Question

Prove the following trigonometry value

$cos18 \degree = \frac{ \sqrt{10 + 2 \sqrt{5} } }{4}$
​

Answer 1

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

Let assume that

$\rm \: x = 18\degree$

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

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

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

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

$\rm \: sin2x = cos3x$

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

As, cosx is not equals to 0, so cancelled out cosx on both sides, we get

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

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

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

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

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

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

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

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

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

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

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

$\rm \: sinx = \dfrac{2( - 1 \: \pm \: \sqrt{5} )}{8}$

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

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

Or

$\rm\implies \:sinx = \dfrac{ - \sqrt{5} - 1}{4} \: \: \: \: \{rejected \: as \: sinx > 0 \}$

So,

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

Now, we know that

$\rm \: {sin}^{2}18\degree + {cos}^{2}18\degree = 1$

$\rm \: \bigg(\dfrac{ \sqrt{5} - 1}{4} \bigg)^{2} + {cos}^{2} 18 = 1$

$\rm \: \dfrac{5 + 1 - 2 \sqrt{5} }{16} + {cos}^{2}18\degree = 1$

$\rm \: \dfrac{6 - 2 \sqrt{5} }{16} + {cos}^{2}18\degree = 1$

$\rm \: {cos}^{2}18\degree = 1 - \dfrac{6 - 2 \sqrt{5} }{16}$

$\rm \: {cos}^{2}18\degree = \dfrac{16 - 6 + 2 \sqrt{5} }{16}$

$\rm \: {cos}^{2}18\degree = \dfrac{10 + 2 \sqrt{5} }{16}$

$\rm \: {cos}18\degree = \sqrt{ \dfrac{10 + 2 \sqrt{5} }{16}}$

[ - ve sign rejected as cos18° > 0 ]

$\rm\implies \: \: \boxed{\sf{ \:\:cos18\degree \: = \: \dfrac{ \sqrt{10 + 2 \sqrt{5} } }{4} \: \: }}$

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Additional Information

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

$\boxed{\sf{ \:cos72\degree = \frac{ \sqrt{5} - 1}{4} \: }}$

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

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

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

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

Answer 2

Step-by-step explanation:

$\color{red} \tt\[ \begin{array}{l} \text { \color{lime}\tiny \boxed{ \dfrac{ \nwarrow}{ \swarrow} \dfrac{ \nearrow}{ \searrow}} } \text { since } \sin 18^{\circ}=\dfrac{\sqrt{5}-1}{4} \\ \\ \color{blue} \Rightarrow \cos ^{2} 18^{\circ}=1-\sin ^{2} 18^{\circ} \\ \\ \color{magenta} =1-\left[\dfrac{6-2 \sqrt{5}}{16}\right]=\dfrac{10+2 \sqrt{5}}{16} \\ \\ \color{olive} \Rightarrow \cos 18^{\circ}=\pm \sqrt{\dfrac{10+2 \sqrt{5}}{16}} \end{array} \]$

But $18^{\circ}$lies in tex]\sf 1 ^{st}[/tex] quadrant.

Therefore $\cos 18^{\circ}$ is positive

$\color{Turquoise} \tt\[ \Rightarrow \cos 18^{\circ}=\sqrt{\frac{10+2 \sqrt{5}}{16}} \\ \color{palegreen} \qquad \qquad \: \: \: \: =\frac{\sqrt{10+2 \sqrt{5}}}{4} \]$