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If I have the following equation:

$$\cos(x+a)=\cos(x+y+z)$$

Can I take the inverse cos of both sides?

$$\begin{align} \cos^{-1}(\cos(x + a))&=\cos^{-1}(\cos(x+y+z)) \\ x+a&=x+y+z \\ a&=y+z \end{align}$$

Is this correct?

Blue
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user367640
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    No, you can hardly ever take inverse cosine of both sides, and there appears to be no reason for this to be one of those. –  Dec 10 '18 at 00:10

2 Answers2

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No it'is wrong, let instead consider the trigonometric circle to derive that

$$\cos \theta = \cos \alpha \implies \theta = \alpha+2k\pi \quad \lor \quad \theta = -\alpha+2k\pi$$

Refer also to the related

user
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    Or if you want a more elementary version of what gimusi is saying, OP - in case you're not familiar with the logic notation and such - it's because:

    $$\cos(x) = \cos(x + 2\pi) = \cos(x + 4\pi) = ... = \cos(x + 2k\pi)$$

    for any integer $k$. That is to say, it's because the cosine function is periodic, with period $2\pi$. We also have this because cosine is an even function, i.e.

    $$\cos(-x) = \cos(x)$$

     – PrincessEev Dec 10 '18 at 00:16
  • @EeveeTrainer That's not complete we also have $$\cos(x) = \cos(-x + 2\pi) = \cos(-x + 4\pi) = ... = \cos(-x + 2k\pi)$$ – user Dec 10 '18 at 00:17
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    I suggest simply to draw the trigonometric circle to check that. it is always the best way to see those simple equalities or inequalities. – user Dec 10 '18 at 00:18
  • The answer is $a=y+x$. I tried converting to exponential functions and taking the natural logarithm of both sides but I just get zero. – user367640 Dec 10 '18 at 00:18
  • Yeah, my bad gimusi, I forgot to add in that bit. Should be fixed now. – PrincessEev Dec 10 '18 at 00:19
  • @EeveeTrainer Perfect that's fine now but I also suggest to consider the interpretation directly on the trigonometric circle. – user Dec 10 '18 at 00:21
  • I think I do get it. $a$ has to be equal to $y+z$ no matter what right? Picturing it on the circle helps. If $a$ was $2\pi$ then that just means $cos(x+a) = cos(x)$. Since it is just one complete rotation. Also meaning that $y+z$ has to equal $2\pi$ to satisfy the equation. – user367640 Dec 10 '18 at 00:25
  • @user367640 That's one (set of) solutions. Can you find the other one? – user Dec 10 '18 at 00:29
  • infinite more values, correct? If $cos(x)$ experiences a phase shift, $a$ then no matter what that phase shift is, $y+z$ is equal to it. – user367640 Dec 10 '18 at 00:32
  • @user367640 Yes a set of solution is $a=y+z+2k\pi \quad \forall k\in \mathbb{Z}$. But what about the second case? – user Dec 10 '18 at 00:34
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Notice that due to $\cos\theta$ having a period of $2\pi$, for any $\lambda, \mu\in\Bbb Z, \theta\in \Bbb R$, we have:

$$\cos(\theta+2\lambda\pi)=\cos(\theta+2\mu\pi)$$

$\lambda=\mu$ is not a necessity to this end.

Effectively in your expression we have that $a=y+z+2\lambda\pi, \lambda \in \Bbb Z$.

$\lambda=0$ is a possibility, and there $a=y+z$, but there are infinitely many other possibilities as well.

Rhys Hughes
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