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given-that-sinA-12-13-and-sinB-4-5-where-A-and-B-are-acute-angles-find-cos-A-B-and-sin-A-B-




Question Number 10837 by okhema last updated on 27/Feb/17
given that sinA=((12)/(13))and sinB=(4/5),  where A and B are acute angles,  find cos(A−B) and sin(A+B)
$${given}\:{that}\:{sinA}=\frac{\mathrm{12}}{\mathrm{13}}{and}\:{sinB}=\frac{\mathrm{4}}{\mathrm{5}}, \\ $$$${where}\:{A}\:{and}\:{B}\:{are}\:{acute}\:{angles}, \\ $$$${find}\:{cos}\left({A}−{B}\right)\:{and}\:{sin}\left({A}+{B}\right) \\ $$$$ \\ $$
Answered by ridwan balatif last updated on 27/Feb/17
sinA=12/13        sinB=4/5  cosA=5/13          cosB=3/5  cos(A−B)=cosA×cosB+sinA×sinB                           =((5/(13)))×((3/5))+(((12)/(13)))×((4/5))                           =((15)/(65))+((48)/(65))=((63)/(65))  sin(A+B)=sinA×cosB−sinB×cosA                          =((12)/(13))×(3/5)−(4/5)×(5/(13))                          =((36)/(65))−((20)/(65))=((16)/(65))
$$\mathrm{sinA}=\mathrm{12}/\mathrm{13}\:\:\:\:\:\:\:\:\mathrm{sinB}=\mathrm{4}/\mathrm{5} \\ $$$$\mathrm{cosA}=\mathrm{5}/\mathrm{13}\:\:\:\:\:\:\:\:\:\:\mathrm{cosB}=\mathrm{3}/\mathrm{5} \\ $$$$\mathrm{cos}\left(\mathrm{A}−\mathrm{B}\right)=\mathrm{cosA}×\mathrm{cosB}+\mathrm{sinA}×\mathrm{sinB} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{5}}{\mathrm{13}}\right)×\left(\frac{\mathrm{3}}{\mathrm{5}}\right)+\left(\frac{\mathrm{12}}{\mathrm{13}}\right)×\left(\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{15}}{\mathrm{65}}+\frac{\mathrm{48}}{\mathrm{65}}=\frac{\mathrm{63}}{\mathrm{65}} \\ $$$$\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{sinA}×\mathrm{cosB}−\mathrm{sinB}×\mathrm{cosA} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{12}}{\mathrm{13}}×\frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{5}}{\mathrm{13}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{36}}{\mathrm{65}}−\frac{\mathrm{20}}{\mathrm{65}}=\frac{\mathrm{16}}{\mathrm{65}} \\ $$$$ \\ $$$$ \\ $$

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