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Question Number 34064 by math1967 last updated on 30/Apr/18

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

let sin^2 α=t (t^2 /a)+(((1−t)^2 )/b)=(1/(a+b)) bt^2 +a(1−2t+t^2 )=((ab)/(a+b)) t^2 (a+b)+t(−2a)+a−((ab)/(a+b))=0 (a+b)^2 t^2 +t(−2a)(a+b)+a^2 =0 {(a+b)t−a}^2 =0 t=(a/(a+b)) sin^2 α=(a/(a+b )) ,cos^2 α=(b/(a+b)) ((sin^(12) α)/a^5 )+((cos^(12) α)/b^5 ) (a^6 /((a+b)^6 .a^5 ))+(b^6 /((a+b)^6 .b^5 )) ((a+b)/((a+b)^6 )) (1/((a+b)^5 )) Tanmay chaudhury

$${let}\:{sin}^{\mathrm{2}} \alpha={t} \\ $$$$\frac{{t}^{\mathrm{2}} }{{a}}+\frac{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{{b}}=\frac{\mathrm{1}}{{a}+{b}} \\ $$$${bt}^{\mathrm{2}} +{a}\left(\mathrm{1}−\mathrm{2}{t}+{t}^{\mathrm{2}} \right)=\frac{{ab}}{{a}+{b}} \\ $$$${t}^{\mathrm{2}} \left({a}+{b}\right)+{t}\left(−\mathrm{2}{a}\right)+{a}−\frac{{ab}}{{a}+{b}}=\mathrm{0} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} {t}^{\mathrm{2}} +{t}\left(−\mathrm{2}{a}\right)\left({a}+{b}\right)+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left\{\left({a}+{b}\right){t}−{a}\right\}^{\mathrm{2}} =\mathrm{0} \\ $$$${t}=\frac{{a}}{{a}+{b}} \\ $$$${sin}^{\mathrm{2}} \alpha=\frac{{a}}{{a}+{b}\:}\:\:,{cos}^{\mathrm{2}} \alpha=\frac{{b}}{{a}+{b}} \\ $$$$\frac{{sin}^{\mathrm{12}} \alpha}{{a}^{\mathrm{5}} }+\frac{{cos}^{\mathrm{12}} \alpha}{{b}^{\mathrm{5}} } \\ $$$$\frac{{a}^{\mathrm{6}} }{\left({a}+{b}\right)^{\mathrm{6}} .{a}^{\mathrm{5}} }+\frac{{b}^{\mathrm{6}} }{\left({a}+{b}\right)^{\mathrm{6}} .{b}^{\mathrm{5}} } \\ $$$$ \\ $$$$\frac{{a}+{b}}{\left({a}+{b}\right)^{\mathrm{6}} } \\ $$$$\frac{\mathrm{1}}{\left({a}+{b}\right)^{\mathrm{5}} \:}\:\:{Tanmay}\:{chaudhury} \\ $$

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