Question

In a $\Delta ABC,{\sin }^{4}A+{\sin }^{4}B+{\sin }^{4}C=$

• A. $\frac{3}{2}+2\cos A\cos B\cos C+\frac{1}{2}\cos 2A\cos 2B\cos 2C$
• B. $2\cos A\cos B\cos C+\frac{1}{2}\cos 2A\cos 2B\cos 2C$
• C. $2\cos A\cos B\cos C+\frac{1}{2}\cos 2A\cos 2B$
• D. $2\cos A\cos B\cos C+\cos 2A\cos 2B$

Answer 1

The correct option is A $\frac{3}{2}+2\cos A\cos B\cos C+\frac{1}{2}\cos 2A\cos 2B\cos 2C$

$A+B+C=180\Rightarrow C=180-(A+B)$

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

Direct expansion gives,

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

Consider,

$({\cos }^{2}2A+{\cos }^{2}2B+{\cos }^{2}2C)=\left(\frac{1+\cos 4A}{2}\right)+\left(\frac{1+\cos 4B}{2}\right)+\left(\frac{1+\cos 4C}{2}\right)$

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

$=\frac{3}{2}+\frac{1}{2}(2\cos \left(2\right(A+B\left)\right)\cos \left(2\right(A-B\left)\right)+2{\cos }^{2}2(A+B)-1)$

$=1+\cos \left(2\right(A+B\left)\right)(\cos \left(2\right(A+B\left)\right)+\cos 2(A-B))=1+\cos 2(A+B)(2\cos 2A\cos 2B)$

$=1+2\cos 2A\cos 2B\cos 2C\cos 2A+\cos 2B+\cos 2C=\cos 2A+\cos 2B+\cos 2(A+B)$

$=2\cos (A+B)\cos (A-B)+2{\cos }^2(A+B)-1$

$=-4\cos A\cos B\cos C-1$

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

$=\frac{3}{2}+2\cos A\cos B\cos C+\frac{1}{2}\cos 2A\cos 2B\cos 2C$