1. ## Complete the Identity

1-cos(2[theta])+cos(6[theta])-cos(8[theta]) = ?

I got up to...
1 - ( cos^2 ([theta]) - sin^2 ([theta]) ) + cos(6[theta]) - cos(8[theta])

Thank you,
Jason

2. Hello, Jason!

Are there more instructions?
Otherwise, we could stop anywhere . . .

"Complete the identity" (?)

. . $\displaystyle 1-\cos2\theta + \cos 6\theta - \cos8\theta \;=\; ?$

We have: .$\displaystyle 1 - \cos2\theta + \cos6\theta - \cos8\theta$

. . . $\displaystyle = \;1 - \overbrace{(\cos^2\!\theta - \sin^2\!\theta)} + \cos6\theta - \cos8\theta$

. . . $\displaystyle =\; \underbrace{1 - \cos^2\!\theta} + \sin^2\!\theta + \cos6\theta - \cos8\theta$

. . . $\displaystyle = \quad \underbrace{\sin^2\!\theta + \sin^2\!\theta} + \cos6\theta - \cos8\theta$

. . . $\displaystyle = \qquad 2\sin^2\!\theta + \underbrace{\cos6\theta - \cos8\theta}$

. . . $\displaystyle = \qquad 2\sin^2\!\theta - 2\sin7\theta\sin(\text{-}\theta)$

. . . $\displaystyle =\qquad 2\sin^2\!\theta + 2\sin7\theta\sin\theta$

. . . $\displaystyle = \qquad 2\sin\theta\underbrace{(\sin\theta + \sin7\theta)}$

. . . $\displaystyle = \qquad 2\sin\theta \cdot 2\sin4\theta\cos3\theta$

. . . $\displaystyle = \qquad 4\sin\theta\sin4\theta\cos3\theta$

. . . . . . . . . . . . . $\displaystyle \vdots$

. . . . . . . . . . and so on . . .

3. Perfect! The problem I have is how to change:

to

Thanks again!