Eleven Standard >> Trigonometric ratios of angle 18 degrees

Value of \(sin18^{0}\), \(cos18^{0}\), \(tan18^{0}\)

Let \(\theta=18^{o}\)
\(\Rightarrow\) \(5\theta=90^{o}\)
\(\Rightarrow\) \(2\theta=90^{o}-3\theta\) 
\(\Rightarrow\) \(\sin2\theta=\sin(90^{o}-3\theta)\)
\(\Rightarrow\) \(2\sin\theta \cos\theta=4\cos^{3}\theta-3\cos\theta)\)
\(\Rightarrow\) \(2\sin\theta=4\cos^{2}\theta-3)\)
\(\Rightarrow\) \(2\sin\theta=4(1-\sin^{2}\theta)-3)\)
\(\Rightarrow\) \(4\sin^{2}\theta+2\sin\theta-1=0\)
\(\therefore\) \(\sin\theta=\frac{-2\pm\sqrt{4-4.4.(-1)}}{2.4}\)
\(\therefore\) \(\sin\theta=\frac{-2\pm\sqrt{20}}{8}\)
         =\(\frac{-1\pm\sqrt{5}}{4}\)

As \(\sin\theta\)=\(\sin18^{o}\)>0
      \(\sin18^{o}\)=\(\frac{\sqrt{5}-1}{4}\)

 \(\cos18^{o}\)=\(\sqrt{1-\sin^{2}18^{o}}\)
          = \(\sqrt{1-\left(\frac{\sqrt{5}-1}{4}\right)^{2}}\)
          = \(\frac{1}{4}\sqrt{16-(5+1-2\sqrt{5})}\)
          = \(\frac{1}{4}\sqrt{10+2\sqrt{5}}\)

 \(\tan18^{o}\)=\(\frac{\sin18^{o}}{\cos18^{o}}\)
           =\(\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\)
             [\(\because\) \(\sin18^{o}=\frac{\sqrt{5}-1}{4}\) and \(\cos18^{o}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)]