Nice trigonometric identities

by henderson, May 30, 2016, 6:51 PM

$\bf\color{red}Nice\ trigonometric\ identities\$ $=================================================================================================$
$1\blacktriangleright$ Identity: $\cos A+\cos B+\cos C=1+\frac{r}{R}$ $\Longrightarrow$ Result: $\boxed{\cos A+\cos B+\cos C \leq \frac{3}{2}}.$

$2\blacktriangleright$ $2\left(\sin^2x-\sin^2y\right)=(1-\cos 2x)-(1-\cos 2y)=$ $\cos 2y-\cos 2x=$ $2\sin(x+y)\sin (x-y)\implies$ $\boxed{\sin^2x-\sin^2y=\sin(x+y)\sin (x-y)}\ .$
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This post has been edited 10 times. Last edited by henderson, Jun 12, 2016, 8:11 AM

Trigonometric Identities

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$\bf\color{green}Proof\ of\ the\ 1^{st} \ identity \$ $===================================================================$
We have $\begin{align*}\cos{\alpha}+\cos{\beta}+\cos{\gamma}-1&=\frac{a^2+b^2-c^2}{2ab}+\frac{a^2+c^2-b^2}{2ac}+\frac{b^2+c^2-a^2}{2bc}-1 \\ &=\frac{(a+b-c)(b+c-a)(c+a-b)}{2abc} \\ &=\frac{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}{2abc(a+b+c)}.\end{align*}$ If $K$ is the area of $\triangle{ABC}$ , we have $K=\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{4},$ which means that $\frac{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}{2abc(a+b+c)}=\frac{8K^2}{abc(a+b+c)}.$ However, we also have $K=\frac{r(a+b+c)}{2}\implies r=\frac{2K}{a+b+c}.$ Furthermore, $K=\frac{abc}{4R}\implies R=\frac{abc}{4K}.$ Thus, we have $\frac{r}{R}=\frac{8K^2}{abc(a+b+c)}.$ Equating, we have $\cos\alpha +\cos\beta+\cos\gamma=1+\frac{r}{R}$ , as desired.
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This post has been edited 1 time. Last edited by henderson, Sep 12, 2016, 2:30 PM

by henderson, Jun 12, 2016, 8:09 AM

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Nice trigonometric identities

by henderson, May 30, 2016, 6:51 PM

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