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 32159 by rahul 19 last updated on 20/Mar/18

Express the following in a+ib form:  (((cos x+isin x)(cos y+isin y))/((cosa+isin a)(cosb+isinb))).

$$\boldsymbol{{E}}{xpress}\:{the}\:{following}\:{in}\:{a}+{ib}\:{form}: \\ $$$$\frac{\left(\mathrm{cos}\:{x}+{i}\mathrm{sin}\:{x}\right)\left(\mathrm{cos}\:{y}+{i}\mathrm{sin}\:{y}\right)}{\left({cosa}+{i}\mathrm{sin}\:{a}\right)\left({cosb}+{isinb}\right)}. \\ $$

Commented by abdo imad last updated on 20/Mar/18

E =((e^(ix)  e^(iy) )/(e^(ia)  e^(ib) )) =e^(i(x+y))  e^(−i(a+b)) =e^(i(x+y−a−b))   =cos(x+y−a−b) +i sin(x+y −a −b) .

$${E}\:=\frac{{e}^{{ix}} \:{e}^{{iy}} }{{e}^{{ia}} \:{e}^{{ib}} }\:={e}^{{i}\left({x}+{y}\right)} \:{e}^{−{i}\left({a}+{b}\right)} ={e}^{{i}\left({x}+{y}−{a}−{b}\right)} \\ $$$$={cos}\left({x}+{y}−{a}−{b}\right)\:+{i}\:{sin}\left({x}+{y}\:−{a}\:−{b}\right)\:. \\ $$

Commented by rahul 19 last updated on 20/Mar/18

thank you !

$${thank}\:{you}\:! \\ $$

Answered by Tinkutara last updated on 20/Mar/18

=((e^(ix) e^(iy) )/(e^(ia) e^(ib) ))=e^(i(x+y−a−b))   =cos (x+y−a−b)+isin (x+y−a−b)

$$=\frac{{e}^{{ix}} {e}^{{iy}} }{{e}^{{ia}} {e}^{{ib}} }={e}^{{i}\left({x}+{y}−{a}−{b}\right)} \\ $$$$=\mathrm{cos}\:\left({x}+{y}−{a}−{b}\right)+{i}\mathrm{sin}\:\left({x}+{y}−{a}−{b}\right) \\ $$

Commented by rahul 19 last updated on 20/Mar/18

thank you !

$${thank}\:{you}\:! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com