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$\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 2\cos\alpha \cos\beta \cos\gamma = 1$

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

Em.
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John Rawls
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  • The only thing you know is that $\alpha + \beta + \gamma = \pi$, so you can express everything in terms of two angles, but not one. Nonetheless, that would be a good strategy here. – user49640 May 25 '17 at 02:14
  • You can use MathJax to display Greek letters. Most follow the pattern \Name \name for the upper and lower cases, like \Gamma \gamma $\Gamma, \gamma$. It's outlined in the tutorilal. – Em. May 25 '17 at 02:19
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    It's difficult to prove a false statement. Note, for instance, that if $\alpha = \beta = \gamma = \pi/3$ then $\cos\alpha = \cos\beta = \cos\gamma = 1/2$ and $\cos 2\alpha = \cos 2\beta = \cos 2\gamma = -1/2$, so that the left-hand side is $-5/4$, not $1$. Perhaps you meant $\cos^2 \alpha$ for $\cos 2\alpha$, etc. In that case, though, this question becomes a duplicate of one you posted in January. (I think there are other duplicates around here, as well, but yours was the first one I found.) – Blue May 25 '17 at 02:23
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    @Blue This would explain why I was having trouble writing out a full solution! – B. Mehta May 25 '17 at 02:25
  • Seee https://in.answers.yahoo.com/question/index?qid=20120228220904AAiTPyt – lab bhattacharjee May 25 '17 at 04:57

3 Answers3

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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 $-1=1$.

Quang Hoang
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I think you mean the following problem.

Let $\alpha+\beta+\gamma=\pi$. Prove that: $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1$$

We need to prove that $$\cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta\cos(\alpha+\beta)=\sin^2(\alpha+\beta)$$ or $$\cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)^2$$ or $$\cos^2\alpha+\cos^2\beta-2\cos^2\alpha\cos^2\beta=\sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta$$ or $$\cos^2\alpha(1-\sin^2\beta)+\cos^2\beta(1-\sin^2\alpha)-2\cos^2\alpha\cos^2\beta=0,$$ which is obvious.

Done!

Michael Rozenberg
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Your first thought is good, but it's not possible to write all the angles in terms of just one angle. Instead, we can express $\gamma = \pi - \alpha - \beta$, which might help!

$$\begin{align}&\cos (2\alpha) + \cos (2 \beta) + \cos (2\pi - 2\alpha - 2\beta) + 2 \cos (\alpha) \cos (\beta) \cos (\pi - \alpha - \beta) \\=& \cos (2\alpha) + \cos (2 \beta) + \cos (2\alpha + 2\beta) - 2 \cos (\alpha) \cos (\beta) \cos ( \alpha +\beta) \end{align}$$ Then, expand out using standard identities to finish off.

B. Mehta
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