in $\Delta ABC $ not An equilateral triangle, let $$\begin{vmatrix} 2\sin{A}\sin{C}-\cos{B}&2\sin{A}\sin{B}-\cos{C}\\ 4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+\cos{B}&4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+\cos{C} \end{vmatrix}=0$$

prove that $$\dfrac{1}{\cos{A}}-\dfrac{\sin{\dfrac{A}{2}}}{\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}}=4$$

my idea: use this well known $$4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+1=\cos{A}+\cos{B}+\cos{C}$$ then we have $$\Longrightarrow \begin{vmatrix} 2\sin{A}\sin{C}-\cos{B}&2\sin{A}\sin{B}-\cos{C}\\ \cos{A}+\cos{C}+2\cos{B}-1&\cos{A}+\cos{B}+2\cos{C}-1 \end{vmatrix}=0$$

then $$(2\sin{A}\sin{C}-\cos{B})(\cos{A}+\cos{B}+2\cos{C}-1)-(2\sin{A}\sin{B}-\cos{C})(\cos{A}+\cos{C}+2\cos{B}-1)=0$$ folowing I can't work,Thank you everyone can help

  • $\begingroup$ If $A=B=C=\pi/3$, then your the left-hand side of your "prove that" equation is zero. Should there be a "$+$" in that equation instead of a "$-$"? BTW: What's your reason for believing that the equation holds, anyway? Is this a textbook exercise? Or did you come up with this on your own and are hoping it works? $\endgroup$ – Blue Jul 22 '13 at 4:45
  • $\begingroup$ Hello:if $A=B=C$ can't such contition. $\endgroup$ – math110 Jul 22 '13 at 4:49
  • $\begingroup$ If $A=B=C=\pi/3$, then isn't each entry in the original determinant $1$, so that the determinant itself is zero (so that the condition holds)? $\endgroup$ – Blue Jul 22 '13 at 4:59
  • $\begingroup$ oh,This problem is my frend ask me,I don't kown this problem is true.Thank you $\endgroup$ – math110 Jul 22 '13 at 5:11
  • $\begingroup$ @Blue,I have edit, and $+$ replace $-$, $\endgroup$ – math110 Jul 22 '13 at 5:13

I'm getting the result without using the formula.

Note that, since $\Delta ABC$ is a triangle, $$2\sin A\sin C-\cos B=\cos (A-C)-\cos(A+C)-\cos C=\cos(A-C)$$ Similarly, $$2\sin A\sin B-\cos C=\cos (A-B)$$

Now, observe that if $B=C$, the determinant condition is automatically satisfied and in that case $$\large \cos A=-\cos 2B,\ \sin \frac{A}{2}=\cos B$$ So, if the answer holds then we will be able to find a solution for $B$ and hence for $C,A$! So the result is not a general one if $B=C$. So I'm assuming $B\ne C$.

Now, from the determinant, and with the assumption stated above along with the above equalities, we get,\begin{equation} \begin{split} \ &4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\left(\cos(A-C)-\cos(A-B)\right)\\ \ =&\cos B\cos(A-B)-\cos C \cos(A-C)\\ \ \Rightarrow & 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}2\sin\left(\frac{C-B}{2}\right)\sin\left(\frac{3A-\pi}{2}\right)\\ \ =&\frac{1}{2}\left(\cos A+\cos(A-2B)-\cos A-\cos (A-2C)\right) \ =&\sin(C-B)\sin 2A\\ \ \Rightarrow & 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos\left(\frac{3A}{2}\right)\\ \ =&-\cos\left(\frac{B-C}{2}\right)\sin 2A\\ \ \Rightarrow & \sin \frac{B}{2}\sin \frac{C}{2}\cos\left(\frac{3A}{2}\right)\\ \ =& -\cos\left(\frac{B-C}{2}\right)\cos\left(\frac{A}{2}\right)\cos A\\ \end{split} \end{equation}

So, now we get \begin{equation} \begin{split} \large \frac{\cos \frac{3A}{2}}{\cos \frac{A}{2} \cos A}=& -\frac{\cos \frac{B-C}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}\\ \Rightarrow \frac{\cos \frac{3A}{2}-\cos \frac{A}{2} \cos A}{\cos \frac{A}{2} \cos A}=& \frac{-\cos \frac{B-C}{2}-\sin \frac{B}{2}\sin \frac{C}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}\\ \Rightarrow -\frac{\sin A \sin \frac{A}{2}}{\cos \frac{A}{2} \cos A}=&\frac{-\cos \frac{B}{2}\cos \frac{C}{2}-2\sin \frac{B}{2}\sin \frac{C}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}\ \\ \Rightarrow \frac{2\sin^2 \frac{A}{2}}{\cos A}= &\frac{\cos \frac{B}{2}\cos \frac{C}{2}}{\sin \frac{B}{2} \sin \frac{C}{2}}+2\\ \Rightarrow \frac{1}{\cos A}=3+\cot \frac{B}{2} \cot \frac{C}{2} \end{split} \end{equation} Now, observe that $$\sin \frac{A}{2}=\cos \frac{B+C}{2}=\cos \frac{B}{2}\cos \frac{C}{2}-\sin \frac{B}{2}\sin \frac{C}{2}\\ \Rightarrow \frac{\sin \frac{A}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}=\cot \frac{B}{2} \cot \frac{C}{2}-1$$ Hence we get $$\Rightarrow \frac{1}{\cos A}=4+\frac{\sin \frac{A}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}$$

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  • $\begingroup$ Welcome! @math110 $\endgroup$ – Samrat Mukhopadhyay Jul 24 '13 at 12:36
  • $\begingroup$ Since you eventually divided the determinant equation by $\sin((C-B)/2)$, you're restricting yourself to the case $B\neq C$. $\endgroup$ – Blue Jul 24 '13 at 14:00
  • $\begingroup$ Correct @Blue. I'll add that part also. $\endgroup$ – Samrat Mukhopadhyay Jul 24 '13 at 14:04


Your identity is equivalent to:

$$\frac{\sin(A/2)\sin(B/2)\sin(C/2)}{\cos A} + \sin^2(A/2) = 4\sin(A/2)\sin(B/2)\sin(C/2)$$ $$\frac{1}{2}\left(1-\cos(A)\right)=\left(\cos(A)+\cos(B)+\cos(C)-1\right)\left(1 - \frac{1}{\cos(A)} \right)$$

Notice that this form is quite close to the fraction you get from the determinant.

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  • $\begingroup$ The $\sin(A/2) \sin(B/2)\sin(C/2)$ on the left-hand side of the first equation becomes $\frac{1}{4}\left( \cos A + \cos B + \cos C - 1 \right)$. You lost the "$1/4$". $\endgroup$ – Blue Jul 22 '13 at 6:10
  • $\begingroup$ Thank you, I have ask my frend, $\Delta ABC$ not equilateral triangle.sorry. $\endgroup$ – math110 Jul 22 '13 at 6:53
  • $\begingroup$ @math110 - I never said it was :) $\endgroup$ – nbubis Jul 22 '13 at 7:44

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