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I have been struggling with this equation. Is it unsolvable?

$$\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=c$$

$\beta$ and $c$ are constant. How can I solve for the angle $\alpha$?

PS: As you might imagine, I'm still new to this, so please be easy on me. Thanks for your help.

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  • $\begingroup$ Try the angle sum formula? $\endgroup$ Dec 5, 2020 at 15:48
  • $\begingroup$ i've already tried that and it led me nowhere. That's just me though. $\endgroup$
    – Yniq
    Dec 5, 2020 at 15:56
  • $\begingroup$ @Yniq: Welcome to Math.SE! What struggles i.e. tries have you tried? We may point out your mistakes so as you have a better understanding of the concepts. $\endgroup$ Dec 5, 2020 at 15:57
  • $\begingroup$ primarily i tried using trigonometric identities to ultimately simplify enough to use an arcfunction. I don't see any other way really. $\endgroup$
    – Yniq
    Dec 5, 2020 at 15:59
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    $\begingroup$ "i've already tried that ..." Please always include what you've tried as part of your question. (Comments are easily overlooked.) This will help people avoid wasting time (theirs and yours) suggesting things you already attempted or explaining things you already understand. By showing your work, someone may be able to point out that something as simple as a sign error is holding you back. ... Oh, and Welcome to Math.SE! :) $\endgroup$
    – Blue
    Dec 5, 2020 at 16:02

2 Answers 2

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Rewrite $$\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=c$$ as $$ (1-c)\cos\alpha\cos\beta + (1+c)\sin\alpha\sin\beta = 0$$ Taking $a = (1-c)\cos\beta ~$ and $~b = -(1+c)\sin\beta$, we get $$a\cos\alpha = b\sin\alpha$$ Now, the case $~\cos\alpha = 0~$ is not interesting. Assume $\cos \alpha \ne 0$ and divide by it to get $$b\tan \alpha = a$$ Hence, if either of $a$ and $b$ is $0$, then so is another. Assume, neither is zero and divide by $b$ to get $$\tan \alpha = \frac ab = \frac{c - 1}{c + 1} \cot \beta$$

Now, you can decide if it is solvable or not.

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  • $\begingroup$ I think you accidentally swapped the angle sum formulas for sine and cosine. Your comment was of big help anyway. Thank you! $\endgroup$
    – Yniq
    Dec 5, 2020 at 18:04
  • $\begingroup$ @Yniq I'm glad that it helps and sorry for that error. $\endgroup$
    – VIVID
    Dec 5, 2020 at 18:28
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I think I've found a solution to this problem, solving for $\alpha$. Following VIVID's answer i arrived at: $$\cos(\alpha)\cdot\cos(\beta)(1-c)-\sin(\alpha)\cdot\sin(\beta)(1+c)=0$$ following the angle sum formula: $$\cos(x+y)=\cos(x)\cdot\cos(y)-\sin(x)\cdot\sin(y)$$ finally i can express $\alpha$ in terms of $\beta$ and c, which gives...

$$\alpha(\beta,c) = \arctan\Biggr(\frac{1-c}{1+c}\cdot\frac{1}{\tan(\beta)}\Biggr)$$

I hope there is no big oversight and this is correct. Also, I've never used LaTeX before, so writing this was painfully slow, so I skipped some steps. I thank everyone for their help.

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