Question

prove that 4 ( sin 24° + cos 6°) =√3+√15

Answer 1

Answer:

4 ( sin 24° + cos 6°) =√3+√15

Step-by-step explanation:

Prove that 4 ( sin 24° + cos 6°) =√3+√15

sin 24° + sin84°

2Sin( (24 + 84)/2) Cos((84 - 24)/2)

= 2 Sin 54° Cos30°

= 2 Sin 54° (√3 / 2)

= √3 Sin 54°

Sin 54°  = Cos36° = 1 - 2Sin²18°

SinA = Sin18°

A = 18°

Therefore, 5A = 90°

⇒ 2A + 3A = 90˚

⇒ 2A= 90˚ - 3A

Taking sin on both sides, we get

sin 2A = sin (90˚ - 3A) = cos 3A as we know , sin(90 - A) = cos A

⇒ 2sinAcos A = 4cos³A - 3 cos A using formulas for sin 2A and cos 3A

⇒ 2sinAcos A - 4cos³A + 3 cos A = 0

⇒ cos A (2 sin A - 4 cos² A + 3) = 0

Dividing both sides by cos A

⇒ 2 Sin A - 4 (1 - sin² A) + 3 = 0

=>  4 sin² A + 2 sin A - 1 = 0, which is a quadratic in sin A

SinA = (-2 ± √(4 + 16) )/(2*4)

=> SinA = (-1 ± √5)/4

as A = 18° => SinA is +ve

=> SinA =  (-1 + √5)/4

=> Sin18° = (-1 + √5)/4

Sin 54°  =  1 - 2Sin²18°

= 1 - 2 ( (-1 + √5)/4)²

= 1 - (1/8) ( 1 + 5 - 2√5)

= ( 8 - 6 + 2√5)/8

= (2 + 2√5)/8

= ( 1 + √5)/4

4 ( sin 24° + cos 6°) = 4 √3 Sin 54°

= 4 √3  ( 1 + √5)/4

= √3  ( 1 + √5)

= √3  + √15

= RHS

QED

Proved

4 ( sin 24° + cos 6°) =√3+√15

Answer 2

Answer:

LHS=4(sin24°+cos6°)

Let's take (sin24°+cos6°)

(Cos6°=sin84°)

ApplysinC+sinD

Then[(sin24°+cos6°) =root over 15+Root over 3/4

=4(root15+root3/4)=root15+root3

LHS=RHS,hence proved

Step-by-step p explanaion: