On question 1: 

The simplest solution is to multiply the integrand by
(1-sin(x))/(1-sin(x)). A more systematic approach is the trigonometric
substitution u=tan(x/2) which can be used for any rational function of
sin(x) and cos(x).


On question 2:

If any of f, f', f'', or f''' ever vanish we are done.

Suppose no such a exists. Then f, f', f'', or f'''
all have constant sign.

If f is a counterexample, then so is f(-x) as well as c f(x) for
constants c \neq 0. So if there is a counterexample, there must be one
with f''<0 and f'''<0. Now WLOG assume f is such a counterexample.

Such an f has f' always decreasing.

If f'(0) > 0, the equation

          f'(x) = \int_{0}^{x} f''(t) dt + f'(0)

then shows lim_{x \to \infty} f''(x)=0 which contradicts f''
decreasing and f''(0)<0.

Hence if there is a counterexample, there must be one with f'<0,
f''<0, f'''<0.

Now

    f(x) = \int_{0}^{x} f'(t) dt + f(0)

shows that f(0) > 0 requires lim_{x \to \infty} f'(x)=0 which
contradicts f' decreasing and f'(0)<0.

So f<0 and no counterexample exists.







On question 3: 

One writes a_n^{n/n+1} = a_n a_n^{n/n+1 - 1} and then one observes
that there are two regimes: either a_n^{-1/(n+1)} < M, or not (M being
a fixed big constant, say 10). In the first case, one gets a_n >
(1/M)^{n+1} while in the second, the opposite inequality a_n <
(1/M)^{n+1}. But this basically end the argument, since one observes
that in the first case one can use comparison principle and majorize
our series with the convergent one "\sum a_n", while in the second the
series that we want to understand can be majorized by a geometric
series, which is also convergent.

On question 4:

(a) Let X be the expected number of tosses. With probability p, X is 1
and with probability 1-p, the first toss is wasted, and so the
expected value of X is 1 more than it was before we started. Hence if
E is the expected value of X

E = p(1)+(1-p)(E+1)

giving E=1/p.

(b) Let E2 be the expected number of tosses required. There are three
possible results of the first toss.

If it is a head, we need to then wait for a tail. The expected number
of extra tosses needed to finish is 1/q by (a).

Similarly if it is a tail, we need to then wait for a head. The
expected number of extra tosses needed to finish is 1/p by (a).

If the first toss is on edge, the expected total (since the first was
a waste) is one more than it was at the beginning.

So

E2=p(1/q+1) + q(1/p+1)+(1-p-q)(E2+1)

Solving for E2 gives

 E2 = (1/(p+q))(p/q +q/p +1)


On Question 5:

The answer is YES:

Arguing by contradicition, assume that every two points one inch apart
have different collors.

Then, let's take two points say A, B in the plane one inch apart. They
must have different collors. Let's then take two other points U, V one
"to the RHS of the segment AB" and the other "to the LHS of AB" so
that the triangles ABU, ABV are equilateral. It is then easy to see
that both U, V must have been painted in the third collor so they have
the same collor. This argument shows that ANY two points in the
situation of U,V have the same collor (i.e. which are \sqrt{3} inch
apart).

But then, one can take a triangle UVV' so that UV=UV' = \sqrt{3} while
VV'=1 and since all three points have the same collor (as a
consequence of the above observation) we obtain a contradiction.

On question 6:

Factor sqrt(x^10) out of the radical, complete the square on the rest
under the radical and and notice dx/x^6 is very convenient.

On question 7: 

Let O be the center of the circle and E the other endpoint of the
diameter extending radius AO. Then triangle AEB is a right triangle
with hypotenuse 1 and angle AEB = angle C. So side AB has length
sin(C) (the first result) and BE has length cos(C).

To finish proving the second result, first observe that CH is parallel
to BE since the lines containing both are perpendicular to AB. And
since AEC is also a right triangle, we have CE is parallel to BH. Thus
HBEC is a parallelogram.  We already found that BE has length cos(C),
so the requested length of the opposite side HC is also cos(C).

(This problem is based on a neat article by Jerzy Kocik and Andrzej
Solecki in the March 2009 issue of the American Mathematical Monthly.)