Question and Answers Forum

All Questions      Topic List

Vector Calculus Questions

Previous in All Question      Next in All Question      

Previous in Vector Calculus      Next in Vector Calculus      

Question Number 201947 by Mingma last updated on 16/Dec/23

Answered by Frix last updated on 18/Dec/23

$$\sin \alpha \sin \beta \cos (\pi -\alpha -\beta )+$$$$+\sin \beta \sin (\pi -\alpha -\beta )\cos \alpha +$$$$+\sin (\pi -\alpha -\beta )\sin \alpha \cos \beta =$$$$=\frac{3-(\cos 2\alpha +\cos 2\beta +\cos 2(\alpha +\beta )}{4}=$$$$[\mathrm{Let}\alpha =u-v\wedge \beta =u+v]$$$$=\frac{3-(\cos 4u+\cos 2(u-v)+\cos 2(u-v)}{4}$$$$\frac{d}{dv}\left[\frac{3-(\cos 4u+\cos 2(u-v)+\cos 2(u-v)}{4}\right]=0$$$$\frac{\sin 2(u+v)-\sin 2(u-v)}{2}=0$$$$\sin 2(u+v)=\sin 2(u-v)$$$$0⩽v<\frac{\pi }{2}\Rightarrow v=0\Rightarrow$$$$\frac{3-(\cos 4u+\cos 2(u-v)+\cos 2(u-v)}{4}=$$$$=\frac{3-(\cos 4u+2\cos 2u)}{4}$$$$\frac{d}{du}\left[\frac{3-(\cos 4u+2\cos 2u)}{4}\right]=0$$$$\sin 4u+\sin 2u=0$$$$-2\cos u\sin u(1-2\cos u)(1+2\cos u)=0$$$$0<u⩽\frac{\pi }{2}\Rightarrow u=\frac{\pi }{3}[\vee u=\frac{\pi }{2}\Rightarrow \alpha =\beta =\frac{\pi }{2}\mathrm{impossible}]$$$$\Rightarrow \alpha =\beta =\frac{\pi }{3}\Rightarrow$$$$\max \left(\frac{3-(\cos 2\alpha +\cos 2\beta +\cos 2(\alpha +\beta )}{4}\right)=\frac{9}{8}$$

Commented by Mingma last updated on 18/Dec/23

Perfect ��

Terms of Service

Privacy Policy

Contact: info@tinkutara.com