Question

Find six trigonometric functions of ${540}^{°}$

Answer 1

Find trigonometric ratios for the given expression:

Given, ${540}^{°}$

1. $\sin \left(540°+x\right)=\sin \left(3\mathrm{\pi }+\mathrm{x}\right)=-\sin \left(x\right)$ [$\therefore \sin \left(n\pi +x\right)=-\sin \left(x\right)$ for $n$odd and $540°=3·180°=3\pi$]
2. $\cos \left({540}^{°}+x\right)=\cos \left(3\pi +x\right)=-\cos \left(x\right)$[$\therefore \cos \left(n\pi +x\right)=-\cos \left(x\right)$ for $n$odd and $540°=3·180°=3\pi$]
3. $\tan \left(540°+x\right)=\tan \left(3\mathrm{\pi }+x\right)=\tan \left(x\right)$ [$\therefore \tan \left(n\pi +x\right)=\tan \left(x\right)$ for $n$odd and $540°=3·180°=3\pi$]
4. $\cot \left(540°+x\right)=\frac{1}{\tan \left(540°\right)}=\frac{1}{\tan \left(3\pi +x\right)}=\frac{1}{\tan \left(x\right)}=\cot \left(x\right)$[$\because \cot \left(x\right)=\frac{1}{\tan \left(x\right)}$,$,\tan \left(n\pi +x\right)=\tan \left(x\right)$ for $n$odd and $540°=3·180°=3\pi$]
5. $\sec \left(540°+x\right)=\frac{1}{\cos \left(540°+x\right)}=-\frac{1}{\cos \left(x\right)}=-\sec \left(x\right)$[$\because \sec \left(x\right)=\frac{1}{\cos \left(x\right)}$,$\cos \left(n\pi +x\right)=-\cos \left(x\right)$ for $n$odd and $540°=3·180°=3\pi$]
6. $\mathrm{cosec}\left(540°+x\right)=\frac{1}{\sin \left(540°+x\right)}=-\frac{1}{\sin \left(x\right)}=-\mathrm{cosec}\left(x\right)$ [$\because \mathrm{cosec}\left(x\right)=\frac{1}{\sin \left(x\right)},$$\sin \left(n\pi +x\right)=-\sin \left(x\right)$ for $n$odd and $540°=3·180°=3\pi$]

Hence, six trigonometric functions of ${540}^{°}$ is $\sin \left(540°+x\right)=-\sin \left(x\right),\cos \left(540°+x\right)=-\cos \left(x\right),\tan \left(540°+x\right)=\tan \left(x\right),\cot (540°+x)=\cot \left(x\right),\sec (540°+x)=-\sec \left(x\right)$and $\mathrm{cosec}(540°+x)=-\mathrm{cosec}\left(x\right)$.