103 Trigonometry Problems

Hence
4. Solutions to Introductory Problems 109
tan 2x + tan 2y + tan 2z = tan 2x tan 2y tan 2z,
using a similar argument to the one in Problem 20(a). This implies that
and the conclusion follows.
2 tan x
1 − tan 2 x + 2 tan y
1 − tan 2 y + 2 tan z
1 − tan 2 z
= 2 tan x
1 − tan 2 x ·
2 tan y
1 − tan 2 y ·
2 tan z
1 − tan 2 z ,
37. Prove that a triangle ABC is isosceles if and only if
a cos B + b cos C + c cos A = a + b + c .
2
Solution: By the extended law of sines, a = 2R sin A, b = 2R sin B, and
c = 2R sin C. The desired identity is equivalent to
or
2 sin A cos B + 2 sin B cos C + 2 sin C cos A = sin A + sin B + sin C,
sin(A + B) + sin(A − B) + sin(B + C)
+ sin(B − C) + sin(C + A) + sin(C − A)
= sin A + sin B + sin C.
Because A + B + C = 180 ◦ , sin(A + B) = sin C, sin(B + C) = sin A,
sin(C + A) = sin B. The last equality simplifies to
which in turn is equivalent to
sin(A − B) + sin(B − C) + sin(C − A) = 0,
4 sin A − B
2
sin B − C
2
by Problem 15. The conclusion now follows.
sin C − A
2
= 0,
38. Evaluate
where a = 2π
1999 .
cos a cos 2a cos 3a ···cos 999a,