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Question Number 7514 by Tawakalitu. last updated on 01/Sep/16

Answered by Yozzia last updated on 01/Sep/16

$$sin\left(\pi cos\alpha \right)=cos\left(\pi sin\alpha \right).$$$$Usingsina=cos(\frac{\pi }{2}-a),weget$$$$cos(\frac{\pi }{2}-\pi cos\alpha )=cos\left(\pi sin\alpha \right)$$$$\Rightarrow \frac{\pi }{2}-\pi cos\alpha =2n\pi \pm \pi sin\alpha (n\in \mathbb{Z})$$$$\mp \pi sin\alpha -\pi cos\alpha =2n\pi -\frac{\pi }{2}$$$$\mp sin\alpha -cos\alpha =2n-\frac{1}{2}$$$$------------------------------$$$$Supposewewrite+inplaceof\mp .$$$$\therefore sin\alpha -cos\alpha =\frac{4n-1}{2}$$$$\sqrt{2}sin(\alpha -\frac{\pi }{4})=\frac{4n-1}{2}$$$$sin(\alpha -\frac{\pi }{4})=\frac{4n-1}{2\sqrt{2}}$$$$Since\mid sin(\alpha -\frac{\pi }{4})\mid ⩽1$$$$\Rightarrow \mid \frac{4n-1}{2\sqrt{2}}\mid ⩽1\Rightarrow \mid 4n-1\mid ⩽2\sqrt{2}<4$$$$\Rightarrow -4<4n-1<4$$$$-3<4n<5$$$$-\frac{3}{4}<n<\frac{5}{4}forn\in \mathbb{Z}\Rightarrow n=0or1.$$$$Forn=0,sin(\alpha -\frac{\pi }{4})=\frac{-1}{2\sqrt{2}}$$$$\Rightarrow {sin}^2(\alpha -\frac{\pi }{4})=\frac{1}{8}$$$$\Rightarrow \frac{1-cos(2\alpha -\frac{\pi }{2})}{2}=\frac{1}{8}$$$$\Rightarrow cos(-(2\alpha -\frac{\pi }{2}))=\frac{3}{4}$$$$\Rightarrow sin2\alpha =\frac{3}{4}$$$$\Rightarrow \alpha =\frac{1}{2}{sin}^{-1}\frac{3}{4}$$$$Supposen=1.\therefore sin(\alpha -0.25\pi )=\frac{3}{2\sqrt{2}}$$$$But,{sin}^2(\alpha -0.25\pi )=\frac{9}{8}>1.$$$$Hence,n\ne 1.$$$$-------------------------$$$$Supposewewrite-inplaceof\mp .$$$$Wemusthaven=0.$$$$\therefore -sin\alpha -cos\alpha =\frac{-1}{2}$$$$\sqrt{2}sin(\alpha +\frac{\pi }{4})=\frac{1}{2}$$$$\Rightarrow {sin}^2(\alpha +\frac{\pi }{4})=\frac{1}{8}$$$$\Rightarrow \frac{1-cos(2\alpha +\pi /2)}{2}=\frac{1}{8}$$$$\therefore cos(2\alpha +\frac{\pi }{2})=\frac{3}{4}$$$$-sin2\alpha =\frac{3}{4}$$$$\Rightarrow \alpha =\frac{-1}{2}{sin}^{-1}\frac{3}{4}$$$$---------------------$$$$\therefore \alpha =\pm \frac{1}{2}{sin}^{-1}\frac{3}{4}ifsin\left(\pi cos\alpha \right)=cos\left(\pi sin\alpha \right).$$$$Generally,\alpha =2n\pi \pm \frac{1}{2}{sin}^{-1}\frac{3}{4}since$$$$sinu=sin(u+2n\pi )andcosk=cos(k+2n\pi )$$$$foranyu,k\in \mathbb{R}.$$

Commented by Tawakalitu. last updated on 01/Sep/16

$$Wow,thankssomuch.$$

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