Question

EXpress Cos6X interm of CosX​

Answer 1

Expressing
`\( \cos(6x) \)` in terms of
`\( \cos(x) \)` yields
`\( 32\cos^6(x) - 48\cos^4(x) + 18\cos^2(x) - 1 \).` This simplification utilizes multiple angle formulas for cosine and double-angle identities.

`\[ \cos(6x) = 4(2\cos^2(x) - 1)^3 - 3(2\cos^2(x) - 1) \]`

let's express
`\( (2\cos^2(x) - 1) \) as \( \cos(2x) \)` using the double-angle formula:
`\( \cos(2x) = 2\cos^2(x) - 1 \)`

Substituting
`\( \cos(2x) \)` into the equation for
`\( \cos(6x) \)`:

`\[ \cos(6x) = 4(\cos(2x))^3 - 3\cos(2x) \]`

let's express
`\( \cos(2x) \)` again in terms of
`\( \cos(x) \)` :

`\[ \cos(2x) = 2\cos^2(x) - 1 \]`

Substitute this back into the equation for
`\( \cos(6x) \):`

`\[ \cos(6x) = 4(2\cos^2(x) - 1)^3 - 3(2\cos^2(x) - 1) \]`

Expanding the cube term
`\( (2\cos^2(x) - 1)^3 \):`

`\[ \cos(6x) = 4(8\cos^6(x) - 12\cos^4(x) + 6\cos^2(x) - 1) - 6\cos^2(x) + 3 \]`

Simplifying further:

`\[ \cos(6x) = 32\cos^6(x) - 48\cos^4(x) + 24\cos^2(x) - 4 - 6\cos^2(x) + 3 \]`

Combine like terms:

`\[ \cos(6x) = 32\cos^6(x) - 48\cos^4(x) + 18\cos^2(x) - 1 \]`

Hence,
`\( \cos(6x) \) in terms of \( \cos(x) \) is \( 32\cos^6(x) - 48\cos^4(x) + 18\cos^2(x) - 1 \).`

complete the question

Simplify the expression
`\( \cos(6x) \)` in terms of
`\( \cos(x) \)`