Question

if sin3B/sinB=(a^2-c^2/2ac)^2 then a^2,b^2,c^2 are in

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{\dfrac{sin3B}{sinB}=\left(\dfrac{a^2-c^2}{2ac}\right)^2}$

$\underline{\textsf{To find:}}$

$\mathsf{a^2,b^2,c^2\;are\;in\,A.P,G.P\;(or)H.P}$

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{sin3B}{sinB}=\left(\dfrac{a^2-c^}{2ac}\right)^2}$

$\mathsf{\dfrac{3\,sinB-4\,sin^3B}{sinB}=\left(\dfrac{a^2-c^}{2ac}\right)^2}$

$\mathsf{3-4\,sin^2B=\left(\dfrac{a^2-c^2}{2ac}\right)^2}$

$\mathsf{3-4(1-cos^2B)=\left(\dfrac{a^2-c^2}{2ac}\right)^2}$

$\mathsf{-1+4\,cos^2B=\left(\dfrac{a^2-c^2}{2ac}\right)^2}$

$\mathsf{4\,cos^2B=1+\left(\dfrac{a^2-c^2}{2ac}\right)^2}$

$\mathsf{Using,\;}$

$\boxed{\mathsf{COSINE\;FORMULA:\;cosB=\dfrac{c^2+a^2-b^2}{2ac}}}$

$\mathsf{4\left(\dfrac{c^2+a^2-b^2}{2ac}\right)^2=\dfrac{4a^2c^2+(a^2-c^2)^2}{4a^2c^2}}$

$\mathsf{\dfrac{2^2(c^2+a^2-b^2)^2}{4a^c^2}=\dfrac{4a^2c^2+a^2+c^2-2a^2c^2}{4a^2c^2}}$

$\mathsf{(2c^2+2a^2-2b^2)^2=(a^2+c^2)^2}$

$\mathsf{(2c^2+2a^2-2b^2)^2-(a^2+c^2)^2=0}$

$\mathsf{(2c^2+2a^2-2b^2+a^2+c^2)(2c^2+2a^2-2b^2-a^2-c^2)=0}$

$\mathsf{(3c^2+3a^2-2b^2)(c^2+a^2-2b^2)=0}$

$\implies\mathsf{3c^2+3a^2-2b^2=0\;\;(or)\;\;c^2+a^2-2b^2=0}$

$\implies\mathsf{a^2+c^2=2b^2}$

$\implies\mathsf{b^2-a^2=c^2-b^2}$

$\therefore\mathsf{a^2,b^2,c^2\;are\;in\;A.P}$

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