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If $A+B+C=π$, prove that $$\cos^2A+\cos^2B-\cos^2C=-2\cos A\cdot\cos B\cdot\cos C.$$

ATTEMPT:
Given $$A+B+C=π,$$ $$A+B=π-C$$ Taking "cos" on both sides $$\cos(A+B)=-\cos C.$$

Now, \begin{align*} L.H.S&=\cos^2A+\cos^2B-\cos^2C\\ &=\frac{1+\cos 2A}{2}+\frac{1+\cos 2B}{2}-\cos^2 C\\ &=\frac{2+\cos 2A+\cos 2B}{2}-\cos^2C. \end{align*}

How should I complete now?

Em.
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Iaamuser user
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    Let $A=0$, $B=0$, $C=π$. Then, $$\cos^2A+\cos^2B-\cos^2C=1+1-(-1)=3 \neq 2 = -2\cos A\cdot\cos B\cdot\cos C$$ – KonKan Feb 08 '16 at 02:09
  • Hint . Let $A=(m+d)/2$ and $ B=(m-d)/2.$ For the RHS, note that $\cos A \cos B=(1/2) (\cos (A+B)+\cos (A-B))=(1/2)(\cos 2 m+\cos 2 d).$ – DanielWainfleet Feb 08 '16 at 02:27
  • the question is wrong once I used Wolfram expands polynomials. – runaround Feb 08 '16 at 02:33
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    The statement as given is false. It's a near miss of this identity $$1 -\cos^2 A - \cos^2 B - \cos^2 C-2\cos A\cos B\cos C= 0$$ – Blue Feb 08 '16 at 02:34
  • Changing the $\cos^2C$ on the left-hand side to $\sin^2C$ gives a valid identity. – Blue Feb 08 '16 at 02:42

4 Answers4

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I edited to prove $$1 -\cos^2 A - \cos^2 B - \cos^2 C-2\cos A\cos B\cos C= 0$$

I believe you need cosine law. Let a, b, c be respective sides of angle A, B, and C, then $$-cos C = \frac{c^2 - a^2 - b^2}{2ab} $$ $$-cos B = \frac{b^2 - a^2 - c^2}{2ac} $$ $$-cos A = \frac{a^2 - b^2 - c^2}{2bc} $$

so we have $$cosA cosB cos C = -\frac{(c^2 - a^2 - b^2)(b^2 - a^2 - c^2)(a^2 - b^2 - c^2)}{8(abc)^2}$$ $$1 -cos^2 C - cos^2 A - cos^2 B = -\frac{4(abc)^2 -c^2(c^2 - a^2 - b^2)^2 - b^2(b^2 - a^2 - c^2)^2 - a^2(a^2 - b^2 - c^2)^2}{4(abc)^2}$$

To provde original identity, we just need to prove: $$4(abc)^2 -c^2(c^2 - a^2 - b^2)^2 - b^2(b^2 - a^2 - c^2)^2 - a^2(a^2 - b^2 - c^2)^2 = -(c^2 - a^2 - b^2)(b^2 - a^2 - c^2)(a^2 - b^2 - c^2)$$

Expand both side should verify the identity (From Wolfram, identity holds)

runaround
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There is definitely something wrong with the identity as you have entered it. First, as user Konstantinos Kanak0glou has noted in the comments, the identity fails to hold for $A=B=0$ and $C=\pi$.

A second piece of evidence that the "identity" is false is that the condition $A+B+C=\pi$ is symmetric in the three variables, as is the RHS of the purported identity, but the LHS is not.

mweiss
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$2\cos A\cos B\cos C$

$=\{\cos(A-B)+\cos(A+B)\}\cos C$

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

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

$\implies2\cos A\cos B\cos C=1-(\cos^2A+\cos^2B+\cos^2C)$

Community
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lab bhattacharjee
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The identity

$$1 - \cos^2 A - \cos^2 B - \cos^2 C - 2 \cos A \cos B \cos C = 0$$

can be seen as a re-writing of the relation $\cos C = - \cos(A+B)$ without sines of $A$ and $B$:

$$\cos C = - \cos(A+B) = \sin A \sin B - \cos A \cos B$$ $$\cos C + \cos A \cos B = \sin A \sin B$$ $$\left(\;\cos C + \cos A \cos B\;\right)^2 = \sin^2 A \sin^2 B = \left(1-\cos^2 A\right)\left(1-\cos^2 B\right)$$ $$\cos^2 C + 2 \cos A \cos B \cos C \color{red}{+ \cos^2A \cos^2 B} = 1 - \cos^2 A - \cos^2 B \color{red}{+ \cos^2 A \cos^2 B}$$ and the result follows.

Blue
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