$\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 2\cos\alpha \cos\beta \cos\gamma = 1$
I really didn't know how to solve this problem and I am very unused to the utilization of trigonometric identities, I was wondering if I may have some assistance in this problem with detailed explanations
I was thinking more along the lines of making all of the angles in terms of one angle
Best Answer
It looks like you got a wrong identity.
Similar to/Extend Mehta's answer, use the identities: $$\begin{aligned}\cos(2\alpha)+\cos(2\beta) &= 2\cos(\alpha+\beta)\cos(\alpha-\beta),\\ &= -2\cos(\gamma)\cos(\alpha-\beta) \end{aligned}$$ and $$\cos(2\gamma)\color{\red}+1 = 2 \cos^2\gamma.$$ Adding them up we obtain $$\begin{aligned} \cos(2\alpha)+\cos(2\beta)+\cos(2\gamma) +1 &= -2\cos\gamma\cos(\alpha-\beta)+2\cos^2\gamma\\ &= -2\cos\gamma(\cos(\alpha-\beta)+\cos(\alpha+\beta))\\ &= -4\cos\gamma\cos\alpha\cos\beta. \end{aligned}$$
So the identity we have here is $$\cos(2\alpha)+\cos(2\beta)+\cos(2\gamma)+1 = - 4 \cos\alpha\cos\beta\cos\gamma,\tag{1}$$ which is different from what you ask for. We can check by specific values of $(\alpha,\beta,\gamma)$. For example $(\pi/2,\pi/4,\pi/4)$ turns (1) into $0=0$ while your identity would be $-1=1$.