Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 211989 by Spillover last updated on 26/Sep/24

Answered by som(math1967) last updated on 26/Sep/24

$$cos\alpha cos\beta +sin\alpha sin\beta +cos\beta cos\gamma$$$$+sin\beta sin\gamma +cos\gamma cos\alpha +sin\gamma sin\alpha$$$$=-\frac{3}{2}$$$$2(cos\alpha cos\beta +cos\beta cos\gamma +cos\gamma cos\alpha )$$$$+2(sin\alpha sin\beta +sin\beta sin\gamma +sin\alpha sin\gamma )$$$$=-1-1-1$$$${cos}^2\alpha +{cos}^2\beta +{cos}^2\gamma$$$$+2(cos\alpha cos\beta +cos\beta cos\gamma +cos\gamma cos\alpha )$$$$+{sin}^2\alpha +{sin}^2\beta +{sin}^2\gamma$$$$+2(sin\alpha sin\beta +sin\beta sin\gamma +sin\alpha sin\gamma )=0$$$${(cos\alpha +cos\beta +cos\gamma )}^2+{(sin\alpha +sin\beta +sin\gamma )}^2$$$$=0$$$$\therefore cos\alpha +cos\beta +cos\gamma =sin\alpha +sin\beta +sin\gamma )=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com