite Mr Goski.

=(b-c)(c-a)(a-b)(a+b+c) by factor theorem e.g. in the original exp, replace b by c gives zero, .: b-c is a factor. The original expression is cyclic, .: c-a and a-b are also factors, .....

very likely with a hinting question beforehand

Einstein once said there is only one genius left in this world, who can tell whether you are one or not.

Factorise: 1-(sinx)^10=(1-(sinx)^2)(1+(sinx)^2+(sinx)^4+(sinx)^6+(sinx)^8) =(cosx)^2 (1+(sinx)^2+(sinx)^4+(sinx)^6+(sinx)^8) integ (cosx)(1+(sinx)^2+(sinx)^4+(sinx)^6+(sinx)^8) dx let u=sinx etc. integ (1+u^2+u^4+u^6+u^8) du etc.

This question does not require the use of those formulas.

New question: Show (sinx+sin3x+sin5x+sin7x+sin9x)/(cosx+cos3x+cos5x+cos7x+cos9x) = tan5x

meant to be tan^-1 4/3 = cos^-1 3/5

P(1)=a+b+1=0 P'(1)=7a+6b=0 a=6, b=-7

Anchored a particular boy. Arrange the remaining 9 boys. There are 10 gaps between the boys. Place the 5 girls (at most one in a gap) in the gaps. 9!10P5 = 10973491200

Another approach 10973491200?

30046464000?

Still working on it.

14!

Thanks I checked my working, should be just 2i.

You passed your subtraction test.

1-z=(1/2)sin(@/2)cis(@/2) 1-z^n=(1/2)sin(n@/2)cis(n@/2) (1-z^n)/(1-z)=result

or be a proud father of 3.