Answers
To find the value of sin 18°
Let A = 18°
Therefore, 5A = 90°
⇒ 2A + 3A = 90˚
⇒ 2A= 90˚ - 3A
Taking sine on both sides, we get
sin 2A = sin (90˚ - 3A) = cos 3A we know , sin(90 - A) = cos A
⇒ 2 sin A cos A = 4cos^3 A - 3 cos A using formulas for sin 2A and cos 3A
⇒ 2 sin A cos A - 4cos^3A + 3 cos A = 0
⇒ cos A (2 sin A - 4 cos^2 A + 3) = 0
Dividing both sides by cos A = cos 18˚ ≠ 0, we get
⇒ 2 sin θ - 4 (1 - sin^2 A) + 3 = 0
⇒ 4 sin^2 A + 2 sin A - 1 = 0, which is a quadratic in sin A
Using SHRI DHAR ACHRYA formula
x = [-b ± √(b^2 - 4ac)] / 2a
18° lies in 1st quadrant,and sine is positive in 1st quadrant, so we take only positive value.
Therefore, sin 18° = sin A = −1+5–√4 = 0.30901699437
Thank you .
The explanation in video :
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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
Now how I get the Cos3x value?
Cox3x = Cos(2x+x)
Cox3x = Cos2xCosx - Sin2xSinx
Cox3x = (2Cos²x - 1)Cosx - (2SinxCosx)Sinx
Cox3x = 2Cos³x -Cosx - 2Sin²xCosx
Cox3x = 2Cos³x -Cosx - 2Cosx(1-Cos²x)
Cox3x = 2Cos³x -Cosx - 2Cosx + 2Cos³x
Cox3x = 4Cos³x - 3Cosx
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Let A = 18°
Therefore, 5A = 90°
⇒ 2A + 3A = 90˚
⇒ 2θ = 90˚ - 3A
Taking sine on both sides, we get
sin 2A = sin (90˚ - 3A) = cos 3A
⇒ 2 sin A cos A = 4cos^3 A - 3 cos A
⇒ 2 sin A cos A - 4cos^3A + 3 cos A = 0
⇒ cos A (2 sin A - 4 cos^2 A + 3) = 0
Dividing both sides by cos A = cos 18˚ ≠ 0, we get
⇒ 2 sin θ - 4 (1 - sin^2 A) + 3 = 0
⇒ 4 sin^2 A + 2 sin A - 1 = 0, which is a quadratic in sin A
Therefore, sin θ = −2±−4(4)(−1)√2(4)−2±−4(4)(−1)2(4)
⇒ sin θ = −2±4+16√8−2±4+168
⇒ sin θ = −2±25√8−2±258
⇒ sin θ = −1±5√4−1±54
Now sin 18° is positive, as 18° lies in first quadrant.
Therefore, sin 18° = sin A = −1±5√4