Question

3 sin 6 degrees - 4cos cube 84 degrees

Answer 1

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

Given expression is

$\rm \: 3sin6\degree - 4 {cos}^{3}84\degree$

can be rewritten as

$\rm \: = \: 3sin6\degree - 4 {cos}^{3}(90\degree - 6\degree)$

We know,

$\boxed{\sf{ \:cos(90\degree - x) = sinx \: }}$

So, using this identity, we get

$\rm \: = \: 3sin6\degree - 4 {sin}^{3}6\degree$

We know,

$\boxed{\sf{ \:3sinx - {4sin}^{3}x = sin3x \: }}$

So, using this identity, we get

$\rm \: = \: sin3(6\degree )$

$\rm \: = \: sin18\degree$

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

Hence,

$\rm\implies \:\boxed{\sf{ \:\rm \: 3sin6\degree - 4 {cos}^{3}84\degree = \frac{ \sqrt{5} - 1}{4} \: }}$

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Additional Information

Value of sin18°

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

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

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

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

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

$\rm \: sin2x = cos3x$

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

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

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

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

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

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

So, its a quadratic equation in sinx, so using Quadratic Formula, we get

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

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

$\rm \: sinx = \dfrac{ - 2 \: \pm \: \sqrt{ 20} }{2(4)}$

$\rm \: sinx = \dfrac{ - 2 \: \pm \: \sqrt{ 2 \times 2 \times 5} }{2(4)}$

$\rm \: sinx = \dfrac{ - 2 \: \pm \: 2 \sqrt{5} }{2(4)}$

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

$\rm \: sinx = \dfrac{ - 1 +\sqrt{5} }{4} \: \: or \:\:\dfrac{ - 1 - \sqrt{5} }{4} \: \{rejected\:as\: x\: is \: acute \}$

Hence,

$\rm\implies \:\boxed{\sf{ \:\rm \: sin18\degree = \dfrac{\sqrt{5} - 1}{4} \: }} \: \:$

Answer 2

Given,

$3 \sin{6}^{o} - 4 \cos3 {84}^{o}$

We can write this expression as,

$\boxed{3 \sin {6}^{o} - 4 { \cos}^{3} {90}^{o} - {6}^{o} }$

Using the identity,

$\underline{ \cos( {90}^{o} - x) = \sin \: x}$

So, after using we get,

$3 \sin {6}^{o} - 4 { \sin}^{3} {6}^{o}$

Another identity which we know,

$\underline{3 \sin x \: - 4 { \sin}^{3} x = \sin3x}$

So, after using the identity we get,

$= \: \sin3( {6}^{o} )$

$= \: \sin {18}^{o}$

$\large = \: \frac{ \sqrt{5} - 1 }{4}$

Therefore,

$\leadsto \: \boxed{3 \sin {6}^{o} - 4 { \cos}^{3} {84}^{o} = \frac{ \sqrt{5} - 1 }{4} }$