Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 39588 by Rio Mike last updated on 08/Jul/18

if cos A= (3/5) and tan B = ((12)/5)  where A and B are reflex angles  find without using tables,the  value of  a) sin (A − B) b) tan(A−B)  c) cos (A + B).

$${if}\:{cos}\:{A}=\:\frac{\mathrm{3}}{\mathrm{5}}\:{and}\:{tan}\:{B}\:=\:\frac{\mathrm{12}}{\mathrm{5}} \\ $$$${where}\:{A}\:{and}\:{B}\:{are}\:{reflex}\:{angles} \\ $$$${find}\:{without}\:{using}\:{tables},{the} \\ $$$${value}\:{of} \\ $$$$\left.{a}\left.\right)\:{sin}\:\left({A}\:−\:{B}\right)\:{b}\right)\:{tan}\left({A}−{B}\right) \\ $$$$\left.{c}\right)\:{cos}\:\left({A}\:+\:{B}\right). \\ $$

Answered by MJS last updated on 08/Jul/18

I see two Pythagorean triples  3^2 +4^2 =5^2  and 5^2 +12^2 =13^2   cos α=(3/5) ⇒ sin α=(4/5) ⇒ tan α=(4/3)  tan β=((12)/5) ⇒ cos β=(5/(13)) ⇒ sin β=((12)/(13))    sin(α−β)=sin α cos β −cos α sin β=       =(4/5)×(5/(13))−(3/5)×((12)/(13))=−((16)/(65))  tan(α−β)=((tan α −tan β)/(1+tan α tan β))=       =(((4/3)−((12)/5))/(1+(4/3)×((12)/5)))=−((16)/(63))  cos(α+β)=cos α cos β −sin α sin β=       =(3/5)×(5/(13))−(4/5)×((12)/(13))=−((33)/(65))

$$\mathrm{I}\:\mathrm{see}\:\mathrm{two}\:\mathrm{Pythagorean}\:\mathrm{triples} \\ $$$$\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} =\mathrm{13}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{3}}{\mathrm{5}}\:\Rightarrow\:\mathrm{sin}\:\alpha=\frac{\mathrm{4}}{\mathrm{5}}\:\Rightarrow\:\mathrm{tan}\:\alpha=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{12}}{\mathrm{5}}\:\Rightarrow\:\mathrm{cos}\:\beta=\frac{\mathrm{5}}{\mathrm{13}}\:\Rightarrow\:\mathrm{sin}\:\beta=\frac{\mathrm{12}}{\mathrm{13}} \\ $$$$ \\ $$$$\mathrm{sin}\left(\alpha−\beta\right)=\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta\:−\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta= \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{5}}{\mathrm{13}}−\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{12}}{\mathrm{13}}=−\frac{\mathrm{16}}{\mathrm{65}} \\ $$$$\mathrm{tan}\left(\alpha−\beta\right)=\frac{\mathrm{tan}\:\alpha\:−\mathrm{tan}\:\beta}{\mathrm{1}+\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta}= \\ $$$$\:\:\:\:\:=\frac{\frac{\mathrm{4}}{\mathrm{3}}−\frac{\mathrm{12}}{\mathrm{5}}}{\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{12}}{\mathrm{5}}}=−\frac{\mathrm{16}}{\mathrm{63}} \\ $$$$\mathrm{cos}\left(\alpha+\beta\right)=\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta= \\ $$$$\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{5}}{\mathrm{13}}−\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{12}}{\mathrm{13}}=−\frac{\mathrm{33}}{\mathrm{65}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com