Question-220745

Question Number 220745 by Spillover last updated on 18/May/25

Answered by som(math1967) last updated on 18/May/25

cosx+cosy=a ⇒2cos(((x+y)/2))cos(((x−y)/2))=a...(1) sinx+siny=b ⇒2sin(((x+y)/2))cos(((x−y)/2))=b ....(2) ∴ tan(((x+y)/2))=(b/a) [ (2)÷(1) ] (cosx+cosy)(sinx+siny)=ab cosxsinx+cosxsiny+cosysinx +sinycosy=ab 2sinxcosx+2sin(x+y)+2sinycosy=2ab sin2x+sin2y+2×((2tan(((x+y)/2)))/(1+tan^2 (((x+y)/2))))=2ab [∵ sinx=((2tan((x/2)))/(1+tan^2 (x/2)))] sin2x+sin2y+2×(((2b)/a)/(1+(b^2 /a^2 )))=2ab sin2x+sin2y=2ab−((4ab)/(a^2 +b^2 )) ∴sin2x+sin2y=2ab[1−(2/(a^2 +b^2 ))]

$${cosx}+{cosy}={a} \\ $$$$\Rightarrow\mathrm{2}{cos}\left(\frac{{x}+{y}}{\mathrm{2}}\right){cos}\left(\frac{{x}−{y}}{\mathrm{2}}\right)={a}…\left(\mathrm{1}\right) \\ $$$$\:{sinx}+{siny}={b} \\ $$$$\Rightarrow\mathrm{2}{sin}\left(\frac{{x}+{y}}{\mathrm{2}}\right){cos}\left(\frac{{x}−{y}}{\mathrm{2}}\right)={b}\:\:….\left(\mathrm{2}\right) \\ $$$$\therefore\:{tan}\left(\frac{{x}+{y}}{\mathrm{2}}\right)=\frac{{b}}{{a}}\:\:\:\:\:\left[\:\:\left(\mathrm{2}\right)\boldsymbol{\div}\left(\mathrm{1}\right)\:\right] \\ $$$$\:\left({cosx}+{cosy}\right)\left({sinx}+{siny}\right)={ab} \\ $$$$\:{cosxsinx}+{cosxsiny}+{cosysinx} \\ $$$$\:\:\:\:\:\:+{sinycosy}={ab} \\ $$$$\mathrm{2}{sinxcosx}+\mathrm{2}{sin}\left({x}+{y}\right)+\mathrm{2}{sinycosy}=\mathrm{2}{ab} \\ $$$${sin}\mathrm{2}{x}+{sin}\mathrm{2}{y}+\mathrm{2}×\frac{\mathrm{2}{tan}\left(\frac{{x}+{y}}{\mathrm{2}}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}+{y}}{\mathrm{2}}\right)}=\mathrm{2}{ab} \\ $$$$\:\left[\because\:{sinx}=\frac{\mathrm{2}{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}\right] \\ $$$${sin}\mathrm{2}{x}+{sin}\mathrm{2}{y}+\mathrm{2}×\frac{\frac{\mathrm{2}{b}}{{a}}}{\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}=\mathrm{2}{ab} \\ $$$${sin}\mathrm{2}{x}+{sin}\mathrm{2}{y}=\mathrm{2}{ab}−\frac{\mathrm{4}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\therefore{sin}\mathrm{2}{x}+{sin}\mathrm{2}{y}=\mathrm{2}{ab}\left[\mathrm{1}−\frac{\mathrm{2}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right] \\ $$$$ \\ $$

Commented by Spillover last updated on 18/May/25