Heya !!!!! Find the value of Sin(4π/5) ----------#"NOT GIVE SPAM ANS ". Created by Aasthakatheriya1. Math

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HELLO DEAR,

sin(4π/5)

=> sin(π - π/5)

=> sinπ/5
[ sin(180 - theta) = sintheta]

=> sin180/5

=> sin36°

NOW,

Let A = 18°

THEN, 5A = 90°

⇒ 2A + 3A = 90˚

⇒ 2θ = 90˚ - 3A

[ take "sine" both sides,]

we get,

sin 2A = sin (90˚ - 3A) = cos 3A

⇒ 2 sin A cos A = 4 cos³ A - 3 cos A

⇒ 2 sin A cos A - 4 cos³ A + 3 cos A = 0

⇒ cos A (2 sin A - 4 cos² A + 3) = 0

Dividing both sides by cos A = cos 18˚ ≠ 0,

we get

⇒ 2 sinA - 4 (1 - sin² A) + 3 = 0

⇒ 4sin² A + 2 sin A - 1 = 0,

now using Quadratic Equation,

where,

a = 4

b = 2

c = -1

D = b² - 4ac

⇒ D = 2² - 4 × 4 ×(-1)

⇒D = 4 + 16

⇒D = 20

now,

⇒√D = 2√5

now,

sinA = (- 2 ± √D) / 2a

sinA = ( - 2 ± 2√5)/2×4

sinA = 2( - 1 ± √5)/4

now,

sin 18° is positive, as 18° lies in first quadrant.

Therefore, sin 18° = sin A = ( - 1 ± √5)/4

Now, cos 36° = cos 2 ∙ 18°

⇒ cos 36° = 1 - 2sin² 18°

⇒ cos36° = 1 - 2{( √5 - 1)/4}²

⇒ cos36° = (8 - 5 - 1 + 2√5)/8

⇒cos36° = (2 + 2√5)/8

⇒ cos36° = (1 + √5)/4

now,

sin36° = √(1 - cos²36)

⇒ sin36° = √[( 16 - 1 - 5 - 2√5)/16]

⇒ sin36° = √(10 - 2√5)/16}

⇒sin36° = √(10 - 2√5) / 4

I HOPE ITS HELP YOU,
THANKS

rohitkumargupta :-)

Roshan4tech Superb!!

Swarup1998 The answer is given below :

Given,

sin(4π/5)

= sin{(4×180)/5}°, since π radian = 180°

= sin144°

= sin(180° - 36°)

= sin36°, since sin(180° - θ) = sinθ

So, we need to find the value of sin36°.

Let us consider,

θ = 18°

Then, 5θ = 90°

=> 2θ + 3θ = 90°

=> 2θ = 90° - 3θ

Now, taking sine to both sides, we get

sin2θ = sin(90° - 3θ)

=> sin2θ = cos3θ, since sin(90° - θ) = cosθ

=> 2 sinθ cosθ = 4 cos³θ - 3 cosθ

=> 2 sinθ cosθ - 4 cos³θ + 3 cosθ = 0

=> cosθ (2sinθ - 4 cos²θ + 3) = 0

So, 2 sinθ - 4 cos²θ + 3 = 0,
since cosθ = cos18° ≠ 0

=> 2 sinθ - 4(1 - sin²θ) + 3 = 0,
since sin²θ + cos²θ = 1

=> 4 sin²θ + 2 sinθ - 1 = 0

Using Sridhar Acharya's formula, we get

sinθ = [-2 ± √{2² - 4×4×(-1)}]/(2×4)

= (-2 ± √20)/8

= (-2 ± 2√5)/8

= (-1 ± √5)/4

Since, θ = 18°, sin18° > 0

Therefore, sin18° = (√5 - 1)/4

Now, cos36°

= cos(2 × 18°)

= 1 - 2 sin²18°, since cos2θ = 1 - 2 sin²θ

= 1 - 2 [(√5 - 1)/4]²

= 1 - [2 × (5 - 2√5 + 1)/16]

= 1 - (6 - 2√5)/8

= (8 - 6 + 2√5)/8

= (2 + 2√5)/8

= (1 + √5)/4

So, sin36°

= √[1 - cos²36°], since sin²θ + cos²θ = 1

= √[1 - {(1 + √5)/4}²]

= √[1 - {(1 + 2√5 + 5)/16}

= √[1 - (6-2√5)/16]

= √[(16 - 6 + 2√5)/16]

= √[(10 - 2√5)/16]

= 1/4 × √(10 - 2√5), since √(4²) = 4

Hence,

sin(4π/5)

= sin144°

= sin36°

= 1/4 × √(10 - 2√5) [Ans.]

Thank you for your question.

Roshan4tech Nice

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