Question

The numerical value of $3(\sin x-\cos x{)}^4+6(\sin x+\cos x{)}^2+4({\sin }^{6}x+{\cos }^{6}x)$ is​​​​​​​

Answer 1

$3(\sin x-\cos x{)}^4+6(\sin x+\cos x{)}^2+4({\sin }^{6}x+{\cos }^{6}x)$
Let
$A=3(\sin x-\cos x{)}^4\Rightarrow A=3[(\sin x-\cos x{)}^2{]}^2\Rightarrow A=3(1-2\sin x\cos x{)}^2\Rightarrow A=3(1-4\sin x\cos x+4{\sin }^{2}x{\cos }^{2}x)\Rightarrow A=3-12\sin x\cos x+12{\sin }^{2}x{\cos }^{2}x\cdots \left(1\right)$

$B=6(\sin x+\cos x{)}^2\Rightarrow B=6(1+2\sin x\cos x)\Rightarrow B=6+12\sin x\cos x\cdots (2)$

$C=4({\sin }^{6}x+{\cos }^{6}x)\Rightarrow C=4\left(\right({\sin }^{2}x{)}^3+\left({\cos }^{2}x{)}^3\right)\Rightarrow C=4\left\{\right({\sin }^{2}x+{\cos }^{2}x{)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)\left\}\right[\because x^3+y^3=(x+y{)}^3-3xy(x+y\left)\right]\Rightarrow C=4(1-3{\sin }^{2}x{\cos }^{2}x)\Rightarrow C=4-12{\sin }^{2}x{\cos }^{2}x\cdots \left(3\right)$
So,
$3(\sin x-\cos x{)}^4+6(\sin x+\cos x{)}^2+4({\sin }^{6}x+{\cos }^{6}x)=A+B+C\text{From equation (1), (2) and (3)}\therefore A+B+C=13$