A) \[\frac{\sqrt{5}-1}{4}\]
B) \[\frac{\sqrt{10+2\sqrt{5}}}{2}\]
C) \[\frac{\sqrt{5}+1}{4}\]
D) \[\frac{\sqrt{10-2\sqrt{5}}}{4}\]
Correct Answer: C
Solution :
Let \[\theta =36{}^\circ \] \[\therefore \]\[5\theta =180{}^\circ \] \[\therefore \]\[2\theta +3\theta =180{}^\circ \Rightarrow 2\theta =(180{}^\circ -3\theta )\] \[\sin 2\theta =\sin (180{}^\circ -3\theta )\] \[\Rightarrow \]\[2\sin \theta \cos \theta =\sin 3\theta \] \[\Rightarrow \]\[2\sin \theta \cos \theta =3\sin \theta -4{{\sin }^{3}}\theta \] \[\Rightarrow \]\[2\cos \theta =3-4{{\sin }^{2}}\theta \] \[\Rightarrow \]\[2\cos \theta =3-4(1-{{\cos }^{2}}\theta )\] \[\Rightarrow \]\[2\cos \theta =3-4+4{{\cos }^{2}}\theta \] \[\Rightarrow \]\[4{{\cos }^{2}}\theta -2\cos \theta -1=0\] \[\therefore \]\[\cos \theta =\frac{2\pm \sqrt{4+16}}{8}=\frac{1\pm \sqrt{5}}{4}\] Again, as \[\theta =36{}^\circ \]and in 1st quadrant, so \[\cos \theta \]is positive. \[\Rightarrow \] \[\cos 36{}^\circ =\frac{\sqrt{5}+1}{4}\]
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