Question

The value of $\sin {27}^o-\cos {27}^o$ is equal to ? ; where $\cos \theta =\frac{\sqrt{5}+1}{4}$

• 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

Answer: (D)

$\frac{\sqrt{3-\sqrt{5}}}{2}$

We know that

${(\sin {27}^{\circ }-\cos {27}^{\circ })}^2={\sin }^2{27}^{\circ }+{\cos }^2{27}^{\circ }-2\sin {27}^{\circ }\cos {27}^{\circ }$

$=1-\sin 2\times {27}^{\circ }$

$=1-\sin {54}^{\circ }$

$=1-\sin ({90}^{\circ }-{36}^{\circ })$

$=1-\cos {36}^{\circ }$

$=1-\frac{\sqrt{5}+1}{4}$

$=\frac{3-\sqrt{5}}{4}$

Hence,

$(\sin {27}^{\circ }-\cos {27}^{\circ })=\sqrt{\frac{3-\sqrt{5}}{4}}\Rightarrow \frac{\sqrt{3-\sqrt{5}}}{2}$

Hence, this is the answer.