$1 + \cos2C - \cos2A - \cos2B=4\sin A\sin B\sin C$.

Question

How can I prove this equation?

$1 + \cos2C - \cos2A - \cos2B=4\sin A\sin B\sin C$

if we know that $A$, $B$, $C$ are a triangle's angles.

I have come to the point where on the left side I have

$-4 \cos A \cos B \cos C$.

On the other side I have

$4\sin A\sin B\sin C$.

Am I correct till now?

Answer 1

It's wrong! Try $C=90^{\circ}$ and $A=B=45^{\circ}.$

By the way, $$1+\cos2C-\cos2A-\cos2B$$

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

$$=2\cos C[\cos(A-B)-\cos(A+B)]$$ $$=4\cos C\sin A\sin B.$$

Answer 2

$$C=\pi-A-B$$

$$\sin(C)=\sin(A)\cos(B)+\cos(A)\sin(B)$$

$$4\sin(C)\sin(A)\sin(B)=2\sin^2(A)\sin(2B)+2\sin^2(B)\sin(2A)$$

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

$$=\sin(2B)+\sin(2A)-\sin(2B+2A)$$

$$=\sin(2B)+\sin(2A)-\sin(2C)$$

Answer 3

We have,

$\mathsf{1+cos(2C)-cos(2A)-cos(2B)}$

$\mathsf{=2\,cos^2(C)-\big[cos(2A)+cos(2B)\big]}$

$\mathsf{=2\,cos^2(C)-\left[2\,cos\left(\dfrac{2A+2B}{2}\right)\,cos\left(\dfrac{2A-2B}{2}\right)\right]}$

$\mathsf{=2\,cos^2(C)-\left[2\,cos(A+B)\,cos(A-B)\right]}$

Since A, B, C are the angles of a triangle, so,

$\mathsf{A+B+C=\pi\,\,\,\,\,\,\,\,\,\,...(1)}$

$\mathsf{\implies\,A+B=\pi-C}$

So,

$\mathsf{=2\,cos^2(C)-2\,cos(\pi-C)\,cos(A-B)}$

$\mathsf{=2\,cos^2(C)+2\,cos(C)\,cos(A-B)}$

$\mathsf{=2\,cos(C)\big[cos(C)+cos(A-B)\big]}$

Again, from (1),

$\mathsf{=2\,cos(C)\big[cos\left(\pi-(A+B)\right)+cos(A-B)\big]}$

$\mathsf{=2\,cos(C)\big[-cos(A+B)+cos(A-B)\big]}$

$\mathsf{=2\,cos(C)\big[cos(A-B)-cos(A+B)\big]}$

$\mathsf{=2\,cos(C)\cdot2\,sin(A)\,sin(B)}$

$\mathsf{=4\,sin(A)\,sin(B)\,cos(C)}$