prove that sin π/10=√5-1/4
Answers
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:
sinθ= 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=
Hence proved.