Question

If $A+B+C=\pi$, then prove that ${\sin }^{4}A+{\sin }^{4}B+{\sin }^{4}C=\frac{7}{8}+\cos A\cos B\cos C+\frac{3}{2}\cos 2A\cos 2B\cos 2C$

Answer 1

We have,

$A+B+C=\pi$ …….. (1)

L.H.S

$={\sin }^{4}A+{\sin }^{4}B+{\sin }^{4}C$

$={\left({\sin }^{2}A\right)}^2+{\left({\sin }^{2}B\right)}^2+{\left({\sin }^{2}C\right)}^2$

$={\left(\frac{1-\cos 2A}{2}\right)}^2+{\left(\frac{1-\cos 2B}{2}\right)}^2+{\left(\frac{1-\cos 2C}{2}\right)}^2$

$=\frac{1}{4}[1+{\cos }^{2}2A-2\cos 2A+1+{\cos }^{2}2B-2\cos 2B+1+{\cos }^{2}2C-2\cos 2C]$

$=\frac{1}{4}[3+{\cos }^{2}2A+{\cos }^{2}2B+{\cos }^{2}2C-2(\cos 2A+\cos 2B+\cos 2C)]$

$=\frac{1}{4}[3+\left(\frac{\cos 4A+1}{2}\right)+\left(\frac{\cos 4B+1}{2}\right)+\left(\frac{\cos 4C+1}{2}\right)-2(\cos 2A+\cos 2B+\cos 2C)]$

$=\frac{1}{4}[3+(\frac{\cos 4A}{2}+\frac{1}{2})+(\frac{\cos 4B}{2}+\frac{1}{2})+(\frac{\cos 4C}{2}+\frac{1}{2})-2(\cos 2A+\cos 2B+\cos 2C)]$

$=\frac{1}{4}[3+\frac{3}{2}(\cos 4A+\cos 4B+\cos 4C)-2(\cos 2A+\cos 2B+\cos 2C)]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos \left(\frac{4A+4B}{2}\right)\cdot \cos \left(\frac{4A-4B}{2}\right)+\cos 4C]-2[2\cos \left(\frac{2A+2B}{2}\right)\cdot \cos \left(\frac{2A-2B}{2}\right)+\cos 2C]]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos \left(2(A+B)\right)\cdot \cos \left(2(A-B)\right)+\cos 4C]-2(2\cos (A+B)\cdot \cos (A-B)+\cos 2C)]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos \left(2(\pi -C)\right)\cdot \cos \left(2(A-B)\right)+\cos 4C]-2(2\cos (\pi -C)\cdot \cos (A-B)+\cos 2C)]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos 2C\cdot \cos \left(2(A-B)\right)+2{\cos }^{2}2C-1]-2(-2\cos C\cdot \cos (A-B)+\cos 2C)]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos 2C(\cos \left(2(A-B)\right)+\cos 2C)-1]-2(\cos 2C-2\cos C\cdot \cos (A-B))]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos 2C(\cos \left(2(A-B)\right)+\cos 2(\pi -(A+B)))-1]-2(2{\cos }^{2}C-1-2\cos C\cdot \cos (A-B))]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos 2C(\cos \left(2(A-B)\right)+\cos (2\pi -2(A+B)))-1]-4{\cos }^{2}C+2+4\cos C\cdot \cos (A-B)]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos 2C(\cos \left(2(A-B)\right)+\cos \left(2(A+B)\right))-1]-4\cos C[\cos C-\cos (A-B)]+2]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos 2C(\cos \left(2(A-B)\right)+\cos 2(A+B))-1]-4\cos C[\cos C-\cos (A-B)]+2]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos 2C(\cos \left(2(A+B)\right)+\cos 2(A-B))-1]-4\cos C[\cos C-\cos (A-B)]+2]$

$=\frac{1}{4}[3+\frac{3}{2}[2\cos 2C\left(2\cos 2A\cos 2B\right)-1]-4\cos C[\cos C-\cos (A-B)]+2]$

$=\frac{1}{4}[3+3\cos 2C\left(2\cos 2A\cos 2B\right)-\frac{3}{2}-2\cos C(\cos (\pi -(A+B))-\cos (A-B))+2]$

$=\frac{1}{4}[5+6\cos 2A\cos 2B\cos 2C-\frac{3}{2}-2\cos C(-\cos (A+B)-\cos (A-B))]$

$=\frac{1}{4}[\frac{7}{2}+6\cos 2A\cos 2B\cos 2C+2\cos C(\cos (A+B)+\cos (A-B))]$

$=\frac{1}{4}[\frac{7}{2}+6\cos 2A\cos 2B\cos 2C+2\cos C\left(2\cos A\cos B\right)]$

$=\frac{1}{4}[\frac{7}{2}+6\cos 2A\cos 2B\cos 2C+4\cos A\cos B\cos C]$

$=\frac{7}{8}+\cos A\cos B\cos C+\frac{3}{2}\cos 2A\cos 2B\cos 2C$

Hence, proved.