Question

show that sin 18=√5-1/4​

Answer 1

Step-by-step explanation:

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–√4−1+54 = 0.30901699437

Thank you .

Answer 2

Question:

Show that-

$\sf{sin\: {18}^{\circ}=\dfrac{sqrt{5}-1}{4}}$

Solution:

❍ Let A equals to 18°. That is,

$\: \: \: \longmapsto\sf{A={18}^{\circ}}$

~Multiplying 5 on both the side,

$\: \: \: \longmapsto\sf{5\times A=18\times 5}$

$\: \: \: \longmapsto\sf{5A=90}$

We know, that 5A can also be written as 2A+3A. So,

$\: \: \: \longmapsto\sf{2A+3A=90}$

$\: \: \: \longmapsto\sf{2A=90-3A}$

• Taking 'sin' on both the side,

$\: \: \: \longmapsto\sf{sin\: 2A=sin\: (90-3A)}$

We know, $\sf{sin\: (90-A)=cos\: A}$. So,

$\: \: \: \longmapsto\sf{sin\: 2A=cos\: 3A}$

We know,

$\: \: \: :\implies\sf{sin\: 2\theta=2\: sin \theta\times cos\theta}$

$\: \: \: :\implies\sf{cos\: 3\theta=4\: {cos}^{3}\theta-3\: cos\theta}$

So,

$\: \: \: \longmapsto\sf{2sin\: A\times cos\: A=4{cos}^{3}\: A-3cos\: A}$

$\: \: \: \longmapsto\sf{2sin\: A\times cos\: A=cos\: A(4{cos}^{2}\: A-3)}$

$\: \: \: \longmapsto\sf{2sin\: A=4{cos}^{2}\: A-3}$

We know, $\sf{{cos}^{2}\theta=1-{sin}^{2}\theta}$. So,

$\: \: \: \longmapsto\sf{2sin\: A=4(1-{sin}^{2}\: A)-3}$

$\: \: \: \longmapsto\sf{2sin\: A=4-4{sin}^{2}\: A-3}$

$\: \: \: \longmapsto\sf{4{sin}^{2}\: A+2sin\: A-1=0}$

• Taking 'sinA' as 'x'. Then,

$\: \: \: \longmapsto\sf{4{x}^{2}+2x-1=0}$

~We observe that the equation formed; is a Quadratic equation. Where,

$\rightarrow\sf{a=4}$

$\rightarrow\sf{b=2}$

$\rightarrow\sf{c=-1}$

We know,

$\: \: \: \longrightarrow\sf{x=\dfrac{- b\frac{+}{}\sqrt{{b}^{2} - 4ac}}{2a}}$

So,

$\: \: \: \longrightarrow\sf{x=\dfrac{-2\frac{+}{}\sqrt{{2}^{2} - 4\times 4\times -1}}{2\times 4}}$

$\: \: \: \longrightarrow\sf{x=\dfrac{-2\frac{ + }{}\sqrt{4+16}}{8}}$

$\: \: \: \longrightarrow\sf{x=\dfrac{-2\frac{+}{}\sqrt{20}}{8}}$

$\: \: \: \longrightarrow\sf{x=\dfrac{-2\frac{+}{}2\sqrt{5}}{8}}$

$\: \: \: \longrightarrow\sf{x=\dfrac{-1\frac{+}{}\sqrt{5}}{4}}$

But, $\sf{sin\: {18}^{\circ}>0}$. So,

$\: \: \: \longrightarrow\bold{sin\: {18}^{\circ}=\dfrac{\sqrt{5}-1}{4}}$

Hence Proved !