$Case=%v$

⇒\left\lbrace \begin{array}{ll} sin(90°-θ)=cosθ \ cos(90°-θ)=sinθ \ tan(90°-θ)=cotθ\cot(90°-θ)=tanθ\sec(90°-θ)=cscθ\csc(90°-θ)=secθ \end{array}  \right.\ 

⇒\left\lbrace \begin{array}{ll} sin(90°+θ)=cosθ \ cos(90°+θ)=-sinθ \ tan(90°+θ)=-cotθ\cot(90°+θ)=-tanθ\sec(90°+θ)=-cscθ\csc(90°+θ)=secθ \end{array}  \right.\ 

⇒\left\lbrace \begin{array}{ll} sin(270°-θ)=-cosθ \ cos(270°-θ)=-sinθ \ tan(270°-θ)=cotθ\cot(270°-θ)=tanθ\sec(270°-θ)=-cscθ\csc(270°-θ)=-secθ \end{array}  \right.\ 

⇒\left\lbrace \begin{array}{ll} sin(270°+θ)=-cosθ \ cos(270°+θ)=sinθ \ tan(270°+θ)=-cotθ\cot(270°+θ)=-tanθ\sec(270°+θ)=cscθ\csc(270°+θ)=-secθ \end{array}  \right.\ 

⇒\left\lbrace \begin{array}{ll} sin(180°-θ)=sinθ \ cos(180°-θ)=-cosθ \ tan(180°-θ)=-tanθ\cot(180°-θ)=-cotθ\sec(180°-θ)=-secθ\csc(180°-θ)=cscθ \end{array}  \right.\ 

⇒\left\lbrace \begin{array}{ll} sin(180°+θ)=-sinθ \ cos(180°+θ)=-cosθ \ tan(180°+θ)=tanθ\cot(180°+θ)=cotθ\sec(180°+θ)=-secθ\csc(180°+θ)=-cscθ \end{array}  \right.\ 

⇒\left\lbrace \begin{array}{ll} sin(-θ)=-sinθ \ cos(-θ)=cosθ \ tan(-θ)=-tanθ\cot(-θ)=-cotθ\sec(-θ)=secθ\csc(-θ)=-cscθ \end{array}  \right.\ 

⇒\left\lbrace \begin{array}{ll} sin(360°+θ)=sinθ \ cos(360°+θ)=cosθ \ tan(360°+θ)=tanθ\cot(360°+θ)=cotθ\sec(360°+θ)=secθ\csc(360°+θ)=cscθ \end{array}  \right.\ 

$⇒90°-θ\;在第一象限$

$Step=%v$

$⇒90°+θ\;在第二象限$

$⇒270°-θ\;在第三象限$

$⇒270°+θ\;在第四象限$

$⇒180°-θ\;在第二象限$

$⇒\left\lbrace \begin{array}{ll} sinθ⇔sinθ\cosθ⇔cosθ \ tanθ⇔tanθ\cotθ⇔cotθ \secθ⇔ secθ\cscθ⇔cscθ \end{array}  \right.\ $

$⇒\left\lbrace \begin{array}{ll} sinθ⇔sinθ\cosθ⇔cosθ \ tanθ⇔tanθ\cotθ⇔cotθ \secθ⇔ secθ\cscθ⇔cscθ \end{array}  \right.\ $

$⇒\left\lbrace \begin{array}{ll} sinθ>0,cscθ>0 \ cosθ>0,secθ>0 \ tanθ>0,cotθ>0 \end{array}  \right.\ $

$⇒\left\lbrace \begin{array}{ll} sinθ⇔cosθ \ tanθ⇔cotθ \ secθ⇔cscθ \end{array}  \right.\ $

$⇒90°-θ\;⇒取餘函數$

$⇒\left\lbrace \begin{array}{ll} sinθ>0,cscθ>0 \ cosθ>0,secθ>0 \ tanθ>0,cotθ>0 \end{array}  \right.\ $

$⇒\left\lbrace \begin{array}{ll} sinθ⇔cosθ \ tanθ⇔cotθ \ secθ⇔cscθ \end{array}  \right.\ $

$⇒90°+θ\;⇒取餘函數$

$⇒\left\lbrace \begin{array}{ll} sinθ>0,cscθ>0 \ cosθ<0,secθ<0 \ tanθ<0,cotθ<0 \end{array}  \right.\ $

$⇒\left\lbrace \begin{array}{ll} sinθ⇔cosθ \ tanθ⇔cotθ \ secθ⇔cscθ \end{array}  \right.\ $

$⇒270°-θ\;⇒取餘函數$

$⇒\left\lbrace \begin{array}{ll} sinθ<0,cscθ<0 \ cosθ<0,secθ<0 \ tanθ>0,cotθ>0 \end{array}  \right.\ $

$⇒\left\lbrace \begin{array}{ll} sinθ⇔cosθ \ tanθ⇔cotθ \ secθ⇔cscθ \end{array}  \right.\ $

