cos4x again.....

[small1bc]

Verify the identity \(\displaystyle \L\\cos(4x)=8cos^{4}(x)-8cos^{2}(x)+1\)

Here is what we have so far...

$\displaystyle cos 4x = 2(cos^{2}x-1)$

Now What?

Response 1:
\(\displaystyle \L\\sin^{2}(x)cos^{2}(x)=\frac{1-cos(4x)}{8}\)


i am still really confused....how do i plug that in... we havent been given that identity yet....here is wut i have so far now...
\(\displaystyle \L\\cos(x+y)=cos(x)cos(y)-sin(x)sin(y)\)
so \(\displaystyle \L\\cos2x=cos(x+x)=cos(x)cos(x)-sin(x)sin(x)\)
so \(\displaystyle \L\\cos4x=cos(2x+2x)=cos(2x)cos(2x)-sin(2x)sin(2x)\)
\(\displaystyle \L\\cos(2x+2x)=(2cos^{2}x-1)(2cos^{2}x-1)-(2sin(x)cos(x))(2sin(x)cos(x))\)
\(\displaystyle \L\\cos(2x+2x)=(4cos^{4}x-4cos^{2}x+1)-(4sin^{2}xcos^{2}x)\)
\(\displaystyle \L\\cos(2x+2x)=(4cos^{4}x-4cos^{2}x+1)-(4((1-cosx)(1-cosx))cos^{2}x)\)
\(\displaystyle \L\\cos(2x+2x)=(4cos^{4}x-4cos^{2}x+1)-(4(1-2cosx+cos^{2}x)cos^{2}x)\)
\(\displaystyle \L\\cos(2x+2x)=(4cos^{4}x-4cos^{2}x+1)-(4cos^{2}x-8cos^{3}x+4cos^{4}x)\)
\(\displaystyle \L\\cos(2x+2x)=4cos^{4}x-4cos^{2}x+1-4cos^{2}x+8cos^{3}x-4cos^{4}x\)
but that is \(\displaystyle \L\\cos4x=8cos^{3}x-8cos^{2}x+1\)
not \(\displaystyle \L\\cos4x=8cos^{4}x-8cos^{2}x+1\)

So I'm almost thinking there is an error on the quiz, or the answer is N/A beings that the identity is not verifiable??? Please HELP!!!

 

[k9fireman]

cos4x

I always think of it as cos 2(2x) which equals cos (4x).... see..... that way i use the cos 2x identity... just that 2x is the variable instead of just x.

cos 4x = cos 2(2x)

= 2 cos^2(2x) - 1
which is equal to : = 2 (cos 2x)^2 - 1 , moving the ^2 outside the term... see?
use double ident. on the cos2x term again...

= 2 (2 cos^2x - 1)^2 - 1
(2 cos^2x - 1)(2 cos^2x - 1)....

= 2 (4 cos^4x - 4 cos^2x + 1) - 1

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

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

 

[soroban]

Hello, small1bc!

Verify the identity: \(\displaystyle \L\,\cos(4x)\:=\:8\cos^4(x)\,-\,8\cos^2(x)\,+\,1\)

Your first derivation is correct: \(\displaystyle \L\:\cos(2x) \:=\:\cos^2(x)\,-\,\sin^2(x)\)

Since $\displaystyle \,\sin^2(x) \:=\:1\,-\,\cos^2(x)$, the identity becomes: $\displaystyle \:\cos(2x)\:=\:\cos^2(x)\,-\,\left[1\,-\,\cos^2(x)\right]$
. . and we have: \(\displaystyle \L\:\cos(2x) \:=\:2\cos^2(x)\,-\,1\)

Since $\displaystyle \,\cos^2(x)\:=\:1\,-\,\sin^2(x)$, the identity becomes: $\displaystyle \:\cos(2x)\:=\:\left[1\,-\,\sin^2(x)\right]\,-\,\sin^2(x)$
. . and we have: \(\displaystyle \L\:\cos(2x)\:=\:1\,-\,2\sin^2(x)\)


So we have three forms of the identity!

. . \(\displaystyle \L\cos(2x) \:=\:\begin{Bmatrix}\cos^2(x)\,-\,\sin^2(x) \\ 1\,-\,2\sin^2(x) \\ 2\cos^2(x)\,-\,1\end{Bmatrix}\;\begin{array}{ccc}[1] \\ [2] \\ [3]\end{array}\)$\displaystyle \;\;\;\;\begin{array}{cc}\text{Memorize them}\\ \text{or keep them on a list}\end{array}$

[3] is used when the result is to be in terms of cosine only.

[2] is used when the result is to be in terms of sine only.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Your problem should have taken only a few steps.


$\displaystyle \cos(4x) \;= \;2\cdot\cos^2(2x) \,-\,1$

. . . . . . $\displaystyle = \;2\left[\cos(2x)\right]^2\,-\,1$
. . . . . . . . . . . .$\displaystyle \downarrow$
. . . . $\displaystyle = \;2\left[2\cos^2(x)\,-\,1\right]^2\,-\,1$

. . . . $\displaystyle = \;2\left[4\cos^4(x) \,-\,4\cos(x) \,+\,1\right] \,-\,1$

. . . . $\displaystyle = \;8\cos^4(x)\,-\,8\cos^2(x)\,+\,2\,-\,1$

. . . . $\displaystyle = \;8\cos^4(x)\,-\,8\cos^2(x) \,+\,1$