Trigonometric Identities Question 140
Question: If $ \sin 18^{0}=\frac{\sqrt{5}-1}{4}, $ then what is the value of $ \sin 18{}^\circ $ ?
Options:
A) $ \frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4} $
B) $ \frac{\sqrt{3+\sqrt{5}}+\sqrt{5+\sqrt{5}}}{4} $
C) $ \frac{\sqrt{3-\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4} $
D) $ \frac{\sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}}}{4} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \because \sin 18{}^\circ =\frac{\sqrt{5}-1}{4} $ $ x^{2}=4^{2}-{{( \sqrt{5}-1 )}^{2}} $
$ \Rightarrow x=\sqrt{10+2\sqrt{5}} $
$ \Rightarrow \cos 18{}^\circ =\frac{\sqrt{10+2\sqrt{5}}}{4} $
$ \Rightarrow ,2,{{\cos }^{2}}9-1=\frac{\sqrt{10+2\sqrt{5}}}{4} $ $ {{\cos }^{2}}9=\frac{\sqrt{10+2\sqrt{5}}+4}{8} $
$ \Rightarrow ,{{\sin }^{2}}81=\frac{4+\sqrt{10+2\sqrt{5}}}{8} $ After squaring all the options available, we come to a conclusion that option is correct.
BETA
