Question

cos 4x = 1 - 8 cos^2x + 8 cos^4x prove​

Answer 1

❤️Let us consider the LHS

❤️ cos 4x

❤️As we know,

❤️cos 2x = 2 cos2 x – 1

❤️Therefore,

❤️cos 4x

❤️= 2 cos2 2x – 1 =2(2 cos2 2x – 1)2 – 1

❤️= 2[(2 cos2 2x)2 + 12 – 2 × 2 cos2 x] – 1

❤️= 2(4 cos4 2x + 1 – 4 cos2 x) – 1

❤️= 8 cos4 2x + 2 – 8 cos2 x – 1

❤️= 8 cos4 2x + 1 – 8 cos2 x

❤️= RHS

❤️Hence Proved

❤️❤️I Hope it will help you

Thank s..❤️❤️

Answer 2

Answer:

hope it helps you

Step-by-step explanation:

cos4x = 1-8cos^2x + 8cos^4x

solving LHS.

We know

$cos2x = 2cos^{2} x - 1\\So, cos4x = 2cos^22x - 1$

Now again solving

cos2x = 2cos^{2} x - 1

$cos4x = 2(2cos^2x-1)^{2} - 1$  (as their is square above so it will also become whole square)

Now using (a-b)^2 = a^2 + b^2 -2ab

$cos4x = 2[ ((2cos^2x)^2 + 1^2 - 2(2cos^2x)(1)] - 1$

$cos4x = 2[ 4cos^4x + 1 - 4cos^2x] - 1\\cos4x = 8cos^4x + 2 - 8cos^2x - 1\\cos4x = 8cos^4x + 1 - 8cos^2x\\or\\1 - 8cos^2x + 8cos^4x$