If $A+B+C=0$, then prove that the value of the determinant is $0$.

Question

I'll state the question from my textbook below:

If $A+B+C=0$, the prove that $\begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0$.

This is how I tried solving the problem:

$LHS = \begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix}$

$= 1(1- \cos^2 A) - \cos C (\cos C - \cos A \cos B) + \cos B (\cos A \cos C - \cos B)$

$= 1 + 2 \cos A \cos B \cos C - (\cos^2 A +\cos^2 B + \cos^2 C)$

I don't know how to proceed further. I tried using the fact that $A+B+C=0$ but it didn't lead to anything I could solve. I don't know where is it supposed to be used.

Also, I read a solution to this problem somewhere in which the term $(\cos^2 A +\cos^2 B + \cos^2 C)$ was replaced by $1 + 2 \cos A \cos B \cos C$ as $A+B+C=0$. Are these two terms equal for the given condition? Also, is there a way to prove the statement without using this fact?

Any help would be appreciated.

Have you tried substituting $A = -B-C \implies \cos A = \cos (B+C)$ and expanding the associated cosine expression with the angle sum and difference identities (https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities)? — MrYouMath, Feb 28 '18 at 08:42
If you follow MrYouMath clue you will find 0 indeed : https://www.wolframalpha.com/input/?i=1+%2B+2+%5Ccos+A+%5Ccos+B+%5Ccos+C+-+(%5Ccos%5E2+A+%2B%5Ccos%5E2+B+%2B+%5Ccos%5E2+C)+where+A%2BB%2BC%3D0 — Netchaiev, Feb 28 '18 at 08:58
@MrYouMath I did try to apply that but was unable to solve the resulting expression. I actually don't know where to apply it, in the cosine function or the cosine square function. — Shinsekai no Kami, Mar 01 '18 at 10:51

score 4 · Answer 1 · answered Feb 28 '18 at 18:31

When at least one of $A,B$ or $C$ is not an integer multiple of $\pi$, the vector $(\sin A,\sin B,\sin C)$ is nonzero. However, as $A+B+C=0$, the sum-of-angle formula implies that $$ \pmatrix{1&\cos C&\cos B \\ \cos C&1&\cos A\\ \cos B&\cos A& 1} \pmatrix{\sin A\\ \sin B\\ \sin C} =0 $$ and hence the matrix is singular.

In case all of $A,B,C$ are integer multiples of $\pi$, you may complete the proof by a continuity argument, or by noticing that the matrix must have at least two identical rows, because up to a permutation, $(\cos A,\cos B,\cos C)$ is either $(1,1,1)$ or $(1,-1,-1)$.

score 2 · Answer 2 · answered Feb 28 '18 at 17:59

2

It's worth noting that we are dealing with a Gram determinant of three unit vectors in $3$-space. This determinant is $0$ if and only if the vectors are linearly dependent, that is, lying in the same plane. The condition $A+B+C=0$ ensures this.

It is also worth noting that the determinant always equals

$$4 \sin s \sin(s-A) \sin (s-B) \sin (s-C)$$ where $s= \frac{A+B+C}{2}$, so we get a necessary and sufficient condition for the determinant to be $0$, as above.

answered Feb 28 '18 at 17:59

orangeskid

53,909

Why is $A+B+C=0$ necessary? Can't we just have, say, $s=A$? – user1551 Feb 28 '18 at 18:09
@user1551: I did not say it's necessary. Rather, I said that we should have three vectors in the same plane. So $A+ B+C = 2 \pi $ also works. – orangeskid Feb 28 '18 at 18:29
I see. I misunderstood what "as above" refers to in your answer. – user1551 Feb 28 '18 at 18:32
I am not well versed with the study of vectors currently. But can you please explain that how does the determinant always equal $4 \sin s \sin(s-A) \sin (s-B) \sin (s-C)$? – Shinsekai no Kami Mar 01 '18 at 11:17

score 2 · Answer 3 · answered Mar 02 '18 at 04:49

If one of the angles, say $A=0$, then two rows become identical and the determinant becomes zero.

Else we multiply the rows in order by $\sin A, \sin B, \sin C$ and we get

\begin{vmatrix}\sin A & \sin A\cos C & \sin A\cos B \\ \sin B\cos C & \sin B & \sin B\cos A \\ \sin C\cos B & \sin C\cos A & \sin C \end{vmatrix}

Now with $R_1 \rightarrow R_1+R_2+R_3$ we see for instance that $$a_{11} = \sin A +\sin B \cos C+\sin C \cos B = \sin A+\sin (B+C) = \sin A - \sin A =0$$

i.e. the first row becomes null and hence the determinant equals zero.

score 1 · Accepted Answer · answered Mar 01 '18 at 12:41

1

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$\cos^2A+\cos^2B+\cos^2C =1+\cos^2A-\sin^2B+\cos^2C$$

$$=1+\cos(A+B)\cos(A-B)+\cos^2C$$

$$=1-\cos C\cos(A-B)+\cos^2C\text{ as }\cos(A+B)=\cos(\pi-C)=?$$

$$=1-\cos C[\cos(A-B)-\cos C]$$

$\cos(A-B)-\cos C=\cos(A-B)+\cos(A+B)=?$

answered Mar 01 '18 at 12:41

lab bhattacharjee

274,582

Thank you for the answer. I understood how to proceed now. Also, here $A+B+C=0$ not $\pi$. I think you would like to edit accordingly as $\cos^2A+\cos^2B+\cos^2C$ has to be equal to $1 + 2 \cos A \cos B \cos C$ and not $1 - 2 \cos A \cos B \cos C$. – Shinsekai no Kami Mar 01 '18 at 13:04

If $A+B+C=0$, then prove that the value of the determinant is $0$.

4 Answers4