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[Title]
Op֐@藝W

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[Problem]
Op֐̎̌𓱂B

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[Level1]
sin,cos̉@藝

$\sin(\alpha + \beta)$

$\sin(\alpha + \beta)=\sin\alpha \cos\beta + \cos\alpha \sin\beta$

$\sin(\alpha - \beta)$

$\sin(\alpha - \beta)=\sin\alpha \cos\beta - \cos\alpha \sin\beta$

$\cos(\alpha + \beta)$

$\cos(\alpha + \beta)=\cos\alpha \cos\beta - \sin\alpha \sin\beta$

$\cos(\alpha - \beta)$

$\cos(\alpha - \beta)=\cos\alpha \cos\beta + \sin\alpha \sin\beta$


[Level2]
tan̉@藝

$\tan(\alpha - \beta)$

process
$=\dfrac{\sin(\alpha- \beta)}{\cos(\alpha -\beta)}
=\dfrac{\sin\alpha \cos\beta - \cos\alpha \sin\beta}
{\cos\alpha \cos\beta + \sin\alpha \sin\beta}$\\
e$\cos\alpha \cos\beta$@Ŋ\\
$=\dfrac{
\dfrac{\sin\alpha \cos\beta}{\cos\alpha \cos\beta} 
-\dfrac{\cos\alpha \sin\beta}{\cos\alpha \cos\beta}}
{\dfrac{\cos\alpha \cos\beta}{\cos\alpha \cos\beta}
 +\dfrac{\sin\alpha \sin\beta}{\cos\alpha \cos\beta}}$\\
$=\dfrac{
\dfrac{\sin\alpha}{\cos\alpha} \cdot \dfrac11 
-\dfrac11\cdot \dfrac{\sin\beta}{\cos\beta}}
{\dfrac11 \cdot \dfrac11
+\dfrac{\sin\alpha}{\cos\alpha} \cdot 
\dfrac{\sin\beta}{\cos\beta}}$\\
$=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$

$\tan(\alpha - \beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$


$\tan(\alpha + \beta)$

process
$=\dfrac{\sin(\alpha+ \beta)}{\cos(\alpha +\beta)}
=\dfrac{\sin\alpha \cos\beta + \cos\alpha \sin\beta}
{\cos\alpha \cos\beta - \sin\alpha \sin\beta}$\\
e$\cos\alpha \cos\beta$@Ŋ\\
$=\dfrac{
\dfrac{\sin\alpha \cos\beta}{\cos\alpha \cos\beta} 
+\dfrac{\cos\alpha \sin\beta}{\cos\alpha \cos\beta}}
{\dfrac{\cos\alpha \cos\beta}{\cos\alpha \cos\beta}
 -\dfrac{\sin\alpha \sin\beta}{\cos\alpha \cos\beta}}$\\
$=\dfrac{
\dfrac{\sin\alpha}{\cos\alpha} \cdot \dfrac11 
+\dfrac11\cdot \dfrac{\sin\beta}{\cos\beta}}
{\dfrac11 \cdot \dfrac11
 -\dfrac{\sin\alpha}{\cos\alpha} \cdot 
\dfrac{\sin\beta}{\cos\beta}}$\\
$=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$


$\tan(\alpha + \beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$



[Level3]
sin,cos,tan@2{p̌

$\tan 2\alpha$

process
$=\tan(\alpha + \alpha)$\\
$=\dfrac{\tan\alpha+\tan\alpha}{1-\tan\alpha\tan\alpha}$\\
$=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$

$\tan 2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$

$\sin 2\alpha$

process
$=\sin (\alpha + \alpha)$\\
$=\sin \alpha \cos \alpha + \cos \alpha \sin \alpha $\\
$=2\sin \alpha \cos \alpha$

$\sin 2\alpha=2\sin \alpha \cos \alpha$

$\cos 2\alpha$

process
$=\cos(\alpha +\alpha)$\\
$=\cos \alpha \cos \alpha - \sin \alpha \sin \alpha$\\
$=\cos^2 \alpha -\sin^2 \alpha$@\\
$=\cos^2 \alpha -(1-\cos^2 \alpha)$\\
$=2\cos^2 \alpha -1$@\\
$=2(1-\sin^2\alpha)-1$\\
$=1-2\sin^2 \alpha$

