Trigonometric Ratio of some Specific Angles

Mathematics Class XI Chapter 3: Trigonometric Functions Trigonometric Ratio of some Specific Angles

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(For $\sin \text { and } \cos$ )

(i) $\sin 9^o =\sin \dfrac{\pi}{20}=\dfrac{1}{4}\left\{ \sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}} \right\}=\cos \dfrac{9 \pi}{20}=\cos 81^o$

(ii) $\cos 9^o=\cos \dfrac{\pi}{20}=\dfrac{1}{4}\left\{ \sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}} \right\}=\sin \dfrac{9 \pi}{20}=\sin 81^o$

(iii) $\sin 15^o =\sin \dfrac{\pi}{12}=\dfrac{\sqrt{6}-\sqrt{2}}{4}=\cos 75^o=\cos \dfrac{5 \pi}{12}$

(iv) $\cos 15^o=\cos \dfrac{\pi}{12}=\dfrac{\sqrt{6}+\sqrt{2}}{4}=\sin \dfrac{5 \pi}{12}=\sin 75^o$

(v) $\sin 18^o =\sin \dfrac{\pi}{10}=\dfrac{\sqrt{5}-1}{4}=\cos 72^o=\cos \dfrac{2 \pi}{5}$

(vi) $\cos 18^o =\cos \dfrac{\pi}{10}=\dfrac{\sqrt{10+2 \sqrt{5}}}{4}=\sin 72^o=\sin \dfrac{2 \pi}{5}$

(vii) $\sin 36^o =\sin \dfrac{\pi}{5}=\dfrac{\sqrt{10-2 \sqrt{5}}}{4}=\cos 54^o=\cos \dfrac{3 \pi}{10}$

(viii) $\cos 36^o =\cos \dfrac{\pi}{5}=\dfrac{\sqrt{5}+1}{4}-\sin 54^o=\sin \dfrac{3 \pi}{10}$

(For $\tan \text { and } \cot$ )

(i) $\tan 15^o=\tan \dfrac{\pi}{12}=2-\sqrt{3}=\cot \dfrac{5 \pi}{12}=\cot 75^o$

(ii) $\cot 15^o=\cot \dfrac{\pi}{12}-2 + \sqrt{3}=\tan \dfrac{5 \pi}{12}=\tan 75^o$

(iii) $\tan 18^o=\tan \dfrac{\pi}{10}=\sqrt{\dfrac{5-2\sqrt{5}}{5}}=\cot \dfrac{2\pi}{5}=\cot 72^o$

(iv) $\cot 18^o=\cot \dfrac{\pi}{10}=\sqrt{5+2 \sqrt{5}}=\tan \dfrac{2 \pi}{5}=\tan 72^o$

(v) $\tan 36^o=\tan \dfrac{\pi}{5}=\dfrac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}+1}=\cot 54^o=\cot \dfrac{3 \pi}{10}$

(vi) $\cot 36^o=\cot \dfrac{\pi}{5}=\dfrac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}=\tan 54^o=\tan \dfrac{3 \pi}{10}$

(vii) $\tan 9^o=\tan \dfrac{\pi}{20}=\dfrac{\sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}}}{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}=\cot 81^o=\cot \dfrac{9 \pi}{20}$

(viii) $\cot 9^o=\cot \dfrac{\pi}{20}=\dfrac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{\sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}}}=\tan 81^o=\tan \dfrac{9 \pi}{20}$