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Question Number 94522 by Abdulrahman last updated on 19/May/20
if sinA+sinB=n and cosA+cosB=m then sin(A+B)=? a: ((3mn)/(m^2 +n^2 )) b:((2mn)/(m^2 +n^2 )) c:((mn)/(m^2 +n^2 )) d:((2mn)/(m+n)) with steps?
$$\mathrm{if}\:\:\mathrm{sinA}+\mathrm{sinB}=\mathrm{n}\:\mathrm{and}\:\mathrm{cosA}+\mathrm{cosB}=\mathrm{m} \\ $$$$\mathrm{then}\:\:\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)=? \\ $$$$\mathrm{a}:\:\frac{\mathrm{3mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\:\:\:\:\mathrm{b}:\frac{\mathrm{2mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\:\:\mathrm{c}:\frac{\mathrm{mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\:\:\mathrm{d}:\frac{\mathrm{2mn}}{\mathrm{m}+\mathrm{n}} \\ $$$$\mathrm{with}\:\mathrm{steps}? \\ $$
Answered by i jagooll last updated on 19/May/20
squaring eq (1) & (2) then adding
$$\mathrm{squaring}\:\mathrm{eq}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right)\:\mathrm{then}\:\mathrm{adding} \\ $$
Commented by Abdulrahman last updated on 19/May/20
please solve it
$$\mathrm{please}\:\mathrm{solve}\:\mathrm{it} \\ $$
Answered by john santu last updated on 19/May/20
cos (A−B) = ((m^2 +n^2 −2)/2) sin (A+B) = ((2mn−[ sin 2A+sin 2B ])/2) 2sin (A+B) = 2mn −[ 2sin (A+B).cos (A−B)] 2sin (A+B) = 2mn −[ ((m^2 +n^2 −2)/2)×2 sin (A+B)] [ 2+m^2 +n^2 −2 ] sin (A+B) = 2mn sin (A+B) = ((2mn)/(m^2 +n^2 ))
$$\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)\:=\:\frac{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\left(\mathrm{A}+\mathrm{B}\right)\:=\:\frac{\mathrm{2mn}−\left[\:\mathrm{sin}\:\mathrm{2A}+\mathrm{sin}\:\mathrm{2B}\:\right]}{\mathrm{2}} \\ $$$$\mathrm{2sin}\:\left(\mathrm{A}+\mathrm{B}\right)\:=\:\mathrm{2mn}\:−\left[\:\mathrm{2sin}\:\left(\mathrm{A}+\mathrm{B}\right).\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)\right] \\ $$$$\mathrm{2sin}\:\left(\mathrm{A}+\mathrm{B}\right)\:=\:\mathrm{2mn}\:−\left[\:\frac{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}}×\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{A}+\mathrm{B}\right)\right] \\ $$$$\left[\:\mathrm{2}+\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} −\mathrm{2}\:\right]\:\mathrm{sin}\:\left(\mathrm{A}+\mathrm{B}\right)\:=\:\mathrm{2mn} \\ $$$$\mathrm{sin}\:\left(\mathrm{A}+\mathrm{B}\right)\:=\:\frac{\mathrm{2mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\: \\ $$
Commented by Abdulrahman last updated on 19/May/20
bundle of thanks
$$\mathrm{bundle}\:\mathrm{of}\:\mathrm{thanks} \\ $$
Answered by som(math1967) last updated on 19/May/20
sinA+sinB=n 2sin((A+B)/2)cos((A−B)/2)=n cosA+cosB=m 2cos((A+B)/2)cos((A−B)/2)=m ∴tan((A+B)/2)=(n/m) sin((A+B)/2)=(n/( (√(m^2 +n^2 )))) cos((A+B)/2)=(m/( (√(m^2 +n^2 )))) sin(A+B) 2sin((A+B)/2)cos((A+B)/2) =((2mn)/(m^2 +n^2 )) so b)((2mn)/(m^2 +n^2 )) is ans
$$\mathrm{sinA}+\mathrm{sinB}=\mathrm{n} \\ $$$$\mathrm{2sin}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}=\mathrm{n} \\ $$$$\mathrm{cosA}+\mathrm{cosB}=\mathrm{m} \\ $$$$\mathrm{2cos}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}=\mathrm{m} \\ $$$$\therefore\mathrm{tan}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}=\frac{\mathrm{n}}{\mathrm{m}} \\ $$$$\mathrm{sin}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}=\frac{\mathrm{n}}{\:\sqrt{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}=\frac{\mathrm{m}}{\:\sqrt{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }} \\ $$$$\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right) \\ $$$$\mathrm{2sin}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\: \\ $$$$\left.\mathrm{so}\:\mathrm{b}\right)\frac{\mathrm{2mn}}{\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{ans} \\ $$
Commented by peter frank last updated on 19/May/20
thank you
$${thank}\:{you} \\ $$
Commented by Abdulrahman last updated on 19/May/20
$$ \\ $$

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