103 Trigonometry Problems

5. Solutions to Advanced Problems 147
Because T n (cos θ) = cos nθ, P n (2 cos θ) = 2 cos nθ. It follows that for all θ,
P m (P n (2 cos θ)) = P m (2 cos nθ) = 2 cos mnθ
= P n (2 cos mθ) = P n (P m (2 cos θ)).
Thus, P m (P n (x)) and P n (P m (x)) agree at all values of x in the interval [−2, 2].
Because both are polynomials, it follows that they are equal for all x, which
completes the proof.
25. [China 2000, by Xuanguo Huang] In triangle ABC, a ≤ b ≤ c. As a function
of angle C, determine the conditions under which a + b − 2R − 2r is positive,
negative, or zero.
Solution: In Figure 5.6, set ̸ A = 2x, ̸ B = 2y, ̸ C = 2z. Then 0