$\cos 2\alpha=\cos^2 \alpha -\sin^2 \alpha,\ \ 2\cos^2 \alpha -1,\ \ 1-2\sin^2 \alpha$




[Level4]
sin,cos,tan p̌

$\tan^2 \dfrac{\alpha}{2}$

process
$=\dfrac{\sin^2 \dfrac{\alpha}{2}}{\cos^2 \dfrac{\alpha}{2}}
=\dfrac{\dfrac{1-\cos \alpha}{2}}{\dfrac{1+\cos \alpha}{2}}$\\
$=\dfrac{1-\cos \alpha}{1+\cos \alpha}$

$\tan^2 \dfrac{\alpha}{2}=\dfrac{1-\cos \alpha}{1+\cos \alpha}$


$\sin^2 \dfrac{\alpha}{2}$

process
2{p̌$\cos 2\alpha =1-2\sin^2 \alpha$\\
ڍ\\
$2\sin^2 \alpha =1-\cos 2\alpha$@ӂQŊ@\\
$\sin^2 \alpha =\dfrac{1-\cos 2\alpha}{2}$\\
$\alpha  \dfrac{\alpha}{2}$ɒu\\
$\sin^2 \dfrac{\alpha}{2}=\dfrac{1-\cos \alpha}{2}$

$\sin^2 \dfrac{\alpha}{2}=\dfrac{1-\cos \alpha}{2}$


$\cos^2 \dfrac{\alpha}{2}$

process
2{p̌$\cos 2\alpha =2\cos^2 \alpha -1$\\
ڍ\\
$2\cos^2 \alpha =1+\cos 2\alpha$ @ӂQŊ@\\
$\cos^2 \alpha =\dfrac{1+\cos 2\alpha}{2}$\\
$\alpha  \dfrac{\alpha}{2}$ɒu\\
$\cos^2 \dfrac{\alpha}{2}=\dfrac{1+\cos \alpha}{2}$

$\cos^2 \dfrac{\alpha}{2}=\dfrac{1+\cos \alpha}{2}$



[Level5]
3{p,Ϙa,a


$\sin 3\alpha$

process
$\ \ \ \sin(2\alpha +\alpha)
=\sin 2\alpha \cos \alpha +\cos 2\alpha \sin \alpha$\\
$=2\sin \alpha \cos \alpha \cdot \cos \alpha
+(1-2\sin^2 \alpha)\cdot \sin \alpha$\\
$=2\sin \alpha(1-\sin^2 \alpha)
+\sin \alpha -2\sin^3 \alpha$\\
$=-4\sin^3 \alpha +3\sin \alpha$

$\sin 3\alpha=-4\sin^3 \alpha +3\sin \alpha$

$\cos 3\alpha$

process
$\ \ \ \cos(2\alpha + \alpha)
=\cos 2\alpha \cos \alpha - \sin 2\alpha \sin \alpha$\\
$=(2\cos^2 \alpha -1)\cdot \cos \alpha
-2\sin \alpha \cos \alpha \cdot \sin \alpha$\\
$=2\cos^3 \alpha -\cos \alpha
-2(1-\cos^2 \alpha)\cdot \cos \alpha$\\
$=2\cos^3 \alpha -\cos \alpha
-2\cos \alpha +2\cos^3 \alpha$\\
$=4\cos^3 \alpha -3\cos \alpha$\\
Õ[RQ܂

$\cos 3\alpha=4\cos^3 \alpha -3\cos \alpha$




$\sin \alpha \cos \beta$

process
\begin{eqnarray*}
 \sin (\alpha+\beta) &=& \sin\alpha \cos\beta + \cos\alpha \sin\beta  \\
+\sin (\alpha-\beta) &=& \sin\alpha \cos\beta - \cos\alpha \sin\beta  \\
\hline
\sin (\alpha+\beta)+\sin (\alpha-\beta)&=&2\sin\alpha \cos\beta
\end{eqnarray*}\\
ӂQŊ

$\sin \alpha \cos \beta=\dfrac12 \{ \sin(\alpha+\beta)+\sin(\alpha-\beta) \}$


$\cos \alpha \sin \beta$

process
\begin{eqnarray*}
 \sin (\alpha+\beta) &=& \sin\alpha \cos\beta + \cos\alpha \sin\beta  \\
-\sin (\alpha-\beta) &=& \sin\alpha \cos\beta - \cos\alpha \sin\beta  \\
\hline
\sin (\alpha+\beta)-\sin (\alpha-\beta)&=&2\cos\alpha \sin\beta
\end{eqnarray*}\\
ӂQŊ

