Menu Close

If-sin-x-sin-2x-a-cos-x-cos-2x-b-show-that-a-2-b-2-a-2-b-2-3-2b-

Tinku Tara 0 Comments
Pin




Question Number 127067 by benjo_mathlover last updated on 26/Dec/20
If { ((sin x+sin 2x = a)),((cos x+cos 2x = b )) :} show that (a^2 +b^2 )(a^2 +b^2 −3)=2b.
$$\:{If}\:\begin{cases}{\mathrm{sin}\:{x}+\mathrm{sin}\:\mathrm{2}{x}\:=\:{a}}\\{\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{2}{x}\:=\:{b}\:}\end{cases} \\ $$$${show}\:{that}\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{2}{b}. \\ $$
Answered by liberty last updated on 26/Dec/20
{ ((a^2 =sin^2 x+sin^2 2x+2sin 2xsin x)),((b^2 =cos^2 x+cos^2 2x+2cos 2xcos x)) :} ⇔ a^2 +b^2 =2+2(cos 2xcos x+sin 2xsin x) a^2 +b^2 = 2+2 cos (2x−x) a^2 +b^2 = 2+2cos x ⇔(a^2 +b^2 )(a^2 +b^2 −3)=(2cos x+2)(2cos x−1) =4cos^2 x+2cos x−2 = 2(2cos^2 x+cos x−1) = 2(cos 2x+cos x)=2b
$$\:\begin{cases}{{a}^{\mathrm{2}} =\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}+\mathrm{2sin}\:\mathrm{2}{x}\mathrm{sin}\:{x}}\\{{b}^{\mathrm{2}} =\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{x}+\mathrm{2cos}\:\mathrm{2}{x}\mathrm{cos}\:{x}}\end{cases} \\ $$$$\:\Leftrightarrow\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}+\mathrm{2}\left(\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:{x}+\mathrm{sin}\:\mathrm{2}{x}\mathrm{sin}\:{x}\right) \\ $$$$\:\:\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:=\:\mathrm{2}+\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{2}{x}−{x}\right) \\ $$$$\:\:\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:=\:\mathrm{2}+\mathrm{2cos}\:{x} \\ $$$$\Leftrightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{3}\right)=\left(\mathrm{2cos}\:{x}+\mathrm{2}\right)\left(\mathrm{2cos}\:{x}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4cos}\:^{\mathrm{2}} {x}+\mathrm{2cos}\:{x}−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:{x}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left(\mathrm{cos}\:\mathrm{2}{x}+\mathrm{cos}\:{x}\right)=\mathrm{2}{b} \\ $$
Answered by mnjuly1970 last updated on 27/Dec/20
{ ((sin^2 (x)+sin^2 (2x)+2sin(x)sin(2x)=a^2 (i) )),((cos^2 (x)+cos^2 (2x)+2cos(x)cos(2x)=b^2 (ii))) :} (i)+(ii):: 1+1+2cos(2x−x)=b^2 +a^2 ⇒cos(x)=((b^2 +a^2 −2)/2) cos(x)+2cos^2 (x)−1=b ((b^2 +a^2 −2)/2)+(((b^2 +a^2 −2)^2 )/2)=b+1 (b^2 +a^2 −2)(b^2 +a^2 −1)=2b+2 (b^2 +a^2 )^2 −3(b^2 +a^2 )+2=2b+2 (b^2 +a^2 )(b^2 +a^2 −3)=2b ✓
$$\begin{cases}{{sin}^{\mathrm{2}} \left({x}\right)+{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)+\mathrm{2}{sin}\left({x}\right){sin}\left(\mathrm{2}{x}\right)={a}^{\mathrm{2}} \left({i}\right)\:\:\:}\\{{cos}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)+\mathrm{2}{cos}\left({x}\right){cos}\left(\mathrm{2}{x}\right)={b}^{\mathrm{2}} \left({ii}\right)}\end{cases} \\ $$$$\left({i}\right)+\left({ii}\right)::\:\mathrm{1}+\mathrm{1}+\mathrm{2}{cos}\left(\mathrm{2}{x}−{x}\right)={b}^{\mathrm{2}} +{a}^{\mathrm{2}} \Rightarrow{cos}\left({x}\right)=\frac{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}}\: \\ $$$${cos}\left({x}\right)+\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)−\mathrm{1}={b} \\ $$$$\:\frac{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}}+\frac{\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}}={b}+\mathrm{1} \\ $$$$\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}\right)\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{2}{b}+\mathrm{2} \\ $$$$\:\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{3}\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)+\mathrm{2}=\mathrm{2}{b}+\mathrm{2} \\ $$$$\:\:\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{2}{b}\:\:\:\checkmark \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 27/Dec/20
thank you
$${thank}\:{you} \\ $$
Commented by liberty last updated on 26/Dec/20
(b^2 +a^2 )(b^2 +a^2 −3)=2b ... ✓✓
$$\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{2}{b}\:…\:\checkmark\checkmark \\ $$
Answered by MJS_new last updated on 26/Dec/20
sin x +2sin x cos x =a cos x +2cos^2 x −1=b t=tan (x/2) a=−((2t(t^2 −3))/((t^2 +1)^2 )) b=−((2(3t^2 −1))/((t^2 +1)^2 )) (a^2 +b^2 )(a^2 +b^2 −3)=(4/(t^2 +1))×−((3t^2 −1)/(t^2 +1))= =−((4(3t^2 −1))/((t^2 +1)^2 ))=2b
$$\mathrm{sin}\:{x}\:+\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:={a} \\ $$$$\mathrm{cos}\:{x}\:+\mathrm{2cos}^{\mathrm{2}} \:{x}\:−\mathrm{1}={b} \\ $$$${t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$$${a}=−\frac{\mathrm{2}{t}\left({t}^{\mathrm{2}} −\mathrm{3}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}=−\frac{\mathrm{2}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{3}\right)=\frac{\mathrm{4}}{{t}^{\mathrm{2}} +\mathrm{1}}×−\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=−\frac{\mathrm{4}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{2}{b} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *