Rory Moran

2023-02-22

Find the value of $\mathrm{cos}{18}^{\circ }$

Samantha Summers

Find the value of $\mathrm{cos}18°$using multiple angle formulae of trigonometric ratios
Let angle $A=18°$
Thus, $5A=90°$
$⇒2A+3A=90˚$
$⇒$$2A=90˚-3A$
Taking sin e on both sides, we get
$\mathrm{sin}2A=\mathrm{sin}\left(90˚-3A\right)=\mathrm{cos}3A$
$⇒$$2\mathrm{sin}A×\mathrm{cos}A=4{\mathrm{cos}}^{3}A-3\mathrm{cos}A$
$⇒2\mathrm{sin}A×\mathrm{cos}A-4{\mathrm{cos}}^{3}A+3\mathrm{cos}A=0$
$⇒$$\mathrm{cos}A\left(2\mathrm{sin}A-4{\mathrm{cos}}^{2}A+3\right)=0$
Dividing both sides by$\mathrm{cos}A=\mathrm{cos}18˚\ne 0$, we get
$2\mathrm{sin}A-4\left(1-{\mathrm{sin}}^{2}A\right)+3=0$
$⇒4{\mathrm{sin}}^{2}A+2\mathrm{sin}A-1=0$, which is a quadratic equation in $\mathrm{sin}A$
$⇒\mathrm{sin}A=\frac{-2±{\left(2\right)}^{2}-4\left(4\right)\left(-1\right)}{2\left(4\right)}$
$=\frac{-2±16+4}{8}=\frac{-2±25}{8}$
$⇒$$\mathrm{sin}A=\frac{-1±5}{4}$
Now the value of $\mathrm{cos}{18}^{\circ }$ is
$⇒$$\mathrm{cos}{18}^{\circ }=1-{\mathrm{sin}}^{2}{18}^{\circ }$
$=1-{\left(\frac{5-1}{4}\right)}^{2}=1-\left(\frac{5+1-25}{16}\right)=\frac{10+25}{16}$
$⇒$$\mathrm{cos}{18}^{\circ }=\frac{10}{+}$
Thus, the value of $\mathrm{cos}{18}^{\circ }=\frac{10}{+}$

Do you have a similar question?