a challenging problem of elementary mathematics?

Question

(i) Show that x^3 - (1+cos@ + sin@) x^2 + (cos@sin@ + cos@ + sin@)x - sin@cos@ = (x-1)(x^2 - (cos@ + sin@)x + cos@sin@)
and deduce that the cubic has roots 1, cos@, sin@

(ii) Give the roots when @ = pi/3

(iii) Find all values of @ in the range 0 <=@<2pi such that two of the three roots are equal

(iv) What is the greatest possible difference between two of the roots, and for what values of @ in the range 0<=@<2pi does this greatest difference occur?

Answer 1

Okay, I will abbreviate S = sin@ and C = cos@, ok?

(i)
x^3 - (1 + C + S) x^2 + (CS + C + S)x - SC =
1 x^3 - 1 x^2 - (C + S) x^2 + (C + S) x + CS x - CS =
(x - 1) (x^2) - (x - 1) (C + S) x + (x - 1) CS =
(x - 1) (x^2 - (C + S) x + CS)

You can factor even further:
(x - 1) (x - C) (x - S)

For this to be zero, any of the factors must be zero, so
x = 1 OR x = C OR x = S.

(ii)
If @ = pi/3, then C = cos@ = 1/2 and S = sin@ = 1/2 sqrt 3, so the roots are 1, 1/2 and 1/2 sqrt 3.

(iii)
Solutions are:
* S = 1; then @ = pi/2.
* C = 1; then @ = 0.
* S = C, then S/C = tan@ = 1, and @ = pi/4 or 3pi/4.

(iv)
The greatest difference possible is between -1 and 1, so we must have S = -1 or C = -1. That happens when @ = 3pi/2 or @ = pi, respectively.

Answer 2

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Answer 3

(i)3588 to the fifth power

(ii)12.45

(iii) 453.2344