Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 94862 by mathocean1 last updated on 21/May/20

$$\mathrm{Show}\mathrm{that}$$$${\cos }^{6}x+{sin}^{6}x=\frac{1}{8}(5+3cos\left(4x\right))$$$$\mathrm{By}\mathrm{using}\mathrm{the}\mathrm{formula}:$$$${\mathrm{a}}^3+{\mathrm{b}}^3=(\mathrm{a}+\mathrm{b})({\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2)$$

Commented by john santu last updated on 21/May/20

$${\left(\cos {}^2\mathrm{x}\right)}^3+{\left(\sin {}^2\mathrm{x}\right)}^3=$$$$1\times (\cos {}^4\mathrm{x}-\cos {}^2\mathrm{xsin}{}^2\mathrm{x}+\sin {}^4\mathrm{x})$$$$=\{{(\sin {}^2\mathrm{x}+\cos {}^2\mathrm{x})}^2-3{\left(\cos \mathrm{xsin}\mathrm{x}\right)}^2\}$$$$=1-3{\left(\frac{1}{2}\sin 2\mathrm{x}\right)}^2$$$$=1-\frac{3}{4}\sin {}^{2}2\mathrm{x}$$$$=1-\frac{3}{4}\{\frac{1}{2}-\frac{1}{2}\cos 4\mathrm{x}\}$$$$=1-\frac{3}{8}+\frac{3}{4}\cos 4\mathrm{x}$$$$=\frac{5}{8}+\frac{3}{4}\cos 4\mathrm{x}$$

Commented by mathocean1 last updated on 21/May/20

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}!$$

Commented by niroj last updated on 21/May/20

����

Commented by i jagooll last updated on 21/May/20

��������

Terms of Service

Privacy Policy

Contact: info@tinkutara.com