a.maths

a) L.H.S. = cos2θ + sin3θ

= cos 2 ( 3π/ 10 ) + sin 3 ( 3π/ 10 )

= cos ( 3π/ 5) + sin ( 9π/ 10 )

= - sin ( π/ 10 ) + sin ( π - 9π / 10 )

= - sin ( π/ 10 ) + sin ( π/ 10 )

= 0

= R.H.S.

So 3π/10 is a root of the equation cos2θ+sin3θ=0





b) cos2θ + sin3θ = 0

cos2θ = - sin3θ

1- 2sin2 θ = - ( 3 sin θ – 4 sin3θ )

1 - 2sin2 θ = 4 sin3θ - 3 sin θ

4 sin3θ + 2 sin2θ – 3 sin θ – 1 = 0

( sin θ + 1 )( 4 sin2 θ – 2 sin θ – 1 ) = 0

Put θ = 3π/ 10,

4 sin2 ( 3π/ 10 ) – 2 sin ( 3π / 10 ) – 1 = 0

sin ( 3π / 10 ) = (1 + √5) / 4 or ( 1 - √5) / 4 ( rejected )

So sin ( 3π / 10 ) = (1 + √5) / 4

(a) cos2θ + sin3θ = 0
　　　　　cos2θ = -sin3θ
　　　　　cos2θ = sin(π + 3θ)
　　　　　cos2θ = cos[π/2 - (π + 3θ)]
　　　　　cos2θ = cos (-π/2 - 3θ)
　　　　　cos2θ = cos (π/2 + 3θ)　　<-- cos -θ = cosθ
　 2θ = πn + π/2 + 3θ　or　2θ = πn - π/2 - 3θ
　　　　-θ = πn + π/2　or　5θ = πn - π/2
　　　　　　　　　　　　　 θ = πn/5 - π/10 .......(1)

Substitute n=2 into (1),
θ = 2π/5 - π/10
　= 4π/10 - π/10
　= 3π/10

Hence, 3π/10 is a root of the equation cos2θ+sin3θ=0.

b)By (a),
cos2(3π/10) +sin3(3π/10) = 0
1 - 2(sin 3π/10)^2 + 3sin3π/10 - 4sin(3π/10)^3 = 0

Let y=sin3π/10 and
let f(y) = 1-2y^2 + 3y - 4y^3
=-4y^3 - 2y^2 + 3y + 1

Apply factor theorem, f(-1) = 0, so (y + 1) is a factor of f(y).
By long division,
f(y) = (y + 1)(-4y^2 + 2y + 1)

Hence, (sin3π/10 + 1)(-4sin3π/10 + 2sin3π/10 + 1) = 0
sin3π/10 = -1 (rejected)
or sin3π/10 = [-2 +-√[4 - 4(-4)(1)] } /2(-4)
　　　　　 = (-2 +- 2√5)/-8
　　　　　 = (1+√5)/4, Ans. or (1-√5)/4 (rejected because sin3π/10 > 0)