Question

show that cos ^3X2Φ+3cos2Φ=(cos^6Φ-sin ^Φ)4

Answer 1

Hello.... ☺


please refer to the attachment above


thank you ☺

Answer 2

$\huge{Hello Friend}$

The answer of u r question is..✌️✌️

Given,

${cos}^{3} 2tita + 3cos \: 2tita = 4( {cos}^{6} - {sin}^{6} )$

So,

RHS

$= 4( {cos}^{2} tita) {}^{3} - { {(sin}^{2} tita)}^{3}$

$= 4 \cos(2 \: tita) (1 - 2 { \sin}^{2} tita { \cos }^{2} tita$

cos2tita(4-sin^2 2 tita)

cos 2tita (3+cos^2 2tuta)

= 3 cos 2tita+ cos^3 2tita



Thank you...⭐️⭐️⭐️⭐️