in $\Delta ABC$, such


prove that $$A=B=C=\dfrac{\pi}{3}$$

My try: use

$$\sin{x}+\sin{y}=2\sin{\dfrac{x+y}{2}}\cos{\dfrac{x-y}{2}}$$ then \begin{align*}&\sin{A}+\cos{B}\\ &=\sin{A}+\sin{(\dfrac{\pi}{2}-B)}=2\sin{\dfrac{\pi+2(A-B)}{4}}\cos{\dfrac{A+B-\dfrac{\pi}{2}}{2}}\\ &=2\sin{\dfrac{\pi+2(A-B)}{4}}\cos{\dfrac{\pi-2C}{4}} \end{align*} my idea is take $\sin{A}+\cos{B}\le f(C)$?and if only if $A=B$,BUt I can't ,Thank you someone can help me


Fix $A \in (0, \frac{\pi}{2}), A \neq \frac{\pi}{3}$. Let $f: [0, \frac{\pi}{2}) \to \mathbb R$ be $$f(x)= \sin(A) +\tan(x)+\cos(\pi-A-x) \,.$$

As $f$ is continuous, $f(0)=\sin(A)+\cos(\pi-A) \leq 2 < \dfrac{3\sqrt{3}+1}{2} $ and $\lim_{x \to \frac{\pi}{2}^-} f(x)= +\infty$, by the Intermediate Value Theorem there exists some $x$ so that

$$f(x)=\dfrac{3\sqrt{3}+1}{2} \,.$$

Then $A=A, B= \pi-A -x, C=x$ is a counterexample to your problem. So you cannot.

P.S. We actually prove something stronger: the angle $A$ can take any value in $(0, \frac{\pi}{2})$.

| cite | improve this answer | |
  • $\begingroup$ Thank you,meaning we can't have $A=B=C$? $\endgroup$ – user94270 Oct 27 '13 at 3:05
  • $\begingroup$ why $f(0)=0$? Thank you,are you define? $\endgroup$ – user94270 Oct 27 '13 at 3:34
  • $\begingroup$ @nanchangjian Sorry typo there, I meant $f(0) < \dfrac{3\sqrt{3}+1}{2} $ but typed it in a hurry. Note that $3\sqrt{3}=\sqrt{27}> \sqrt{16}=4$. $\endgroup$ – N. S. Oct 27 '13 at 3:56
  • $\begingroup$ and use Value theorem, you can only prove $x\in (0,\dfrac{\pi}{2})$, $\endgroup$ – user94270 Oct 27 '13 at 3:57
  • $\begingroup$ @nanchangjian Doesn't matter, you already know that $A \neq \frac{\pi}{3}$. $\endgroup$ – N. S. Oct 27 '13 at 3:59

The plot of $$\sin A + \cos B + \tan \left(\pi - A - B \right)$$ indicates that the expression attains any particular value infinitely often. Consequently, while $A = B = \pi - A - B = \pi/3$ produces the value $\frac{1}{2}\left(3\sqrt{3}+1\right)$, this is far from the only solution to your equation.

enter image description here

| cite | improve this answer | |

Your Answer

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