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Question Number 135895 by liberty last updated on 16/Mar/21
Trigonometry If sin A + sin B = a and cos A + cos B = b , find cos (A+B) in terms of a and b?
$${Trigonometry} \\ $$If sin A + sin B = a and cos A + cos B = b , find cos (A+B) in terms of a and b?
Answered by EDWIN88 last updated on 18/Mar/21
{ ((sin A+sin B=2sin (((A+B)/2))cos (((A−B)/2))= a)),((cos A+cos B=2cos (((A+B)/2))cos (((A−B)/2))=b)) :} we get tan (((A+B)/2))= (a/b) and tan (A+B)=((2tan (((A+B)/2)))/(1−tan^2 (((A+B)/2)))) tan (A+B)= (((((2a)/b)))/(1−((a^2 /b^2 )))) = ((2ab)/(b^2 −a^2 )) then sec^2 (A+B)= 1+((4a^2 b^2 )/((b^2 −a^2 )^2 ))=(((b^2 −a^2 )^2 +4a^2 b^2 )/((b^2 −a^2 )^2 )) so cos^2 (A+B)= (((b^2 −a^2 )^2 )/((b^2 +a^2 )^2 )) ⇒ cos (A+B)= ±(((b^2 −a^2 )/(b^2 +a^2 )))
$$\begin{cases}{\mathrm{sin}\:\mathrm{A}+\mathrm{sin}\:\mathrm{B}=\mathrm{2sin}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)=\:{a}}\\{\mathrm{cos}\:{A}+\mathrm{cos}\:\mathrm{B}=\mathrm{2cos}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)={b}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{tan}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)=\:\frac{{a}}{{b}}\:\mathrm{and}\:\mathrm{tan}\:\left(\mathrm{A}+\mathrm{B}\right)=\frac{\mathrm{2tan}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)} \\ $$$$\mathrm{tan}\:\left(\mathrm{A}+\mathrm{B}\right)=\:\frac{\left(\frac{\mathrm{2}{a}}{{b}}\right)}{\mathrm{1}−\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)}\:=\:\frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\mathrm{then}\:\mathrm{sec}\:^{\mathrm{2}} \left(\mathrm{A}+\mathrm{B}\right)=\:\mathrm{1}+\frac{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${so}\:\mathrm{cos}\:^{\mathrm{2}} \left({A}+\mathrm{B}\right)=\:\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{b}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow\:\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)=\:\pm\left(\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\right)\: \\ $$
Commented by mr W last updated on 17/Mar/21
please check sir: i think ∣...∣ is wrong. example a=(√3), b=1 ⇒((b^2 −a^2 )/(b^2 +a^2 ))=−(1/2) that could mean A=(π/3), B=(π/3) cos (A+B)=cos (((2π)/3))=−(1/2)=((b^2 −a^2 )/(b^2 +a^2 ))≠∣((b^2 −a^2 )/(b^2 +a^2 ))∣
$${please}\:{check}\:{sir}:\: \\ $$$${i}\:{think}\:\mid…\mid\:{is}\:{wrong}.\:{example}\: \\ $$$${a}=\sqrt{\mathrm{3}},\:{b}=\mathrm{1}\:\Rightarrow\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${that}\:{could}\:{mean}\:{A}=\frac{\pi}{\mathrm{3}},\:{B}=\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\left({A}+{B}\right)=\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} }\neq\mid\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} }\mid \\ $$
Commented by EDWIN88 last updated on 17/Mar/21
why sir. If x^2 = (b−a)^2 then x=± (√((b−a)^2 )) x = ∣b−a∣ sir
$$\mathrm{why}\:\mathrm{sir}.\:\mathrm{If}\:\mathrm{x}^{\mathrm{2}} \:=\:\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} \:\mathrm{then}\:\mathrm{x}=\pm\:\sqrt{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} } \\ $$$$\mathrm{x}\:=\:\mid\mathrm{b}−\mathrm{a}\mid\:\mathrm{sir} \\ $$
Commented by mr W last updated on 17/Mar/21
x=± (√((b−a)^2 ))≢x = ∣b−a∣ with x = ∣b−a∣ you fix x >0, but x can also be negative: x=±∣b−a∣
$$\mathrm{x}=\pm\:\sqrt{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} }≢\mathrm{x}\:=\:\mid\mathrm{b}−\mathrm{a}\mid \\ $$$${with}\:\mathrm{x}\:=\:\mid\mathrm{b}−\mathrm{a}\mid\:{you}\:{fix}\:{x}\:>\mathrm{0},\:{but} \\ $$$${x}\:{can}\:{also}\:{be}\:{negative}:\:{x}=\pm\mid{b}−{a}\mid \\ $$
Answered by mr W last updated on 17/Mar/21
sin (((A+B)/2)+((A−B)/2))+sin (((A+B)/2)−((A+B)/2))=a sin ((A+B)/2) cos ((A−B)/2)=(a/2) ...(i) cos (((A+B)/2)+((A−B)/2))+cos (((A+B)/2)−((A+B)/2))=b cos (((A+B)/2))cos (((A−B)/2))=(b/2) ...(ii) (i)/(ii): tan ((A+B)/2)=(a/b) recall: cos 2α=((1−tan^2 α)/(1+tan^2 α)) ⇒cos (A+B)=((1−((a/b))^2 )/(1+((a/b))^2 ))=((b^2 −a^2 )/(b^2 +a^2 ))
$$\mathrm{sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}+\frac{{A}−{B}}{\mathrm{2}}\right)+\mathrm{sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}−\frac{{A}+{B}}{\mathrm{2}}\right)={a} \\ $$$$\mathrm{sin}\:\frac{{A}+{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}=\frac{{a}}{\mathrm{2}}\:\:\:…\left({i}\right) \\ $$$$\mathrm{cos}\:\left(\frac{{A}+{B}}{\mathrm{2}}+\frac{{A}−{B}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{{A}+{B}}{\mathrm{2}}−\frac{{A}+{B}}{\mathrm{2}}\right)={b} \\ $$$$\mathrm{cos}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)=\frac{{b}}{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\mathrm{tan}\:\frac{{A}+{B}}{\mathrm{2}}=\frac{{a}}{{b}} \\ $$$${recall}:\:\mathrm{cos}\:\mathrm{2}\alpha=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha} \\ $$$$\Rightarrow\mathrm{cos}\:\left({A}+{B}\right)=\frac{\mathrm{1}−\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} }=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$

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