You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trigonometric Identities

Given: cos x...

Question

Given: $cos x + {cos}^{2} x = 1$
Prove that:
${sin}^{2} x + {sin}^{4} x = 1$

Open in App

Solution

$cosx + {cos}^{2} x = 1$
$cosx = 1 - {cos}^{2} x$
$cos x = {sin}^{2} x$
${sin}^{4} x - {cos}^{2} x = 0$
${sin}^{4} x + 1 - {cos}^{2} x = 1$
$∴ {sin}^{4} x + {sin}^{2} x = 1$

Suggest Corrections

Similar questions

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

{sin}^{2} x + {sin}^{4} x = 1

\frac{cos 4 x + cos 3 x + cos 2 x}{sin 4 x + sin 3 x + sin 2 x} = cot 3 x

\frac{c o s^{4} x}{c o s^{2} y} + \frac{s i n^{4} x}{s i n^{2} y} = 1

\frac{c o s^{4} y}{c o s^{2} x} + \frac{s i n^{4} y}{s i n^{2} x} = 1

\cos x + \cos 2 x + \cos 3 x = 0

\cos x + \cos 3 x - \cos 2 x = 0

\sin x + \sin 5 x = \sin 3 x

\cos x \cos 2 x \cos 3 x = \frac{1}{4}

\cos x + \sin x = \cos 2 x + \sin 2 x

\sin x + \sin 2 x + \sin 3 = 0

\sin x + \sin 2 x + \sin 3 x + \sin 4 x = 0

\sin 3 x - \sin x = 4 \cos^{2} x - 2

\sin 2 x - \sin 4 x + \sin 6 x = 0

\frac{1 + sin 2 x + cos 2 x}{1 + sin 2 x - cos 2 x} = cot x

Join BYJU'S Learning Program