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$$4\sin\theta +3\cos\theta = 5$$

How would we solve this trigonometric equation using vectors? Since I'm not advanced, I do not truly know where to use vector product.

Regards!

mechanodroid
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Busi
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  • Can you clarify what particular method you want to use. A simple Google search doesn't seem to yield any obvious candidate. – user0 Sep 03 '18 at 08:09
  • Also, you may want to show some if your work or thoughts about the question. Otherwise your post may possibly be downvoted and closed. – user0 Sep 03 '18 at 08:11
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    https://math.stackexchange.com/questions/757497/find-4-cos-theta-3-sin-theta-given-that-4-sin-theta-3-cos-theta-5/757498#757498 – lab bhattacharjee Sep 03 '18 at 12:46

1 Answers1

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Your equation is equivalent to $$\langle (3,4), (\cos\theta, \sin\theta)\rangle = 5$$

Cauchy-Schwarz inequality gives $$|\langle (3,4), (\cos\theta, \sin\theta)\rangle| \le \|(3,4)\|\|(\cos\theta, \sin\theta)\| = 5$$

so $(3,4)$ and $(\cos\theta, \sin\theta)$ are collinear. Therefore

$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac43$$ so $\theta= \arctan\frac43 + 2k\pi$ for $k \in \mathbb{Z}$.

mechanodroid
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  • It's only a congruence modulo 2π. – Bernard Sep 03 '18 at 08:34
  • @Bernard True, thanks. – mechanodroid Sep 03 '18 at 08:37
  • How did you get that $(3,4)$ and $(\cos\theta, \sin\theta)$ collinear? Can you also specify where I can use this method? – Busi Sep 03 '18 at 08:52
  • This step is a bit confusing: $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac43$ If you can explain it further, I would be glad. Sorry for asking multiple questions in a row. This may demand an edit. – Busi Sep 03 '18 at 08:53
  • @Busi The equality $|\langle x,y\rangle| = |x||y|$ in Cauchy-Schwarz inequality holds if and only if vectors $x,y$ are linearly dependent, i.e. there exists a scalar $\lambda$ such that $y = \lambda x$. Here we get $(\cos\theta, \sin\theta) = \lambda(3,4) = (3\lambda, 4\lambda)$ so $$\frac{\sin\theta}{\cos\theta} = \frac{4\lambda}{3\lambda} = \frac43$$ – mechanodroid Sep 03 '18 at 09:09
  • @mechanodroid I still didn't get it properly. I wish you explained your calculation further. As stated in the question, I'm not that advanced. – Busi Sep 03 '18 at 09:14
  • @Busi Ok, do you know how the inner product and norm are defined? For vectors $(x,y), (x', y') \in \mathbb{R}^2$ we define $\langle (x,y), (x',y')\rangle = xx' + yy'$ and $|(x,y)| = \sqrt{x^2 + y^2}, |(x',y')| = \sqrt{x'^2 + y'^2}$. The Cauchy-Schwarz inequality states that $$|\langle (x,y), (x',y')\rangle| \le |(x,y)||(x', y')|$$ and $$|\langle (x,y), (x',y')\rangle| = |(x,y)||(x', y')| \text{ if and only if } \exists \lambda \in \mathbb{R} \text{ such that } (x', y') = \lambda(x,y)$$ – mechanodroid Sep 03 '18 at 11:52
  • I noticed that if $\theta \in \mathbb{R}$ is such that $4\sin\theta +3\cos\theta = 5$, then $$|\langle (3,4), (\cos\theta, \sin\theta)\rangle| = 5 = |(3,4)||(\cos\theta, \sin\theta)|$$ so the equality condition above implies that $\exists \lambda\in\mathbb{R}$ such that $(\cos\theta, \sin\theta) = \lambda(3,4)$. – mechanodroid Sep 03 '18 at 11:54
  • @mechanodroid Can I apply your method to kinda trigonometric equations? – Busi Sep 03 '18 at 12:24
  • @Busi This one was pretty much ad hoc, but you can try it whenever you can write your equality in terms of inner products and norms. – mechanodroid Sep 03 '18 at 12:37