$\cos \alpha \sin \beta=\dfrac12 \{ \sin(\alpha+\beta)-\sin(\alpha-\beta) \}$

$\cos \alpha \cos \beta$

process
\begin{eqnarray*}
 \cos (\alpha+\beta) &=& \cos\alpha \cos\beta - \sin\alpha \sin\beta  \\
+\cos (\alpha-\beta) &=& \cos\alpha \cos\beta + \sin\alpha \sin\beta  \\
\hline
\cos (\alpha+\beta)+\cos (\alpha-\beta)&=&2\cos\alpha \cos\beta
\end{eqnarray*}\\
ӂQŊ

$\cos \alpha \cos \beta=\dfrac12 \{ \cos (\alpha+\beta)+\cos (\alpha-\beta) \}$



$\sin\alpha \sin\beta$

process
\begin{eqnarray*}
 \cos (\alpha+\beta) &=& \cos\alpha \cos\beta - \sin\alpha \sin\beta  \\
-\cos (\alpha-\beta) &=& \cos\alpha \cos\beta + \sin\alpha \sin\beta  \\
\hline
\cos (\alpha+\beta)-\cos (\alpha-\beta)&=&-2\sin\alpha \sin\beta
\end{eqnarray*}\\
ӂQŊ

$\sin\alpha \sin\beta=-\dfrac12 \{ \cos (\alpha+\beta)-\cos (\alpha-\beta) \}$






$\sin A + \sin B$

process
$\alpha+\beta=A,\ \alpha-\beta=B$Ƃ\\
$\alpha=\dfrac{A+B}{2},\ \beta=\dfrac{A-B}{2}$ƂȂB\\
L̎ɑB\\
$\sin (\alpha+\beta)+\sin (\alpha-\beta)=2\sin\alpha \cos\beta$\\
$\sin A+ \sin B= 2\sin \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$

$\sin A+ \sin B= 2\sin \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$


$\sin A - \sin B$

process
$\alpha+\beta=A,\ \alpha-\beta=B$Ƃ\\
$\alpha=\dfrac{A+B}{2},\ \beta=\dfrac{A-B}{2}$ƂȂB\\
L̎ɑB\\
$\sin (\alpha+\beta)-\sin (\alpha-\beta)=2\cos\alpha \sin\beta$\\
$\sin A- \sin B= 2\cos \dfrac{A+B}{2} \sin \dfrac{A-B}{2}$

$\sin A- \sin B= 2\cos \dfrac{A+B}{2} \sin \dfrac{A-B}{2}$



$\cos A + \cos B$

process
$\alpha+\beta=A,\ \alpha-\beta=B$Ƃ\\
$\alpha=\dfrac{A+B}{2},\ \beta=\dfrac{A-B}{2}$ƂȂB\\
L̎ɑB\\
$\cos (\alpha+\beta)+\cos (\alpha-\beta)=2\cos\alpha \cos\beta$\\
$\cos A+\cos B=2\cos \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$

$\cos A+\cos B=2\cos \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$



$\cos A - \cos B$

process
$\alpha+\beta=A,\ \alpha-\beta=B$Ƃ\\
$\alpha=\dfrac{A+B}{2},\ \beta=\dfrac{A-B}{2}$ƂȂB\\
L̎ɑB\\
$\cos (\alpha+\beta)-\cos (\alpha-\beta)=-2\sin\alpha \sin\beta$\\
$\cos A-\cos B=-2\sin \dfrac{A+B}{2} \sin \dfrac{A-B}{2}$

