If $0< \alpha, \beta< \pi$ and $\cos\alpha + \cos\beta-\cos (\alpha + \beta) =3/2$ then prove $\alpha = \beta= \pi/3$

How do I solve for $\alpha$ and $\beta$ when only one equation is given? By simplification I came up with something like $$ \sin\frac{\alpha}{2} \sin\frac{\beta}{2} \cos \frac {\alpha +\beta}{2}=\frac{1}{8}. $$ I don't know if this helps. How to do this?

  • $\begingroup$ are $\alpha,\beta$ angles in a given triangle? $\endgroup$ – Dr. Sonnhard Graubner Jan 14 '17 at 15:38
  • $\begingroup$ @Dr.SonnhardGraubner nope $\endgroup$ – ray Jan 14 '17 at 16:07

Set $p=\cos\beta$ and $q=\sin\beta$; then your equality can be written $$ \cos\alpha+p-p\cos\alpha+q\sin\alpha=\frac{3}{2} $$ or $$ (1-p)\cos\alpha+q\sin\alpha=\frac{3}{2}-p $$ Set $X=\cos\alpha$ and $Y=\sin\alpha$; then the equation becomes $$ \begin{cases} (1-p)X+qY=\frac{3}{2}-p\\[6px] X^2+Y^2=1 \end{cases} $$ and the distance of the line from the origin should be $\le1$, or the line and the circle wouldn't meet: $$ \frac{(\frac{3}{2}-p)^2}{(1-p)^2+q^2}\le1 $$ that becomes $$ \frac{9}{4}-3p+p^2\le 1-2p+p^2+q^2 $$ that is, recalling that $p^2+q^2=1$, $$ p^2-p+\frac{1}{4}\le0 $$ Can you finish up?



Use Prosthaphaeresis Formula on $\cos\alpha,\cos\beta$

and Double angle formula on $$\cos 2\cdot\dfrac{\alpha+\beta}2$$ to get

$$2\cos^2\dfrac{\alpha+\beta}2-2\cos\dfrac{\alpha-\beta}2\cos\dfrac{\alpha+\beta}2+\dfrac12=0\ \ \ \ (1)$$ which is a Quadratic Equation in $\cos\dfrac{\alpha+\beta}2$ whose discriminant must be $\not<0$

i.e., $$\left(2\cos\dfrac{\alpha-\beta}2\right)^2-4=-2\sin^2\dfrac{\alpha-\beta}2\ge0$$

But for real $\dfrac{\alpha-\beta}2,$ $$\sin^2\dfrac{\alpha-\beta}2\ge0$$

So, we have $$\sin^2\dfrac{\alpha-\beta}2=0\iff\sin\dfrac{\alpha-\beta}2=0$$

$\implies\dfrac{\alpha-\beta}2=n\pi$ where $n$ is any integer

But as $0<\alpha,\beta<\pi.n=0\implies\alpha-\beta=0$

So, $(1)$ is reduced to $$0=2\cos^2\beta-2\cos\beta+\dfrac12=\dfrac{(2\cos\beta-1)^2}2$$

So, $\cos\beta=\dfrac12=\cos\dfrac\pi6\implies\beta=2m\pi\pm\dfrac\pi6$ where $m$ is any integer

But as $0<\alpha<\pi$

  • $\begingroup$ with prosthaphaeresis formula, I get a $\cos (\frac {\alpha - \beta}{2})$. How to deal with it? $\endgroup$ – ray Jan 14 '17 at 16:12
  • 1
    $\begingroup$ @ray, Please find the updated answer. $\endgroup$ – lab bhattacharjee Jan 15 '17 at 3:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.