Question

If $\cos x$ $+{\cos }^{2}x=1$ then ${\sin }^{12}x+3{\sin }^{10}x+$$3{\sin }^{8}x+{\sin }^{6}x+{\sin }^{4}x+2{\sin }^{2}x-2$ is equal to

• A. $0$
• B. $\sin x$
• C. $1-\cos x$
• D. $1$

Answer 1

The correct option is C $1-\cos x$

$\cos x+{\cos }^{2}x=1$

$\cos x=1-{\cos }^{2}x$

$\cos x={\sin }^{2}x$

now in the given equation we substitute ${\sin }^{2}x=\cos x$ everywhere this gives

$={\cos }^{6}x+3{\cos }^{5}x+3{\cos }^{4}x+{\cos }^{3}x+2{\cos }^{2}x+\cos x-2$

$={\cos }^{3}x(\cos x+1{)}^3+2{\cos }^{2}x+\cos x-2$

$=({\cos }^{2}x+\cos x{)}^3+\cos x-2(1-{\cos }^{2}x)$

$=(1{)}^3+\cos x-2\cos x$

$=1-\cos x$