$\cos A-\cos B=-2\sin \dfrac{A+B}{2} \sin \dfrac{A-B}{2}$


[Level6]
@藝,2{p,p_



$\sin(\alpha + \beta)$

$\sin(\alpha + \beta)=\sin\alpha \cos\beta + \cos\alpha \sin\beta$

$\sin(\alpha - \beta)$

$\sin(\alpha - \beta)=\sin\alpha \cos\beta - \cos\alpha \sin\beta$

$\cos(\alpha + \beta)$

$\cos(\alpha + \beta)=\cos\alpha \cos\beta - \sin\alpha \sin\beta$

$\cos(\alpha - \beta)$

$\cos(\alpha - \beta)=\cos\alpha \cos\beta + \sin\alpha \sin\beta$




$\tan(\alpha - \beta)$

process
$=\dfrac{\sin(\alpha- \beta)}{\cos(\alpha -\beta)}
=\dfrac{\sin\alpha \cos\beta - \cos\alpha \sin\beta}
{\cos\alpha \cos\beta + \sin\alpha \sin\beta}$\\
e$\cos\alpha \cos\beta$@Ŋ\\
$=\dfrac{
\dfrac{\sin\alpha \cos\beta}{\cos\alpha \cos\beta} 
-\dfrac{\cos\alpha \sin\beta}{\cos\alpha \cos\beta}}
{\dfrac{\cos\alpha \cos\beta}{\cos\alpha \cos\beta}
 +\dfrac{\sin\alpha \sin\beta}{\cos\alpha \cos\beta}}$\\
$=\dfrac{
\dfrac{\sin\alpha}{\cos\alpha} \cdot \dfrac11 
-\dfrac11\cdot \dfrac{\sin\beta}{\cos\beta}}
{\dfrac11 \cdot \dfrac11
+\dfrac{\sin\alpha}{\cos\alpha} \cdot 
\dfrac{\sin\beta}{\cos\beta}}$\\
$=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$

$\tan(\alpha - \beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$


$\tan(\alpha + \beta)$

process
$=\dfrac{\sin(\alpha+ \beta)}{\cos(\alpha +\beta)}
=\dfrac{\sin\alpha \cos\beta + \cos\alpha \sin\beta}
{\cos\alpha \cos\beta - \sin\alpha \sin\beta}$\\
e$\cos\alpha \cos\beta$@Ŋ\\
$=\dfrac{
\dfrac{\sin\alpha \cos\beta}{\cos\alpha \cos\beta} 
+\dfrac{\cos\alpha \sin\beta}{\cos\alpha \cos\beta}}
{\dfrac{\cos\alpha \cos\beta}{\cos\alpha \cos\beta}
 -\dfrac{\sin\alpha \sin\beta}{\cos\alpha \cos\beta}}$\\
$=\dfrac{
\dfrac{\sin\alpha}{\cos\alpha} \cdot \dfrac11 
+\dfrac11\cdot \dfrac{\sin\beta}{\cos\beta}}
{\dfrac11 \cdot \dfrac11
 -\dfrac{\sin\alpha}{\cos\alpha} \cdot 
\dfrac{\sin\beta}{\cos\beta}}$\\
$=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$


$\tan(\alpha + \beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$





$\tan 2\alpha$

process
$=\tan(\alpha + \alpha)$\\
$=\dfrac{\tan\alpha+\tan\alpha}{1-\tan\alpha\tan\alpha}$\\
$=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$

$\tan 2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$

$\sin 2\alpha$

process
$=\sin (\alpha + \alpha)$\\
$=\sin \alpha \cos \alpha + \cos \alpha \sin \alpha $\\
$=2\sin \alpha \cos \alpha$

$\sin 2\alpha=2\sin \alpha \cos \alpha$

$\cos 2\alpha$

process
$=\cos(\alpha +\alpha)$\\
$=\cos \alpha \cos \alpha - \sin \alpha \sin \alpha$\\
$=\cos^2 \alpha -\sin^2 \alpha$@\\
$=\cos^2 \alpha -(1-\cos^2 \alpha)$\\
$=2\cos^2 \alpha -1$@\\
$=2(1-\sin^2\alpha)-1$\\
$=1-2\sin^2 \alpha$

$\cos 2\alpha=\cos^2 \alpha -\sin^2 \alpha,\ \ 2\cos^2 \alpha -1,\ \ 1-2\sin^2 \alpha$






$\tan^2 \dfrac{\alpha}{2}$

process
$=\dfrac{\sin^2 \dfrac{\alpha}{2}}{\cos^2 \dfrac{\alpha}{2}}
=\dfrac{\dfrac{1-\cos \alpha}{2}}{\dfrac{1+\cos \alpha}{2}}$\\
$=\dfrac{1-\cos \alpha}{1+\cos \alpha}$

