Question

If$f\left(x\right)=\sin \left(\log x\right)$ then $f\left(xy\right)+f\left(\frac{x}{y}\right)-2f\left(x\right)\cos \left(\log y\right)$ is equal to

• A. $\cos \left(\log x\right)$
• B. $\sin \left(\log y\right)$
• C. $\cos \left(\log \right(xy\left)\right)$
• D. $0$

Answer 1

The correct option is D $0$

$f\left(xy\right)=\sin \left(\log \left(xy\right)\right)=\sin (\log x+\log y)$

$f\left(\frac{x}{y}\right)=\sin (\log x-\log y)$

$f\left(xy\right)+f\left(\frac{x}{y}\right)=\sin (\log x+\log y)+\sin (\log x-\log y)$

$[\because \sin (A+B)+\sin (A-B)=2\sin A\cos B]$

$=2\sin \left(\log x\right)\cos \left(\log y\right)$

$\therefore f\left(xy\right)+f\left(\frac{x}{y}\right)-2\sin \left(\log x\right)\cos \left(\log y\right)=0$

$f\left(xy\right)+f\left(\frac{x}{y}\right)-2f\left(x\right)\cos \left(\log y\right)=0$