Question

prove that sin 18⁰=root15-1/4​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that,

$\rm :\longmapsto\:x = 18 \degree \:$

So,

$\rm :\longmapsto\:5x = 90\degree \:$

$\rm :\longmapsto\:2x + 3x = 90\degree \:$

$\rm :\longmapsto\:2x = 90\degree \: - \: 3x$

Thus,

$\rm :\longmapsto\:sin2x =sin( 90\degree \: - \: 3x)$

$\rm :\longmapsto\:sin2x = cos3x$

$\rm :\longmapsto\:2sinx \: cosx \: = \: {4cos}^{3}x - 3cosx$

$\rm :\longmapsto \: {4cos}^{3}x - 3cosx - 2sinxcosx = 0$

$\rm :\longmapsto\:cosx( {4cos}^{2}x - 3 - 2sinx) = 0$

$\rm :\longmapsto\:{4cos}^{2}x - 3 - 2sinx = 0 \: \: \: \: as \: cosx \ne \: 0$

$\rm :\longmapsto\:{4(1 - sin}^{2}x) - 3 - 2sinx = 0 \: \: \:$

$\rm :\longmapsto\:{4 - 4sin}^{2}x - 3 - 2sinx = 0 \: \: \:$

$\rm :\longmapsto\:{1 - 4sin}^{2}x - 2sinx = 0 \: \: \:$

$\rm :\longmapsto\: - (4sin^{2}x + 2sinx - 1) = 0 \: \: \:$

$\rm :\longmapsto\: 4sin^{2}x + 2sinx - 1= 0 \: \: \:$

Its a quadratic equation in sinx, whose roots are evaluated by using quadratic formula.

So,

$\rm :\longmapsto\:sinx = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

Here,

$\red{\rm :\longmapsto\:a = 4}$

$\red{\rm :\longmapsto\:b = 2}$

$\red{\rm :\longmapsto\:c = - \: 1}$

So, on substituting these values, we get

$\rm :\longmapsto\:sinx = \dfrac{ - 2 \: \pm \: \sqrt{ {2}^{2} - 4(4)( - 1)} }{2(4)}$

$\rm :\longmapsto\:sinx = \dfrac{ - 2 \: \pm \: \sqrt{ 4 + 16} }{8}$

$\rm :\longmapsto\:sinx = \dfrac{ - 2 \: \pm \: \sqrt{20} }{8}$

$\rm :\longmapsto\:sinx = \dfrac{ - 2 \: \pm \: 2\sqrt{5} }{8}$

$\rm :\longmapsto\:sinx = \dfrac{ - 1 \: \pm \: \sqrt{5} }{4}$

$\bf\implies \:sinx = \dfrac{ \sqrt{5} - 1 }{4} \: \: as \: sinx \: \: \cancel{ < } \: 0$

Additional Information :-

$\red{ \boxed{ \sf{ \:cos18\degree = sin72\degree = \frac{ \sqrt{10 + 2 \sqrt{5} } }{4}}}}$

$\red{ \boxed{ \sf{ \:sin36\degree = cos54\degree = \frac{ \sqrt{10 - 2 \sqrt{5} } }{4}}}}$

$\red{ \boxed{ \sf{ \:sin54\degree = cos36\degree = \frac{ \sqrt{5} + 1}{4}}}}$

Answer 2

Answer:

Solution−

Let assume that,

\rm :\longmapsto\:x = 18 \degree \::⟼x=18°

So,

\rm :\longmapsto\:5x = 90\degree \::⟼5x=90°

\rm :\longmapsto\:2x + 3x = 90\degree \::⟼2x+3x=90°

\rm :\longmapsto\:2x = 90\degree \: - \: 3x:⟼2x=90°−3x

Thus,

\rm :\longmapsto\:sin2x =sin( 90\degree \: - \: 3x):⟼sin2x=sin(90°−3x)

\rm :\longmapsto\:sin2x = cos3x:⟼sin2x=cos3x

\rm :\longmapsto\:2sinx \: cosx \: = \: {4cos}^{3}x - 3cosx:⟼2sinxcosx=4cos3x−3cosx

\rm :\longmapsto \: {4cos}^{3}x - 3cosx - 2sinxcosx = 0:⟼4cos3x−3cosx−2sinxcosx=0

\rm :\longmapsto\:cosx( {4cos}^{2}x - 3 - 2sinx) = 0:⟼cosx(4cos2x−3−2sinx)=0

\rm :\longmapsto\:{4cos}^{2}x - 3 - 2sinx = 0 \: \: \: \: as \: cosx \ne \: 0:⟼4cos2x−3−2sinx=0ascosx=0

\rm :\longmapsto\:{4(1 - sin}^{2}x) - 3 - 2sinx = 0 \: \: \::⟼4(1−sin2x)−3−2sinx=0

\rm :\longmapsto\:{4 - 4sin}^{2}x - 3 - 2sinx = 0 \: \: \::⟼4−4sin2x−3−2sinx=0

\rm :\longmapsto\:{1 - 4sin}^{2}x - 2sinx = 0 \: \: \::⟼1−4sin2x−2sinx=0

\rm :\longmapsto\: - (4sin^{2}x + 2sinx - 1) = 0 \: \: \::⟼−(4sin2x+2sinx−1)=0

\rm :\longmapsto\: 4sin^{2}x + 2sinx - 1= 0 \: \: \::⟼4sin2x+2sinx−1=0

Its a quadratic equation in sinx, whose roots are evaluated by using quadratic formula.

So,

\rm :\longmapsto\:sinx = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}:⟼sinx=2a−b±b2−4ac

Here,

\red{\rm :\longmapsto\:a = 4}:⟼a=4

\red{\rm :\longmapsto\:b = 2}:⟼b=2

\red{\rm :\longmapsto\:c = - \: 1}:⟼c=−1

So, on substituting these values, we get

\rm :\longmapsto\:sinx = \dfrac{ - 2 \: \pm \: \sqrt{ {2}^{2} - 4(4)( - 1)} }{2(4)}:⟼sinx=2(4)−2±22−4(4)(−1)

\rm :\longmapsto\:sinx = \dfrac{ - 2 \: \pm \: \sqrt{ 4 + 16} }{8}:⟼sinx=8−2±4+16

\rm :\longmapsto\:sinx = \dfrac{ - 2 \: \pm \: \sqrt{20} }{8}:⟼sinx=8−2±20

\rm :\longmapsto\:sinx = \dfrac{ - 2 \: \pm \: 2\sqrt{5} }{8}:⟼sinx=8−2±25

\rm :\longmapsto\:sinx = \dfrac{ - 1 \: \pm \: \sqrt{5} }{4}:⟼sinx=4−1±5

\bf\implies \:sinx = \dfrac{ \sqrt{5} - 1 }{4} \: \: as \: sinx \: \: \cancel{ < } \: 0⟹sinx=45−1assinx<0

Additional Information :-

\red{ \boxed{ \sf{ \:cos18\degree = sin72\degree = \frac{ \sqrt{10 + 2 \sqrt{5} } }{4}}}}cos18°=sin72°=4