$\tan^2 \dfrac{\alpha}{2}=\dfrac{1-\cos \alpha}{1+\cos \alpha}$



$\sin^2 \dfrac{\alpha}{2}$

process
2{p̌$\cos 2\alpha =1-2\sin^2 \alpha$\\
ڍ\\
$2\sin^2 \alpha =1-\cos 2\alpha$@ӂQŊ@\\
$\sin^2 \alpha =\dfrac{1-\cos 2\alpha}{2}$\\
$\alpha  \dfrac{\alpha}{2}$ɒu\\
$\sin^2 \dfrac{\alpha}{2}=\dfrac{1-\cos \alpha}{2}$

$\sin^2 \dfrac{\alpha}{2}=\dfrac{1-\cos \alpha}{2}$




$\cos^2 \dfrac{\alpha}{2}$

process
2{p̌$\cos 2\alpha =2\cos^2 \alpha -1$\\
ڍ\\
$2\cos^2 \alpha =1+\cos 2\alpha$ @ӂQŊ@\\
$\cos^2 \alpha =\dfrac{1+\cos 2\alpha}{2}$\\
$\alpha  \dfrac{\alpha}{2}$ɒu\\
$\cos^2 \dfrac{\alpha}{2}=\dfrac{1+\cos \alpha}{2}$

$\cos^2 \dfrac{\alpha}{2}=\dfrac{1+\cos \alpha}{2}$









[Level7]
Lׂă_



$\sin(\alpha + \beta)$

$\sin(\alpha + \beta)=\sin\alpha \cos\beta + \cos\alpha \sin\beta$

$\sin(\alpha - \beta)$

$\sin(\alpha - \beta)=\sin\alpha \cos\beta - \cos\alpha \sin\beta$

$\cos(\alpha + \beta)$

$\cos(\alpha + \beta)=\cos\alpha \cos\beta - \sin\alpha \sin\beta$

$\cos(\alpha - \beta)$

$\cos(\alpha - \beta)=\cos\alpha \cos\beta + \sin\alpha \sin\beta$



$\tan(\alpha - \beta)$

process
$=\dfrac{\sin(\alpha- \beta)}{\cos(\alpha -\beta)}
=\dfrac{\sin\alpha \cos\beta - \cos\alpha \sin\beta}
{\cos\alpha \cos\beta + \sin\alpha \sin\beta}$\\
e$\cos\alpha \cos\beta$@Ŋ\\
$=\dfrac{
\dfrac{\sin\alpha \cos\beta}{\cos\alpha \cos\beta} 
-\dfrac{\cos\alpha \sin\beta}{\cos\alpha \cos\beta}}
{\dfrac{\cos\alpha \cos\beta}{\cos\alpha \cos\beta}
 +\dfrac{\sin\alpha \sin\beta}{\cos\alpha \cos\beta}}$\\
$=\dfrac{
\dfrac{\sin\alpha}{\cos\alpha} \cdot \dfrac11 
-\dfrac11\cdot \dfrac{\sin\beta}{\cos\beta}}
{\dfrac11 \cdot \dfrac11
+\dfrac{\sin\alpha}{\cos\alpha} \cdot 
\dfrac{\sin\beta}{\cos\beta}}$\\
$=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$

$\tan(\alpha - \beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$



$\tan(\alpha + \beta)$

process
$=\dfrac{\sin(\alpha+ \beta)}{\cos(\alpha +\beta)}
=\dfrac{\sin\alpha \cos\beta + \cos\alpha \sin\beta}
{\cos\alpha \cos\beta - \sin\alpha \sin\beta}$\\
e$\cos\alpha \cos\beta$@Ŋ\\
$=\dfrac{
\dfrac{\sin\alpha \cos\beta}{\cos\alpha \cos\beta} 
+\dfrac{\cos\alpha \sin\beta}{\cos\alpha \cos\beta}}
{\dfrac{\cos\alpha \cos\beta}{\cos\alpha \cos\beta}
 -\dfrac{\sin\alpha \sin\beta}{\cos\alpha \cos\beta}}$\\
$=\dfrac{
\dfrac{\sin\alpha}{\cos\alpha} \cdot \dfrac11 
+\dfrac11\cdot \dfrac{\sin\beta}{\cos\beta}}
{\dfrac11 \cdot \dfrac11
 -\dfrac{\sin\alpha}{\cos\alpha} \cdot 
\dfrac{\sin\beta}{\cos\beta}}$\\
$=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$\tan(\alpha + \beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$




