Question

The value of 2 $\cos x-\cos 3x-\cos 5x$ is equal to

• A. 16 ${\cos }^{3}x{\sin }^{2}x$
• B. 16 ${\sin }^{3}x{\cos }^{2}x$
• C. 4 ${\cos }^{3}x{\sin }^{2}x$
• D. 4 ${\sin }^{3}x{\cos }^{2}x$

Answer 1

The correct option is A 16 ${\cos }^{3}x{\sin }^{2}x$

$f\left(x\right)=2cosx-cos3x-cos5x$
$=2cosx-\left(2cos4xcosx\right)$
$=2cosx(1-cos4x)$
$=2cosx[co{s}^{2}2x+si{n}^{2}2x-(co{s}^{2}2x-si{n}^{2}2x\left)\right]$
$=2cosx\times 2si{n}^{2}2x[cos2x=co{s}^{2}x-si{n}^{2}x]$
$=4cosxsi{n}^{2}2x$

$=4cosx(2sinxcosx{)}^2$

$\Rightarrow f\left(x\right)=16si{n}^{2}xco{s}^{3}x$