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Mathematics

Eleven Standard >> Trigonometric ratios of angle 54 degrees

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Derive the value of \(\sin54^{o}\), \(\cos54^{o}\), \(\tan54^{o}\)

 

\(\sin54^{o}=\sin(90^{o}-36^{o})=\cos36^{o}\)
        =\(\frac{\sqrt{5}+1}{4}\)
Thus \(\sin54^{o}=\frac{\sqrt{5}+1}{4}\)

\(\cos54^{o}=\sqrt{1-\sin^{2}54^{o}}\)
        =\(\sqrt{1-\left(\frac{\sqrt{5}+1}{4}\right)^{2}}\)
        =\(\frac{1}{4}\sqrt{16-5-2\sqrt{5}-1}\)
        =\(\frac{1}{4}\sqrt{10-2\sqrt{5}}\)

\(\tan54^{o}\)=\(\frac{\sin54^{o}}{\cos54^{o}}\)
         =\(\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\)
       \([\) \(\because\) \(\sin54^{o}=\frac{\sqrt{5}+1}{4}\) and \(\cos54^{o}=\frac{1}{4}\sqrt{10-2\sqrt{5}}\)\(]\)

\(\sin72^{o}=\sin\left(90^{o}-18^{o}\right)\)
         =\(\cos18^{o}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\)

\(\cos72^{o}=\cos\left(90^{o}-18^{o}\right)\)
         =\(\sin18^{o}=\frac{\sqrt{5}-1}{4}\)

\(\tan72^{o}=\frac{\sin72^{o}}{\cos72^{o}}\)
         =\(\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}\)
         \( [\) \(\because\) \(\sin72^{o}=\frac{1}{4}\sqrt{10+2\sqrt{5}}\) and 
                     \(\cos72^{o}=\frac{\sqrt{5}-1}{4}\)\(]\)

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