if cosx=tany , cosy=tanz , cosz=tanx then find the value of sinx
Dear Student,
We have : cos x = tan y.
∴ cos² x = tan² y = sec² y - 1 ............... (1)
But cos y = tan z ∴ sec y = cot z
∴ from (1),
... cos² x = cot² z - 1
∴ 1 + cos² x = cot² z = cos² z / sin² z = cos² z / ( 1 - cos² z )
But cos z = tan x
∴ 1 + cos² x = tan² x / ( 1 - tan² x )
∴ 1 + ( 1 - sin² x ) = ( sin² x / cos² x ) / [ 1 - ( sin² x / cos² x ) ]
∴ 2 - sin² x = sin² x / ( cos² x - sin² x )
∴ 2 - sin² x = sin² x / ( 1 - 2 sin² x )
∴ ( 2 - sin² x )( 1 - 2 sin² x ) = sin² x
∴ 2 sinā“ x - 6 sin² x + 2 = 0
∴ by Quadratic Formula,
sin² x = [ 3 ± √(9-4) ] / 2 = ( 3 ± √5 ) / 2
But ( 3 + √5 ) /2 > 1 whereas sin² x ≤ 1.
∴ sin² x = ( 3 - √5 ) / 2 = ( 6 - 2√5 ) / 4 = ( √5 - 1 )² / 4
∴ sin x = ( √5 - 1 ) / 2 = 2 sin 18°
We can similarly show that
... sin x = sin y = sin z = 2 sin 18° = ( √5 - 1 ) / 2
We have : cos x = tan y.
∴ cos² x = tan² y = sec² y - 1 ............... (1)
But cos y = tan z ∴ sec y = cot z
∴ from (1),
... cos² x = cot² z - 1
∴ 1 + cos² x = cot² z = cos² z / sin² z = cos² z / ( 1 - cos² z )
But cos z = tan x
∴ 1 + cos² x = tan² x / ( 1 - tan² x )
∴ 1 + ( 1 - sin² x ) = ( sin² x / cos² x ) / [ 1 - ( sin² x / cos² x ) ]
∴ 2 - sin² x = sin² x / ( cos² x - sin² x )
∴ 2 - sin² x = sin² x / ( 1 - 2 sin² x )
∴ ( 2 - sin² x )( 1 - 2 sin² x ) = sin² x
∴ 2 sinā“ x - 6 sin² x + 2 = 0
∴ by Quadratic Formula,
sin² x = [ 3 ± √(9-4) ] / 2 = ( 3 ± √5 ) / 2
But ( 3 + √5 ) /2 > 1 whereas sin² x ≤ 1.
∴ sin² x = ( 3 - √5 ) / 2 = ( 6 - 2√5 ) / 4 = ( √5 - 1 )² / 4
∴ sin x = ( √5 - 1 ) / 2 = 2 sin 18°
We can similarly show that
... sin x = sin y = sin z = 2 sin 18° = ( √5 - 1 ) / 2