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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)?
$$ \\ $$If cos(A+B) =1/2, sinA=1/√2, then what is cos (A – B)?
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)
$$\left(\mathrm{1}\right)\:\mathrm{cos}\:\left({A}+{B}\right)=\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{cos}\:\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{A}+{B}=\:\frac{\pi}{\mathrm{3}}+\mathrm{2}{n}\pi \\ $$$$\Rightarrow{B}\:=\:\frac{\pi}{\mathrm{3}}−{A}+\mathrm{2}{n}\pi \\ $$$$\Rightarrow\mathrm{cos}\:\left({A}−{B}\right)=\mathrm{cos}\:\left({A}−\frac{\pi}{\mathrm{3}}+{A}−\mathrm{2}{n}\pi\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{3}}−\mathrm{2}{A}\right)=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\mathrm{2}{A}\right) \\ $$$$=\:\mathrm{cos}\:\frac{\pi}{\mathrm{3}}.\mathrm{cos}\:\mathrm{2}{A}+\mathrm{sin}\:\frac{\pi}{\mathrm{3}}.\mathrm{sin}\:\mathrm{2}{A} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {A}\right)+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{A}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {A}\right)^{\mathrm{2}} } \\ $$$$=\:\mathrm{0}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Commented by mr 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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