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Question Number 135926 by liberty last updated on 17/Mar/21

$$ \\ $$ If cos(A+B) =1/2, sinA=1/√2, then what is cos (A - B)?\\n

Answered by benjo_mathlover last updated on 17/Mar/21

(1) cos (A+B)=(1/2)=cos (π/3) ⇒A+B= (π/3)+2nπ ⇒B = (π/3)−A+2nπ ⇒cos (A−B)=cos (A−(π/3)+A−2nπ) cos (2nπ+(π/3)−2A)= cos ((π/3)−2A) = cos (π/3).cos 2A+sin (π/3).sin 2A =(1/2)(1−2sin ^2 A)+ ((√3)/2).(√(1−cos ^2 2A)) =(1/2)(1−2.(1/2))+((√3)/2)(√(1−(1−2sin ^2 A)^2 )) = 0+((√3)/2).(√(1−(1−2.(1/2))^2 )) = ((√3)/2)

$$\left(\mathrm{1}\right)\:\mathrm{cos}\:\left({A}+{B}\right)=\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{cos}\:\frac{\pi}{\mathrm{3}} \\ $$ $$\Rightarrow{A}+{B}=\:\frac{\pi}{\mathrm{3}}+\mathrm{2}{n}\pi \\ $$ $$\Rightarrow{B}\:=\:\frac{\pi}{\mathrm{3}}−{A}+\mathrm{2}{n}\pi \\ $$ $$\color{mathred}{\Rightarrow}\mathrm{\color{mathred}{c}\color{mathred}{o}\color{mathred}{s}}\color{mathred}{\:}\color{mathred}{\left(}{\color{mathred}{A}}\color{mathred}{−}{\color{mathred}{B}}\color{mathred}{\right)}\color{mathred}{=}\mathrm{\color{mathred}{c}\color{mathred}{o}\color{mathred}{s}}\color{mathred}{\:}\color{mathred}{\left(}{\color{mathred}{A}}\color{mathred}{−}\frac{\color{mathred}{\pi}}{\mathrm{\color{mathred}{3}}}\color{mathred}{+}{\color{mathred}{A}}\color{mathred}{−}\mathrm{\color{mathred}{2}}{\color{mathred}{n}}\color{mathred}{\pi}\color{mathred}{\right)} \\ $$ $$\mathrm{\color{mathred}{c}\color{mathred}{o}\color{mathred}{s}}\color{mathred}{\:}\color{mathred}{\left(}\mathrm{\color{mathred}{2}}{\color{mathred}{n}}\color{mathred}{\pi}\color{mathred}{+}\frac{\color{mathred}{\pi}}{\mathrm{\color{mathred}{3}}}−\mathrm{\color{mathred}{2}}{\color{mathred}{A}}\color{mathred}{\right)}\color{mathred}{=}\color{mathred}{\:}\mathrm{\color{mathred}{c}\color{mathred}{o}\color{mathred}{s}}\color{mathred}{\:}\color{mathred}{\left(}\frac{\color{mathred}{\pi}}{\mathrm{\color{mathred}{3}}}−\mathrm{\color{mathred}{2}}{\color{mathred}{A}}\color{mathred}{\right)} \\ $$ $$\color{mathred}{=}\color{mathred}{\:}\mathrm{\color{mathred}{c}\color{mathred}{o}\color{mathred}{s}}\color{mathred}{\:}\frac{\color{mathred}{\pi}}{\mathrm{\color{mathred}{3}}}\color{mathred}{.}\mathrm{\color{mathred}{c}\color{mathred}{o}\color{mathred}{s}}\color{mathred}{\:}\mathrm{\color{mathred}{2}}{\color{mathred}{A}}+\mathrm{\color{mathred}{s}\color{mathred}{i}\color{mathred}{n}}\color{mathred}{\:}\frac{\color{mathred}{\pi}}{\mathrm{\color{mathred}{3}}}\color{mathred}{.}\mathrm{\color{mathred}{s}\color{mathred}{i}\color{mathred}{n}}\color{mathred}{\:}\mathrm{\color{mathred}{2}}{\color{mathred}{A}} \\ $$ $$\color{mathred}{=}\frac{\mathrm{\color{mathred}{1}}}{\mathrm{\color{mathred}{2}}}\color{mathred}{\left(}\mathrm{\color{mathred}{1}}\color{mathred}{−}\mathrm{\color{mathred}{2}\color{mathred}{s}\color{mathred}{i}\color{mathred}{n}}\color{mathred}{\:}^{\mathrm{\color{mathred}{2}}} {\color{mathred}{A}}\color{mathred}{\right)}+\:\frac{\sqrt{\mathrm{\color{mathred}{3}}}}{\mathrm{\color{mathred}{2}}}\color{mathred}{.}\sqrt{\mathrm{\color{mathred}{1}}\color{mathred}{−}\mathrm{\color{mathred}{c}\color{mathred}{o}\color{mathred}{s}}\color{mathred}{\:}^{\mathrm{\color{mathred}{2}}} \mathrm{\color{mathred}{2}}{\color{mathred}{A}}} \\ $$ $$\color{mathred}{=}\frac{\mathrm{\color{mathred}{1}}}{\mathrm{\color{mathred}{2}}}\color{mathred}{\left(}\mathrm{\color{mathred}{1}}\color{mathred}{−}\mathrm{\color{mathred}{2}}\color{mathred}{.}\frac{\mathrm{\color{mathred}{1}}}{\mathrm{\color{mathred}{2}}}\color{mathred}{\right)}+\frac{\sqrt{\mathrm{\color{mathred}{3}}}}{\mathrm{\color{mathred}{2}}}\sqrt{\mathrm{\color{mathred}{1}}\color{mathred}{−}\color{mathred}{\left(}\mathrm{\color{mathred}{1}}\color{mathred}{−}\mathrm{\color{mathred}{2}\color{mathred}{s}\color{mathred}{i}\color{mathred}{n}}\color{mathred}{\:}^{\mathrm{\color{mathred}{2}}} {\color{mathred}{A}}\color{mathred}{\right)}^{\mathrm{\color{mathred}{2}}} } \\ $$ $$\color{mathred}{=}\color{mathred}{\:}\mathrm{\color{mathred}{0}}+\frac{\sqrt{\mathrm{\color{mathred}{3}}}}{\mathrm{\color{mathred}{2}}}\color{mathred}{.}\sqrt{\mathrm{\color{mathred}{1}}\color{mathred}{−}\color{mathred}{\left(}\mathrm{\color{mathred}{1}}\color{mathred}{−}\mathrm{\color{mathred}{2}}\color{mathred}{.}\frac{\mathrm{\color{mathred}{1}}}{\mathrm{\color{mathred}{2}}}\color{mathred}{\right)}^{\mathrm{\color{mathred}{2}}} } \\ $$ $$\color{mathred}{=}\color{mathred}{\:}\frac{\sqrt{\mathrm{\color{mathred}{3}}}}{\mathrm{\color{mathred}{2}}} \\ $$

