Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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? –  Blue Jul 22 '13 at 4:45
    
Hello:if $A=B=C$ can't such contition. –  math110 Jul 22 '13 at 4:49
    
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)? –  Blue Jul 22 '13 at 4:59
    
oh,This problem is my frend ask me,I don't kown this problem is true.Thank you –  math110 Jul 22 '13 at 5:11
    
@Blue,I have edit, and $+$ replace $-$, –  math110 Jul 22 '13 at 5:13

2 Answers 2

up vote 3 down vote accepted

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 $$\Rightarrow \frac{1}{\cos A}=4+\frac{\sin \frac{A}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}$$

share|improve this answer
    
It's nice!,Thank you –  math110 Jul 24 '13 at 11:58
    
Welcome! @math110 –  Samrat Mukhopadhyay Jul 24 '13 at 12:36
    
Since you eventually divided the determinant equation by $\sin((C-B)/2)$, you're restricting yourself to the case $B\neq C$. –  Blue Jul 24 '13 at 14:00
    
Correct @Blue. I'll add that part also. –  Samrat Mukhopadhyay Jul 24 '13 at 14:04

Hint:

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.

share|improve this answer
    
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 "$1/4$". –  Blue Jul 22 '13 at 6:10
    
Thank you, I have ask my frend, $\Delta ABC$ not equilateral triangle.sorry. –  math110 Jul 22 '13 at 6:53
    
@math110 - I never said it was :) –  nbubis Jul 22 '13 at 7:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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