Question

prove that sin π/10=√5-1/4​

Answer 1

Step-by-step explanation:

Question is Sin18° = ?

Let x = 18°

so, 5x = 90°

now we can write

2x + 3x = 90°

so 2x = 90° - 3x

now taking sin both side we can write

sin2x = sin(90°-3x)

sin2x = cos3x [as we know, sin(90°-3x) = Cos3x ]

so evaluating we can write

2sinxcosx =4cos³x - 3cosx [as we know, Cos3x = 4cos³x - 3cosx, This I will explain later ]

Now, 2sinxcosx - 4cos³x + 3cosx = 0

cosx(2sinx - 4cos²x + 3) = 0

Now divding both side by cosx we get,

2sinx - 4cos²x + 3 = 0

2sinx - 4(1-sin²x) + 3 = 0 [as we know, cos²x = (1-sin²x), by sin²x + cos²x = 1 ]

2sinx - 4 + 4sin²x + 3 = 0

2sinx + 4sin²x - 1 = 0

we can write it as,

4sin²x + 2sinx - 1 = 0

Now apply Sridhar Acharya Formula Here,

ax² + bx + c = 0

so, x = (-b ± √(b² - 4ac))/2a

now applying it in the equation

sinx =

(-2 ± √(2² - 44(-1)))/2.(4)

sinx = (-2 ± √(4 +16))/8

sinx = (-2 ± √20)/8

sinx = (-2 ± 2.√5) / 8

sin x = 2(-1 ± √5 ) / 8

sin x = (-1 ± √5)/4

sin18° = (-1 ± √5)/4

Hope this helps :-)

Answer 2

Answer:

sinθ= $\frac{\sqrt{5}-1 }{4}$ is proved.

Step-by-step explanation:

Let θ = 18°

5θ = 5(18) = 90°

2θ + 3θ = 90°

2θ = 90 - 3θ

⇒sin2θ = sin(90 - 3θ)

sin2θ = cos3θ

2sinθ cosθ = 4cos²θ - 3cosθ

4cos²θ -3cosθ - 2sinθcosθ = 0

cosθ(4cos²θ - 3 -2sinθ) = 0

cosθ ≠0

4cos²θ -2sinθ -3 = 0

4(1 - sin²θ)-2sinθ-3 = 0

4 - 4sin²θ - 2sinθ -3 = 0

4sin²θ +2sinθ -1 = 0

sinθ =-2 ±√(-2)²-4(4)(1)/2×4

sinθ = -2±√4+16/8

sinθ = -2±√20/8

       = -1±√5/4

sin18 is 1st quadrant then sin  is positive.

∴sinθ= $\frac{\sqrt{5}-1 }{4}$

Hence proved.