Prove the following:
$c o s 4 x = 1 - 8 s i n^{2} x c o s^{2} x$

Open in App

Solution

$cos 4 x = 2 {cos}^{2} 2 x - 1 = 2 {(2 {cos}^{2} x - 1)}^{2} - 1 = 1 + 2 (4 {cos}^{4} x - 4 {cos}^{2} x) = 1 - 8 {cos}^{2} x (1 - {cos}^{2} x) = 1 - 8 {cos}^{2} x {sin}^{2} x$
Hence proved

Suggest Corrections

Similar questions

c o s 4 x = 1 - 8 s i n^{2} x c o s^{2} x

cos 4 x = 1 - 8 {sin}^{2} x {cos}^{2} x

cos 4x = 1 – 8sin²x cos²x

cos 4 x = 1 - 8 {sin}^{2} x {cos}^{2} x

Prove that $t a n 70^{\circ} = t a n 20^{\circ} + 2 t a n 50^{\circ} .$
Or

Prove that $1 + c o s 2 x + c o s 4 x + c o s 6 x = 4 c o s x c o s 2 x c o s 3 x$

Join BYJU'S Learning Program