If sin x + sin^2 x = 1, then cos^8 x + 2cos^6 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given that $\sin x + \sin^2 x = 1.$This implies $\sin x = 1 - \sin^2 x = \cos^2 x.$Now,we need to evaluate the expression $\cos^8 x + 2\cos^6 x +

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given that $\sin x + \sin^2 x = 1.$This implies $\sin x = 1 - \sin^2 x = \cos^2 x.$Now,we need to evaluate the expression $\cos^8 x + 2\cos^6 x + \cos^4 x.$Substituting $\cos^2 x = \sin x$ into the expression:$= (\cos^2 x)^4 + 2(\cos^2 x)^3 + (\cos^2 x)^2$$= (\sin x)^4 + 2(\sin x)^3 + (\sin x)^2$$= \sin^4 x + 2\sin^3 x + \sin^2 x$$= (\sin^2 x + \sin x)^2.$Since $\sin x + \sin^2 x = 1,$ we have $(1)^2 = 1.$

If $\sin x + \sin^2 x = 1,$ then $\cos^8 x + 2\cos^6 x + \cos^4 x = $

Similar Questions

If $a \sin^2 \theta + b \cos^2 \theta = c$,then $\tan^2 \theta$ is equal to

$\cos ^2 5^{\circ}-\cos ^2 15^{\circ}-\sin ^2 15^{\circ}+\sin ^2 35^{\circ}+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ} = $

The value of $\cos 52^\circ + \cos 68^\circ + \cos 172^\circ$ is

$\sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ = $

If $\tan \theta = \frac{x \sin \phi}{1 - x \cos \phi}$ and $\tan \phi = \frac{y \sin \theta}{1 - y \cos \theta}$,then $\frac{x}{y} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module