cos 4x = 1 - 8 cos^2x + 8 cos^4x prove
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❤️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:
hope it helps you
Step-by-step explanation:
cos4x = 1-8cos^2x + 8cos^4x
solving LHS.
We know
[tex]cos2x = 2cos^{2} x - 1\\So, cos4x = 2cos^22x - 1\\[/tex]
Now again solving
cos2x = 2cos^{2} x - 1
[tex]cos4x = 2(2cos^2x-1)^{2} - 1[/tex] (as their is square above so it will also become whole square)
Now using (a-b)^2 = a^2 + b^2 -2ab
[tex]cos4x = 2[ ((2cos^2x)^2 + 1^2 - 2(2cos^2x)(1)] - 1[/tex]
[tex]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[/tex]