Question Number 84469 by jagoll last updated on 13/Mar/20
$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{sin}\:\mathrm{3b}\:+\:\left(\mathrm{cos}\:\mathrm{b}+\mathrm{sin}\:\mathrm{b}\right)\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{2b}\right) \\ $$$$=\:\mathrm{cos}\:\mathrm{3b} \\ $$
Answered by som(math1967) last updated on 13/Mar/20
$${sin}\mathrm{3}{b}+{cosb}−\mathrm{2}{sin}\mathrm{2}{bcosb}+\mathrm{sin}\:{b}−\mathrm{2sin}\:{b}\mathrm{sin}\:\mathrm{2}{b} \\ $$$$={sin}\mathrm{3}{b}+{cosb}−\left({sin}\mathrm{3}{b}+{sinb}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{sinb}\:−\left({cosb}−{cos}\mathrm{3}{b}\right)\bigstar \\ $$$$={cos}\mathrm{3}{b} \\ $$$$\bigstar\mathrm{2}{sinAcosB}={sin}\left({A}+{B}\right)+{sin}\left({A}−{B}\right) \\ $$$$\mathrm{2}{sinAsinB}={cos}\left({A}−{B}\right)−\mathrm{cos}\:\left({A}+{B}\right) \\ $$
Commented by jagoll last updated on 13/Mar/20
$$\mathrm{thank}\:\mathrm{you}.\:\mathrm{but}\:\mathrm{typo}\:\mathrm{sir} \\ $$$$\mathrm{2sin}\:\mathrm{Asin}\:\mathrm{B}=\:\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)−\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right) \\ $$
Commented by som(math1967) last updated on 13/Mar/20
$${yes}\:{i}\:{fix}\:{it} \\ $$