t(r+ 1)=c6(r)(x)^(6-r)(- 1/√x)^r
=(- 1)^rC6(r)x^(6-r-r/2)
Is a constant term, so
6-r-r/2=0, then r=4.
So (-1) RC6 (r)
=(- 1)^4*C6(4)= 15
So the constant term in the sixth power expansion of binomial (x-1√ x) is 15.
2.sinx+cosx =- 1, where both sides are squared at the same time:
1+2sinxcosx= 1
sin2x=0
{x|x=kπ/2 k is an integer}
3.2 C2+3c 2+4c 2+……+nC2 = 3c 3+3c 2+4c 2+……+nC2
= 4c3+4c2+...+nc2 (keep pushing)
=nC3=n(n- 1)(n-2)/6
The numerator and denominator are divided by n 3 at the same time, and the limit is 1/6.
4. If point P(-2, √3) is on the terminal side of angle α, then sinα=y/r=√3/√7, cosα=x/r=-2/√7,
sin(α+π/3)
= Sina cosπ/3+cosa sinπ/3 =(√3/√7)*( 1/2)+(-2/√7)*(√3/2)=(√3-2√3)/(2√7)=-√3/(2√7)