# Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$

There is a trigonometric identity:

$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\equiv 1\text{ when }A+B+C=\pi$$

It is easy to prove it in an algebraic way, just like that:

$$\quad\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\\=\cos^2A+\cos^2B+\cos^2\left(\pi-A-B\right)+2\cos A\cos B\cos \left(\pi-A-B\right)\\=\cos^2A+\cos^2B+\cos^2\left(A+B\right)-2\cos A\cos B\cos \left(A+B\right)\\=\cos^2A+\cos^2B+\left(\cos A\cos B-\sin A\sin B\right)^2-2\cos A\cos B\left(\cos A\cos B-\sin A\sin B\right)\\=\cos^2A+\cos^2B+\cos^2A\cos^2B+\sin^2A\sin^2B-2\sin A\cos A\sin B\cos B-2\cos^2A\cos^2B+2\sin A\cos A\sin B\cos B\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+\left(1-\cos^2A\right)\left(1-\cos^2B\right)\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+1-\cos^2A-\cos^2B+\cos^2A\cos^2B\\=1$$

Then, I want to find a geometric way to prove this identity, as $$A+B+C=\pi$$ and it makes me think of the angle sum of triangle. However, it is quite hard to prove it in a geometric way. Therefore, I hope there is someone who can help. Thank you!

• For future reference, the algebraic proof can also be found on the site here. Sep 10, 2019 at 20:31
• I've also found this question, for an identity of which yours is a particular case (though you'll still need a bit of algebra to see it). Sep 10, 2019 at 20:33
• I'm still thinking about possibility that the equation will be derived when we divide a triangle into four pieces and consider areas, but I can't find a good partition for that. Sep 11, 2019 at 8:47
• @Isaac YIU Math Studio It means that $A$, $B$ and $C$ are positives? Sep 11, 2019 at 11:41
• @MichaelRozenberg It doesn't mention that, but it's alright if you add this condition. Sep 11, 2019 at 11:45

Since the accent in the OP is put on a purely geometric solution, i can not even consider the chance to write $$\cos^2 =1-\sin^2$$, and rephrase the wanted equality, thus having a trigonometric function which is better suited to geometrical interpretations.

So this answer has two steps, first we reformulate the given identity in a mot-a-mot geometric manner, the geometric framework is introduced, some strictly geometrically transposed equivalent relations are listed, then we give a proof:

In the triangle $$\Delta ABC$$ let $$AA'$$, $$BB'$$, $$CC'$$ be the heights, $$A'\in BC$$, $$B'\in CA$$, $$C'\in AB$$, intersecting in $$H$$, the orthocenter. We assume that the diameter $$2R$$ of the circumcircle is normed to be the unit. Then we have the following situation for the lengths of some segments in the picture:

\begin{aligned} AH &=\cos A\ , \qquad & HA'&=\cos B\cos C\ ,\\ BH &=\cos B\ , \qquad & HB'&=\cos A\cos C\ ,\\ CH &=\cos C\ , \qquad & HC'&=\cos A\cos B\ . \end{aligned}

Proof: We have: $$\sin \hat B =\sin \widehat{C'HA} =\frac{C'A}{AH} =\frac{AC\;\cos A}{AH} =\frac{2R\sin B\; \cos A}{AH} =\frac{\sin B\; \cos A}{AH} \ ,$$ which implies $$AH=\cos A$$, and the similar relations. Then we express twice the area of $$\Delta HBC$$ as $$HA'\cdot BC =2[HBC]=HB\cdot HC\cdot \sin\widehat{BHC}\ ,$$ thus getting $$HA'=\cos B\cos C$$.

