Date: Sat, 20 Sep 1997 10:02:27 -0600 (CST)
Subject: math

Name: Tan

Level: Secondary

Question:
How many distinct acute angles x are there for which cosx cos2x cos4x=1/8?

Hi Tan,
We have two solutions for you. In the first solution we expand your equation to get a seventh degree polynomial in cos(x) and uses Descartes' Rule of Signs to argue that there are three positive solutions. The second solution uses only trigonometry and not only shows there are exactly three solutions but finds them explicitly.

Solution 1.

Let f(x) = cos(x) cos(2x) cos(4x) -1/8 and use the fact that cos(2x) = 2 cos^2(x) -1 to expand the cos(2x) and cos(4x) factors and get

f(x) = 16 cos^7(x) -24 cos^5(x) +10 cos^3(x) - cos(x) -1/8.

Let t=cos(x) and write g(t) = 16 t^7 - 24 t^5 +10 t^3 -t -1/8.

Since cos(x) is non negative for acute angles we are looking for positive zeros of g(t)

Now use Descartes' Rule of Signs which says:

Let P(x) be a polynomial with real coefficients written in descending powers of x. Count the number of sign changes in the signs of the coefficients.

1. The number of positive real zeros is equal to the number of sign changes or is equal to that number decreased by an even number.
2. The number of negative real zeros is equal to the number of sign changes in P(-x) or is equal to that number decreased by an even number.
The number of sign changes in g(t) is three so g(t) has either three positive real zeros or one positive real zero.

• g(1/2)=f(Pi/3)=0 gives one zero.
• g(1)=f(0)>0 and g(sqrt(3)/2)=f(Pi/6)<0 so there is a second zero between sqrt(3)/2 and 1.
Thus, since g(t) has at least two positive real zeros it must have three. Hence there are three acute angles x for which cos(x)cos(2x)cos(4x)=1/8.

Chris and Harley.

Solution 2.

First we show that cos(x) cos(2x) cos(4x) =1/8 if and only if sin(8x) = sin(x) and sin(x) # 0. (Here I use # to mean not equal.)

Proof:

Suppose that cos(x) cos(2x) cos(4x) =1/8 then clearly sin(x) # 0. Multiply both sides of cos(x) cos(2x) cos(4x) =1/8 by 8sin(x) to get 8sin(x)cos(x) cos(2x) cos(4x) =sin(x).

Now
8sin(x)cos(x) cos(2x) cos(4x) =sin(x)
if and only if
4sin(2x)cos(2x)cos(4x)=sin(x)
if and only if
2sin(4x)cos(4x)=sin(x)
if and only if
sin(8x)=sin(x).

Conversely if sin(8x) = sin(x) and sin(x) # 0 then 8sin(x)cos(x) cos(2x) cos(4x) =sin(x) and division by 8sin(x) gives cos(x) cos(2x) cos(4x) =1/8.

Using the trigonometric identity that
sin(p)-sin(q) = 2 sin((p-q)/2)cos((p+q)/2),
sin(8x) - sin(x) =0 if and only if
2cos(9x/2)sin(7x/2)=0

cos(9x/2)=0 gives 9x/2=Pi/2, 3Pi/2, 5Pi/2,...

which, for acute angles, yields x=Pi/9 and x=Pi/3.

sin(7x/2)=0 gives 7x/2=0, Pi, 2Pi,...

which, for acute angles, yields x=0 and x=2Pi/7.

But sin(x)#0 so the solutions are x=Pi/9, Pi/3 and 2Pi/7.

Haragauri.

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