Question

The value of $\sin {27}^{\circ }-\cos {27}^{\circ }$ is equal to

• A. $-\frac{\sqrt{3-\sqrt{5}}}{2}$
• B. $-\frac{\sqrt{5-\sqrt{5}}}{2}$
• C. $-\frac{\sqrt{5}-1}{2\sqrt{2}}$
• D. $\frac{\sqrt{3-\sqrt{5}}}{2}$

Answer 1

The correct option is B $-\frac{\sqrt{5-\sqrt{5}}}{2}$

$\sin {27}^0-\cos {27}^0$

$\Rightarrow \sin {27}^0-\cos ({90}^0-{63}^0)$

$\Rightarrow \sin {27}^0-\sin {63}^0$

$\Rightarrow 2\cos \left(\frac{{27}^0+{63}^0}{2}\right)\sin \left(\frac{{27}^0-{63}^0}{2}\right)$

$\Rightarrow 2\cos {45}^0\sin (-{36}^0)$

$\Rightarrow 2\times \frac{1}{\sqrt{2}}\times -\left[\frac{\sqrt{10-2\sqrt{5}}}{4}\right]$

$\Rightarrow -\frac{1}{\sqrt{2}}\times \sqrt{2}\left(\frac{\sqrt{5-\sqrt{5}}}{2}\right)$

$\Rightarrow -\frac{\sqrt{5-\sqrt{5}}}{2}$

Hence, this is the answer

option $\left(B\right)$ is correct.