Commented bymr W last updated on 17/Mar/21

−((√3)/2) is also possible.

$$−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{is}\:{also}\:{possible}. \\ $$

Answered by mr W last updated on 17/Mar/21

sin (A+B)=±(√(1−((1/2))^2 ))=±((√3)/2) cos A=±(√(1−((1/( (√2))))^2 ))=±(1/( (√2))) cos (A−B)=cos [2A−(A+B)] =cos 2A cos (A+B)+sin 2A sin (A+B) =(1−2 sin^2 A)cos (A+B)+2 sin A cos A sin (A+B) =(1−2×(1/2))×(1/2)+2×(1/( (√2)))×(±(1/( (√2))))×(±((√3)/2)) =±((√3)/2)

$$\mathrm{sin}\:\left({A}+{B}\right)=\pm\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$ $$\mathrm{cos}\:{A}=\pm\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$ $$\mathrm{cos}\:\left({A}−{B}\right)=\mathrm{cos}\:\left[\mathrm{2}{A}−\left({A}+{B}\right)\right] \\ $$ $$=\mathrm{cos}\:\mathrm{2}{A}\:\mathrm{cos}\:\left({A}+{B}\right)+\mathrm{sin}\:\mathrm{2}{A}\:\mathrm{sin}\:\left({A}+{B}\right) \\ $$ $$=\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} {A}\right)\mathrm{cos}\:\left({A}+{B}\right)+\mathrm{2}\:\mathrm{sin}\:{A}\:\mathrm{cos}\:{A}\:\mathrm{sin}\:\left({A}+{B}\right) \\ $$ $$=\left(\mathrm{1}−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\right)×\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\left(\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)×\left(\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$ $$=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

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