If $\cos\beta = −\cos \alpha$ and $\sin \beta =−\sin \alpha $, what must be the relation between $\alpha$ and $\beta$?

Question

If $\cos\beta = −\cos \alpha$ and $\sin \beta =−\sin \alpha $, what must be the relation between $\alpha$ and $\beta$?

I know that then $\alpha = \beta + k \pi$, but I cannot understand why. Would anyone explain this for me please?

Answer 1

Use Prosthaphaeresis Formulas,

$$0=\cos\alpha+\cos\beta=2\cos\dfrac{\alpha+\beta}2\cos\dfrac{\alpha-\beta}2$$

$$0=\sin\alpha+\sin\beta=2\sin\dfrac{\alpha+\beta}2\cos\dfrac{\alpha-\beta}2$$

What if $\cos\dfrac{\alpha-\beta}2$$\ne0$

Answer 2

$(\cos \beta,\sin \beta)$ are coordinates of a point $M$ on unit circle.
The point $P(\cos \alpha, \sin \alpha)$ is symmetrical to $M$ through $0$.
This gives $$\alpha=\beta + (2k+1)\pi, k\in \mathbb{Z}.$$

Answer 3

By the definition we have

$$\cos\beta = −\cos \alpha=\cos (\pi-\alpha) \iff \beta=\pi-\alpha+2k_1\pi \quad \lor \quad -(\pi-\alpha)+2k_2\pi$$

$$\sin\beta = −\sin \alpha=\sin (-\alpha) \iff \beta=-\alpha+2k_3\pi \quad \lor \quad \pi-(-\alpha)+2k_4\pi$$

then we have

• $\beta=\pi-\alpha+2k_1\pi=-\alpha+2k_3\pi\iff \pi =2h\pi$ which is impossible
• $\beta=\pi-\alpha+2k_1\pi=\pi-(-\alpha)+2k_4\pi\iff 2\alpha =2h\pi \iff \alpha =h\pi$

and therefore finally we have

• $\alpha =h\pi$
• $\beta=\pi-h\pi+2k\pi=(1-h)\pi+2k\pi$

or

• $\alpha+\beta=\pi + 2k\pi$

Answer 4

Multiply the first equation by $\sin\beta$ and the second equation by $\cos\beta$ and subtract: $$ \sin\alpha\cos\beta-\cos\alpha\sin\beta=0 $$ that is, $\sin(\alpha-\beta)=0$. This entails $$ \alpha=\beta+2k\pi \qquad\text{or}\qquad \alpha=\beta+\pi+2k\pi $$ ($k$ an integer). Now let's test the first set of solutions: if $\alpha=\beta+2k\pi$, then $\cos\alpha=\cos\beta$ and $\sin\alpha=\sin\beta$: it's impossible that $\cos\beta=-\cos\beta$ and $\sin\beta=-\sin\beta$.

For the second set of solutions, we have \begin{align} \cos\alpha&=\cos(\beta+\pi+2k\pi)=-\cos\beta \\ \sin\alpha&=\sin(\beta+\pi+2k\pi)=-\sin\beta \end{align} for every $\beta$. So the conclusion is that your conditions are equivalent to $$ \alpha=\beta+\pi+2k\pi \qquad\text{($k$ integer)} $$

Answer 5

$$\tan\alpha=\dfrac{-\sin\beta}{-\cos\beta}=?$$

$\implies\alpha=n\pi+\beta$ where $n$ is any integer

Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$