Question

$\left|\begin{array}{ccc}1& a& a^2\\ cos(n-1)x& cosnx& cos(n+1)x\\ sin(n-1)x& sinnx& sin(n+1)x\end{array}\right|=k\sin x$, then $k$ is equal to

• A. $1+a^2-2a$
• B. $1-a^2-2a$
• C. $1+a^2+2acosx$
• D. $1+a^2-2acosx$

Answer 1

The correct option is D $1+a^2-2acosx$

$\left|\begin{array}{ccc}1& a& a^2\\ cos(n-1)x& cosnx& cos(n+1)x\\ sin(n-1)x& sinnx& sin(n+1)x\end{array}\right|$

$C_1\to C_1+C_3$
$=\left|\begin{array}{ccc}1+a^2& a& a^2\\ cos(n-1)x+cos(n+1)x& cosnx& cos(n+1)x\\ sin(n-1)x+sin(n+1)x& sinnx& sin(n+1)x\end{array}\right|$
$=\left|\begin{array}{ccc}1+a^2& a& a^2\\ 2\cos nx\cos x& \cos nx& \cos (n+1)x\\ 2\sin nx\cos x& \sin nx& \sin (n+1)x\end{array}\right|$
$=(1+a^2)[\sin (n+1)x\cos nx-\cos (n+1)x\sin nx]-a\left[2\cos x\right(\cos nx\sin (n+1)x-\cos (n+1)x\sin nx\left)\right]+a^2\left(0\right)$
$=(1+a^2)\sin x-2a\cos x\sin x$
$=(1+a^2-2a\cos x)\sin x$

Therefore, $k=1+a^2-2a\cos x$