Q61 / 160 · TS EAMCET 2019 (04 May Shift 1) · Mathematics

$\cos 18^{\circ}=$

  • A
    $\frac{1}{8}(5-\sqrt{5})$
  • Correct answer
    B
    $\frac{1}{2 \sqrt{2}} \sqrt{5+\sqrt{5}}$
  • C
    $\frac{\sqrt{5}-1}{4}$
  • D
    $\frac{\sqrt{5}+1}{4}$

Solution

We know that, $$ \begin{aligned} \operatorname{sinl} 8^{\circ} & =\frac{\sqrt{5}-1}{4} \\ \operatorname{cosl} 8^{\circ} & =\sqrt{1-\sin ^2 18^{\circ}} \\ & =\sqrt{1-\left(\frac{\sqrt{5}-1}{4}\right)^2}=\sqrt{\frac{16-5-1+2 \sqrt{5}}{16}} \\ \Rightarrow \quad \cos 18^{\circ} & =\sqrt{\frac{10+2 \sqrt{5}}{16}}=\sqrt{\frac{5+\sqrt{5}}{8}}=\frac{\sqrt{5+\sqrt{5}}}{2 \sqrt{2}} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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