We are in position to give a geometric mask to the given equality:

We use the above notations in $$\Delta ABC$$. We denote by $$a,b,c$$ the lenghts of the sides. Let $$M_A, M_B,M_C$$ be the mid points of the sides $$BC$$, $$CA$$, respectively $$AB$$. Let $$G=AM_A\cap BM_B\cap CM_C$$ be the intersection of the medians, the centroid. Let $$A^*, B^*, C^*$$ be the mid points of $$HA$$, $$HB$$, $$HC$$. Let $$N$$ be the center of the Euler circle $$(N)$$ passing through the nine points $$A',B',C'$$; $$M_A,M_B,N_C$$; $$A^*, B^*,C^*$$. It is the mid point of $$OH$$, and $$M_AA^*$$, $$M_BB^*$$, $$M_CC^*$$ are diameters in $$(N)$$, having the lenght $$R=OA=OB=OC$$. (For $$OM_AA^*A$$ is a parallelogram.)

Then we have the following relations: \begin{aligned} HA^2+HB^2+HC^2 + 2 HA\cdot HA' &= 4R^2\ ,\\ HA^*{}^2+HB^*{}^2+HC^*{}^2 + HA^*\cdot HA' &= R^2\ ,\\ 4OM_A^2+4OM_B^2+4OM_C^2 &= 3R^2+OH^2\ ,\\ 9R^2 &= a^2 +b^2 + c^2 +OH^2\\ 9R^2 &= a^2 +b^2 + c^2 +9OG^2\ . \end{aligned}

Proof: The relations above are equivalent:

• $$AH^2=4A^*H^2=4OM_A^2$$, and $$2 HA\cdot HA'=4 HA^*\cdot HA'$$ is the power of $$H$$ in the circle $$(N)$$, so it can be rewritten using its radius $$NA^*=\frac 12 R$$ and the distance to its center, $$NH=\frac 12 OH$$ as $$2 HA\cdot HA'=4 HA^*\cdot HA'=R^2-OH^2$$.

• From the triangle $$OBM_A$$, $$4OM_A^2+BC^2 =4(OM_A^2+BM_A^2)=4OB^2=4R^2$$.

• Note that $$G$$ cuts the median $$AM_A$$ in the proportion $$AG:GM_A=2:1$$, so it projects on $$BC$$ in the same proportion. This also holds for the colinear points $$H,G,O$$, so $$HG:GO=2:1$$, so $$HO=3GO$$.

• The last relation, $$OG^2 = R^2-\frac 13(a^2+b^2+c^2)$$, is a standard formula. We have in general the formula for an arbitrary point $$P$$: $$PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3GP^2\ .$$ We apply it for $$P=O$$, getting $$3R^2=3OG^3+\sum AG^2=3OG^3+\frac 49\sum AM_A^2=3OG^3+\frac 49\sum \left(\frac 12b^2+\frac 12 c^2-\frac 14 a^2\right)=3OG^3+\frac 49\sum \frac 34a^2=3OG^3+\frac 13\sum a^2\ .$$

$$\square$$

• what a fantastic proof! Sep 11, 2019 at 8:55

For positives $$A$$, $$B$$ and $$C$$ there is the following way.

Let $$A=\max\{A,B,C\},$$ $$\pi-A=\alpha,$$ $$\frac{\pi}{2}-B=\beta$$ and $$\frac{\pi}{2}-C=\gamma$$.

Thus, $$\alpha$$, $$\beta$$ and $$\gamma$$ are measured angles of the triangle and let sides-lengths of the triangle be $$a$$, $$b$$ and $$c$$ respectively.

Thus, since by law of sines $$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma},$$ by law of cosine we obtain: $$\sin^2\alpha=\sin^2\beta+\sin^2\gamma-2\sin\beta\sin\gamma\cos\alpha$$ or $$\sin^2(\pi-A)=\sin^2\left(\frac{\pi}{2}-B\right)+\sin^2\left(\frac{\pi}{2}-C\right)-2\sin\left(\frac{\pi}{2}-B\right)\sin\left(\frac{\pi}{2}-C\right)\cos(\pi-A)$$ or $$1-\cos^2A=\cos^2B+\cos^2C+2\cos B\cos C\cos A$$ and we are done!

