270°±θ\;的三角函數值

∵sin(180°+\fgcolor{red}{θ})=-sin\fgcolor{red}{θ},\;\underline{sin(90°+θ)=cosθ}

∴sin(270°+θ)=sin(180°+\fgcolor{red}{(90°+θ)})=\fgcolor{red}{-}\underline{sin\fgcolor{red}{(90°+θ)}}=-\underline{cosθ}

⇒sin(270°-θ)=sin(270°+\fgcolor{669933}{(-θ)})=-cos(\fgcolor{669933}{-θ})=-cosθ

∵cos(180°+\fgcolor{red}{θ})=-cos\fgcolor{red}{θ},\;\underline{cos(90°+θ)}=-\underline{sinθ}

∴cos(270°+θ)=cos(180°+\fgcolor{red}{(90°+θ)})=\fgcolor{red}{-}\underline{cos\fgcolor{red}{(90°+θ)}=sinθ}

⇒cos(270°-θ)=cos(270°+\fgcolor{669933}{(-θ)})=sin(\fgcolor{669933}{-θ})=-sinθ

tan(270°+θ)=\frac{sin(270°+θ)}{cos(270°+θ)}=\frac{-cosθ}{sinθ}=-\frac{1}{tanθ}=-cotθ

tan(270°-θ)=tan(270°+(-θ))=-cot(-θ)=\frac{1}{tanθ}=cotθ

$Step=%v$

A: Corner[4] - (-0.1, 1)
s = 0 (slider from 0 to 8, step 1)