$⇒270°+θ\;⇒取餘函數$

$⇒\left\lbrace \begin{array}{ll} sinθ<0,cscθ<0 \ cosθ>0,secθ>0 \ tanθ<0,cotθ<0 \end{array}  \right.\ $

$⇒\left\lbrace \begin{array}{ll} sinθ⇔sinθ\cosθ⇔cosθ \ tanθ⇔tanθ\cotθ⇔cotθ \secθ⇔ secθ\cscθ⇔cscθ \end{array}  \right.\ $

$⇒180°-θ\;⇒取原函數$

$⇒\left\lbrace \begin{array}{ll} sinθ>0,cscθ>0 \ cosθ<0,secθ<0 \ tanθ<0,cotθ<0 \end{array}  \right.\ $

$⇒180°+θ\;在第三象限$

$⇒\left\lbrace \begin{array}{ll} sinθ<0,cscθ<0 \ cosθ<0,secθ<0 \ tanθ>0,cotθ>0 \end{array}  \right.\ $

$⇒180°+θ\;⇒取原函數$

$⇒-θ\;在第四象限$

$⇒\left\lbrace \begin{array}{ll} sinθ<0,cscθ<0 \ cosθ>0,secθ>0 \ tanθ<0,cotθ<0 \end{array}  \right.\ $

$⇒-θ=0°-θ\;⇒取原函數$

$⇒\left\lbrace \begin{array}{ll} sinθ⇔sinθ\cosθ⇔cosθ \ tanθ⇔tanθ\cotθ⇔cotθ \secθ⇔ secθ\cscθ⇔cscθ \end{array}  \right.\ $

$⇒360°+θ\;⇒取原函數$

$⇒360°+θ\;在第一象限$

請問:\;\left\lbrace \begin{array}{ll} sin(90°-θ)=? \ cos(90°-θ)=? \ tan(90°-θ)=?\cot(90°-θ)=?\sec(90°-θ)=?\csc(90°-θ)=? \end{array}  \right.\ 

請問:\;\left\lbrace \begin{array}{ll} sin(90°+θ)=? \ cos(90°+θ)=? \ tan(90°+θ)=?\cot(90°+θ)=?\sec(90°+θ)=?\csc(90°+θ)=? \end{array}  \right.\ 

請問:\;\left\lbrace \begin{array}{ll} sin(270°-θ)=? \ cos(270°-θ)=? \ tan(270°-θ)=?\cot(270°-θ)=?\sec(270°-θ)=?\csc(270°-θ)=? \end{array}  \right.\ 

請問:\;\left\lbrace \begin{array}{ll} sin(270°+θ)=? \ cos(270°+θ)=? \ tan(270°+θ)=?\cot(270°+θ)=?\sec(270°+θ)=?\csc(270°+θ)=? \end{array}  \right.\ 

請問:\;\left\lbrace \begin{array}{ll} sin(180°-θ)=? \ cos(180°-θ)=? \ tan(180°-θ)=?\cot(180°-θ)=?\sec(180°-θ)=?\csc(180°-θ)=? \end{array}  \right.\ 

請問:\;\left\lbrace \begin{array}{ll} sin(180°+θ)=? \ cos(180°+θ)=? \ tan(180°+θ)=?\cot(180°+θ)=?\sec(180°+θ)=?\csc(180°+θ)=? \end{array}  \right.\ 

請問:\;\left\lbrace \begin{array}{ll} sin(-θ)=? \ cos(-θ)=? \ tan(-θ)=?\cot(-θ)=?\sec(-θ)=?\csc(-θ)=? \end{array}  \right.\ 

請問:\;\left\lbrace \begin{array}{ll} sin(360°+θ)=? \ cos(360°+θ)=? \ tan(360°+θ)=?\cot(360°+θ)=?\sec(360°+θ)=?\csc(360°+θ)=? \end{array}  \right.\ 

$請問你選那一條公式：$
$函數(180°×n±θ)=\fbox{符號}原函數(θ)，\;n為整數$
$函數(\;90°×n±θ\;)=\fbox{符號}餘函數(θ)，\;n為奇數$

n = 1 (slider from 1 to 8, step 1)
A: Corner[4] + (0.15, -0.5)
s = 0 (slider from 0 to 8, step 1)
T1: "$請問你可以判斷\;" + (LaTeX[If[n ≟ 1, "90°-θ", If[n ≟ 2, "90°+θ", If[n ≟ 3, "270°-θ", If[n ≟ 4, "270°+θ", If[n ≟ 5, "180°-θ", If[n ≟ 6, "180°+θ", If[n ≟ 7, "-θ", If[n ≟ 8, "360°+θ"]]]]]]]]]) + "\;的六個三角函數值$
$的正負嗎？$"
T0: "請問:\;" + (LaTeX[If[n ≟ 1, "90°-θ", If[n ≟ 2, "90°+θ", If[n ≟ 3, "270°-θ", If[n ≟ 4, "270°+θ", If[n ≟ 5, "180°-θ", If[n ≟ 6, "180°+θ", If[n ≟ 7, "-θ", If[n ≟ 8, "360°+θ"]]]]]]]]]) + "\;在第幾象限?"