cos y sin x + cos x sin y = 1 , sin y cos x + sin x cos y = 6
Let x , y be real numbers that satisfy the equations above.
Suppose that tan y tan x + tan x tan y = q p for relatively prime positive integers p and q . Find p + q .
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Let a = cos ( y ) sin ( x ) and b = cos ( x ) sin ( y ) . Then we can write the system of equations as
a + b = 1 , a 1 + b 1 = 6 ⟹ a 1 + 1 − a 1 = 6 ⟹ 1 = 6 a ( 1 − a )
⟹ 6 a 2 − 6 a + 1 = 0 ⟹ a = 1 2 6 ± 3 6 − 2 4 = 6 3 ± 3 .
Then b = 1 − a = 6 3 ∓ 3 . Because of this symmetry, we can, without loss of generality, let a = 6 3 + 3 and b = 6 3 − 3 .
This gives us a b = tan ( x ) tan ( y ) = 3 6 6 = 6 1 , and so
tan ( y ) tan ( x ) + tan ( x ) tan ( y ) = tan ( x ) tan ( y ) tan 2 ( x ) + tan 2 ( y ) = 6 ∗ ( sec 2 ( x ) + sec 2 ( y ) − 2 ) , (A).
Now a = cos ( y ) sin ( x ) = 6 3 + 3 ⟹ sin ( x ) = 6 3 + 3 cos ( y ) , and
b = cos ( x ) sin ( y ) = 6 3 − 3 ⟹ cos ( x ) = 3 − 3 6 sin ( y ) = ( 3 + 3 ) sin ( y ) .
Thus sin 2 ( x ) + cos 2 ( x ) = 6 2 + 3 cos 2 ( y ) + ( 1 2 + 6 3 ) sin 2 ( y )
⟹ 1 = 6 2 + 3 cos 2 ( y ) + ( 1 2 + 6 3 ) ( 1 − cos 2 ( y ) )
⟹ cos 2 ( y ) ( 1 2 + 6 3 − 6 2 + 3 ) = 1 1 + 6 3
⟹ sec 2 ( y ) = 6 ( 1 1 + 6 3 ) 3 5 ( 2 + 3 ) = 7 8 3 5 ( 4 − 3 ) .
Now a 2 = sin 2 ( x ) sec 2 ( y ) = 6 2 + 3 , so
sin 2 ( x ) = 6 2 + 3 ∗ 3 5 ( 4 − 3 ) 7 8 = 3 5 1 1 + 6 3 , and thus
cos 2 ( x ) = 1 − sin 2 ( x ) = 3 5 2 4 − 6 3
⟹ sec 2 ( x ) = 6 ( 4 − 3 ) 3 5 = 7 8 3 5 ( 4 + 3 ) .
Thus equation (A) becomes
6 ∗ ( 7 8 3 5 ( 4 − 3 ) + 7 8 3 5 ( 4 + 3 ) − 2 ) = 6 ∗ ( 7 8 2 8 0 − 2 ) = 1 3 1 2 4 ,
and so p + q = 1 2 4 + 1 3 = 1 3 7 .
Flawless as always!Perfect job sir Brian!
Let a = sin y cos x and b = sin x cos y thus from cos y sin x + cos y sin x = 1 and sin y cos x + sin x cos y = 6 . a + b = 6 , a 1 + b 1 = 1 . This leads to b 1 + 6 − b 1 = 1 . thus 6 = 6 b − b 2 and b = 3 ± 3 . Let b = 3 + 3 and a = 3 − 3 .
sin y cos x = 3 − 3 and sin x cos y = 3 + 3
so cos x = ( 3 − 3 ) sin y and cos y = ( 3 + 3 ) sin x
cos 2 x + sin 2 x = 1 , cos 2 y + sin 2 y = 1 . substitute x's for y's and y's for x's to get two equations.
( ( 3 − 3 ) sin y ) 2 + ( 3 + 3 cos y ) 2 = 1 and ( ( 3 + 3 ) sin x ) 2 + ( 3 − 3 cos x ) 2 = 1
Note 3 − 3 1 = 6 1 ( 3 + 3 ) and 3 + 3 1 = 6 1 ( 3 − 3 )
We then simplify these equations and solve for sin x and sin y
( 3 − 3 ) 2 ( sin 2 y + 3 6 1 cos 2 y ) = 1
( 3 + 3 ) 2 ( sin 2 x + 3 6 1 cos 2 x ) = 1
⟹
sin 2 y + 3 6 1 ( 1 − sin 2 y ) = ( 3 − 3 ) 2 1
sin 2 x + 3 6 1 ( 1 − sin 2 x ) = ( 3 + 3 ) 2 1
⟹
3 6 3 5 sin 2 y + 3 6 1 = 6 1 ( 2 − 3 )
3 6 3 5 sin 2 x + 3 6 1 = 6 1 ( 2 + 3 )
⟹
3 6 3 5 sin 2 y = 3 6 1 1 + 6 3
3 6 3 5 sin 2 x = 3 6 1 1 − 6 3
⟹
sin y = 3 5 1 1 + 6 3
sin x = 3 5 1 1 − 6 3
Solve for cos y and cos x
cos x = ( 3 − 3 ) sin y and cos y = ( 3 + 3 ) sin x
cos y = ( 3 + 3 ) 3 5 1 1 − 6 3
cos x = ( 3 − 3 ) 3 5 1 1 + 6 3
These reduce to
cos y = 3 5 6 ( 4 − 3 )
cos x = 3 5 6 ( 4 + 3 ) .
tan y tan x + tan x tan y = q p
Let tan y tan x = z solve for z
z = tan y tan x = cos x sin y sin x cos y = 3 5 1 1 + 6 3 3 5 6 ( 4 + 3 ) 3 5 1 1 − 6 3 3 5 6 ( 4 − 3 ) = 1 1 + 6 3 1 1 − 6 3 ⋅ 4 + 3 4 − 3 = 1 6 9 7 5 1 9 − 4 3 4 0 3 z = 1 3 6 2 − 3 5 3
z + z 1 = q p
q p = 1 3 6 2 − 3 5 3 + 6 2 − 3 5 3 1 3 = 1 3 ( 6 2 − 3 5 3 ) ( 6 2 − 3 5 3 ) 2 + 1 3 2 = 1 3 ( 6 2 − 3 5 3 ) 7 6 8 8 − 4 3 4 0 3 = 1 3 ( 6 2 − 3 5 3 ) 1 2 4 ( 6 2 − 3 5 3 ) = 1 3 1 2 4
p + q = 1 2 4 + 1 3 = 1 3 7
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