Problem: tan 19 x ∘ = cos 9 6 ∘ + sin 9 6 ∘ cos 9 6 ∘ − sin 9 6 ∘ tan 1 9 x ∘ = c o s 9 6 ∘ − s i n 9 6 ∘ c o s 9 6 ∘ + s i n 9 6 ∘
Solution:
The identity
cos A + sin A cos A − sin A = 1 + tan A 1 − tan A = tan 4 5 ∘ + tan A 1 − tan 4 5 ∘ tan A = tan ( 4 5 ∘ + A ) cos A − sin A cos A + sin A = 1 − tan A 1 + tan A = 1 − tan 4 5 ∘ tan A tan 4 5 ∘ + tan A = tan ( 4 5 ∘ + A )
implies that the given equation is equivalent to tan 19 x ∘ = tan ( 4 5 ∘ + 9 6 ∘ ) = tan 14 1 ∘ tan 1 9 x ∘ = tan ( 4 5 ∘ + 9 6 ∘ ) = tan 1 4 1 ∘ 19 x 1 9 x 141 1 4 1 180 1 8 0
19 x = 141 + 180 y = 19 ( 7 + 9 y ) + ( 8 + 9 y ) 1 9 x = 1 4 1 + 1 8 0 y = 1 9 ( 7 + 9 y ) + ( 8 + 9 y )
for some integer y y x x y y 8 + 9 y = 19 z 8 + 9 y = 1 9 z z z y y y = 2 z + z − 8 9 y = 2 z + 9 z − 8 z z 8 8 y = 16 y = 1 6 x = 159 x = 1 5 9
OR OR
Because sin 9 6 ∘ = cos 6 ∘ sin 9 6 ∘ = cos 6 ∘
tan 19 x ∘ = cos 9 6 ∘ + cos 6 ∘ cos 9 6 ∘ − cos 6 ∘ tan 1 9 x ∘ = cos 9 6 ∘ − cos 6 ∘ cos 9 6 ∘ + cos 6 ∘
The identities
cos ( A + B ) + cos ( A − B ) = 2 cos A cos B cos ( A + B ) + cos ( A − B ) = 2 cos A cos B
and
cos ( A + B ) − cos ( A − B ) = − 2 sin A sin B cos ( A + B ) − cos ( A − B ) = − 2 sin A sin B
imply that
cos 9 6 ∘ + cos 6 ∘ cos 9 6 ∘ − cos 6 ∘ = 2 cos 5 1 ∘ cos 4 5 ∘ − 2 sin 5 1 ∘ sin 4 5 ∘ = − sin 3 9 ∘ cos 3 9 ∘ cos 9 6 ∘ − cos 6 ∘ cos 9 6 ∘ + cos 6 ∘ = − 2 sin 5 1 ∘ sin 4 5 ∘ 2 cos 5 1 ∘ cos 4 5 ∘ = − cos 3 9 ∘ sin 3 9 ∘
It follows that the given equation is equivalent to
tan 19 x ∘ = − tan 3 9 ∘ = tan 14 1 ∘ tan 1 9 x ∘ = − tan 3 9 ∘ = tan 1 4 1 ∘
The solution continues as above.
OR OR
Notice that A cos x + B sin x A cos x + B sin x C cos ( x − ϕ ) C cos ( x − ϕ ) C 2 = A 2 + B 2 C 2 = A 2 + B 2 A = C cos ϕ A = C cos ϕ B = C sin ϕ B = C sin ϕ
tan 19 x ∘ = 2 cos ( 9 6 ∘ − 4 5 ∘ ) 2 cos ( 9 6 ∘ + 4 5 ∘ ) = cos 5 1 ∘ cos 14 1 ∘ = sin 14 1 ∘ cos 14 1 ∘ = tan 14 1 ∘ tan 1 9 x ∘ = 2 cos ( 9 6 ∘ + 4 5 ∘ ) 2 cos ( 9 6 ∘ − 4 5 ∘ ) = cos 1 4 1 ∘ cos 5 1 ∘ = cos 1 4 1 ∘ sin 1 4 1 ∘ = tan 1 4 1 ∘
The solution continues as before.
The problems on this page are the property of the MAA's American Mathematics Competitions