Question

Find six trigonometric functions of ${315}^{°}$

Answer 1

Find trigonometric ratios for the given expression

Given, ${315}^{°}$

1. $\sin \left({315}^{°}\right)=\sin \left({360}^{°}-{45}^{°}\right)=-\sin \left({45}^{°}\right)=-\frac{1}{\sqrt{2}};[\because \sin \left(-x\right)=-\sin \left(x\right)\mathrm{and}\sin \left({45}^{°}\right)=\frac{1}{\sqrt{2}}]$
2. $\cos \left({315}^{°}\right)=\cos \left({360}^{°}-{45}^{°}\right)=\cos \left({45}^{°}\right)=\frac{1}{\sqrt{2}};[\because \cos \left(-x\right)=\cos \left(x\right)\mathrm{and}\cos \left({45}^{°}\right)=-\frac{1}{\sqrt{2}}]$
3. $\tan \left({315}^{°}\right)=\tan \left({360}^{°}-{45}^{°}\right)=-\tan \left({45}^{°}\right)=-1;[\because \tan \left(-x\right)=-\tan \left(x\right)\mathrm{and}\tan \left({45}^{°}\right)=\frac{1}{2}]$
4. $\cot \left({315}^{°}\right)=\cot \left({360}^{°}-{45}^{°}\right)=-\cot \left({45}^{°}\right)=-1;\left[\because \cot \left(x\right)=\frac{1}{\tan \left(x\right)}\right]$
5. $\sec \left({315}^{°}\right)=\sqrt{2};[\because \sec (x)=\frac{1}{\cos \left(x\right)}]$
6. $\mathrm{cosec}\left({315}^{°}\right)=-\sqrt{2};[\because \mathrm{cosec}\left(x\right)=\frac{1}{\sin \left(x\right)}]$

Hence, six trigonometric functions of ${315}^{°}$ is $\sin \left({315}^{°}\right)=-\frac{1}{\sqrt{2}},\cos \left({315}^{°}\right)=\frac{1}{\sqrt{2}},\tan \left({315}^{°}\right)=-1,\cot \left({315}^{°}\right)=-1,\sec \left({315}^{°}\right)=\sqrt{2}$and $\mathrm{cosec}\left({315}^{°}\right)=-\sqrt{2}$.