• Nice. This is as “geometric” a proof as any here so far. Sep 11, 2019 at 17:36

I don't know if this counts as a proof, but following your suggestion, I used the cosine laws to obtain your result.

Suppose you have a triangle ABC as in the figure:

Since the angles $$A+B+C=\pi$$, these are the internal angles of a general triangle. Using the law of cosines, you can write:

$$a^2=b^2+c^2-2bc\cos A\\b^2=a^2+c^2-2ac\cos B\\c^2=a^2+b^2-2ab\cos C$$

It follows from here that:

$$\cos A=\frac{a^2-b^2-c^2}{-2bc}\\ \cos B=\frac{b^2-a^2-c^2}{-2ac}\\ \cos C=\frac{c^2-a^2-b^2}{-2ac}$$

Now, to verify your formula, we have:

$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=\\ \left(\frac{a^2-b^2-c^2}{-2bc}\right)^2+\left(\frac{b^2-a^2-c^2}{-2ac}\right)^2+\left(\frac{c^2-a^2-b^2}{-2ac}\right)^2+2\left( \frac{a^2-b^2-c^2}{-2bc} \right)\left( \frac{b^2-a^2-c^2}{-2ac} \right)\left( \frac{c^2-a^2-b^2}{-2ac} \right)$$

It is now a matter of manipulation of the equation to show that this equals 1. Observe that the least common multiple of the first three terms is $$4a^2b^2c^2$$, which is equal to the product of the last term,

$$\frac{a^2(a^2-b^2-c^2)^2+b^2(b^2-a^2-c^2)^2+c^2(c^2-a^2-b^2)^2}{4a^2b^2c^2}-\frac{(a^2-b^2-c^2)(b^2-a^2-c^2)(c^2-a^2-b^2)}{4a^2b^2c^2}$$

expanding the products in the numerator, you can verify that: $$a^2(a^2-b^2-c^2)^2+b^2(b^2-a^2-c^2)^2+c^2(c^2-a^2-b^2)^2-(a^2-b^2-c^2)(b^2-a^2-c^2)(c^2-a^2-b^2)=4a^2b^2c^2$$

so the fraction simplifies to

$$\frac{4a^2b^2c^2}{4a^2b^2c^2}=1$$

• Right, but not a geometric way to prove the claim. Sep 10, 2019 at 18:39
• The critical first step, "Since the angles $A + B + C = \pi$, these are the internal angles of a general triangle. [Therefore] using the law of cosines..." is certainly itself a geometric argument, even if the rest of the solution is algebraic manipulation. Sep 10, 2019 at 18:49
• Though it consists of some geometry, but it is almost using algebra. Sep 11, 2019 at 8:53

A purely geometric way doesn't look likely, because the degrees of the cosine terms (two and three respectively) don't match. For what it's worth, here is an alternative trigonometric derivation.

Writing $$2\cos B\cos C$$ in the second term as $$\cos (B+C)+\cos(B-C)$$ transforms our expression to$$\cos^2A+\cos^2B+\cos^2C+[\cos(B+C)+\cos(B-C)]\cos A.$$Notice that $$\cos(B+C)=-\cos A$$, and use this conversion forwards and backwards to give$$\cos^2B+\cos^2C-\cos(B-C)\cos(B+C).$$Now write the last term as $$-\frac12(\cos2B+\cos2C)$$ and express the first terms also in double-angle format. Then cancellation yields the required result.

• I slightly disagree with the first line since $\cos$'s are dimensionless. Sep 11, 2019 at 3:01
• @SeewooLee : Sure. But I haven't seen any essentially geometric derivations of results like this before. If you try hard enough, you can always work any trigonometric identity back into a geometric diagram, but the proof doesn't generally gain in clarity. Sep 11, 2019 at 6:20
• Sorry, I want a geometric way. Therefore, this is an unavailable answer. Sep 11, 2019 at 8:53