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