$\tan 2\alpha$

process
$=\tan(\alpha + \alpha)$\\
$=\dfrac{\tan\alpha+\tan\alpha}{1-\tan\alpha\tan\alpha}$\\
$=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$

$\tan 2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$

$\sin 2\alpha$

process
$=\sin (\alpha + \alpha)$\\
$=\sin \alpha \cos \alpha + \cos \alpha \sin \alpha $\\
$=2\sin \alpha \cos \alpha$

$\sin 2\alpha=2\sin \alpha \cos \alpha$

$\cos 2\alpha$

process
$=\cos(\alpha +\alpha)$\\
$=\cos \alpha \cos \alpha - \sin \alpha \sin \alpha$\\
$=\cos^2 \alpha -\sin^2 \alpha$@\\
$=\cos^2 \alpha -(1-\cos^2 \alpha)$\\
$=2\cos^2 \alpha -1$@\\
$=2(1-\sin^2\alpha)-1$\\
$=1-2\sin^2 \alpha$

$\cos 2\alpha=\cos^2 \alpha -\sin^2 \alpha,\ \ 2\cos^2 \alpha -1,\ \ 1-2\sin^2 \alpha$




$\tan^2 \dfrac{\alpha}{2}$

process
$=\dfrac{\sin^2 \dfrac{\alpha}{2}}{\cos^2 \dfrac{\alpha}{2}}
=\dfrac{\dfrac{1-\cos \alpha}{2}}{\dfrac{1+\cos \alpha}{2}}$\\
$=\dfrac{1-\cos \alpha}{1+\cos \alpha}$

$\tan^2 \dfrac{\alpha}{2}=\dfrac{1-\cos \alpha}{1+\cos \alpha}$




$\sin^2 \dfrac{\alpha}{2}$

process
2{p̌$\cos 2\alpha =1-2\sin^2 \alpha$\\
ڍ\\
$2\sin^2 \alpha =1-\cos 2\alpha$@ӂQŊ@\\
$\sin^2 \alpha =\dfrac{1-\cos 2\alpha}{2}$\\
$\alpha  \dfrac{\alpha}{2}$ɒu\\
$\sin^2 \dfrac{\alpha}{2}=\dfrac{1-\cos \alpha}{2}$

$\sin^2 \dfrac{\alpha}{2}=\dfrac{1-\cos \alpha}{2}$




$\cos^2 \dfrac{\alpha}{2}$

process
2{p̌$\cos 2\alpha =2\cos^2 \alpha -1$\\
ڍ\\
$2\cos^2 \alpha =1+\cos 2\alpha$ @ӂQŊ@\\
$\cos^2 \alpha =\dfrac{1+\cos 2\alpha}{2}$\\
$\alpha  \dfrac{\alpha}{2}$ɒu\\
$\cos^2 \dfrac{\alpha}{2}=\dfrac{1+\cos \alpha}{2}$

$\cos^2 \dfrac{\alpha}{2}=\dfrac{1+\cos \alpha}{2}$




$\sin 3\alpha$

process
$\ \ \ \sin(2\alpha +\alpha)
=\sin 2\alpha \cos \alpha +\cos 2\alpha \sin \alpha$\\
$=2\sin \alpha \cos \alpha \cdot \cos \alpha
+(1-2\sin^2 \alpha)\cdot \sin \alpha$\\
$=2\sin \alpha(1-\sin^2 \alpha)
+\sin \alpha -2\sin^3 \alpha$\\
$=-4\sin^3 \alpha +3\sin \alpha$

$\sin 3\alpha=-4\sin^3 \alpha +3\sin \alpha$



$\cos 3\alpha$

process
$\ \ \ \cos(2\alpha + \alpha)
=\cos 2\alpha \cos \alpha - \sin 2\alpha \sin \alpha$\\
$=(2\cos^2 \alpha -1)\cdot \cos \alpha
-2\sin \alpha \cos \alpha \cdot \sin \alpha$\\
$=2\cos^3 \alpha -\cos \alpha
-2(1-\cos^2 \alpha)\cdot \cos \alpha$\\
$=2\cos^3 \alpha -\cos \alpha
-2\cos \alpha +2\cos^3 \alpha$\\
$=4\cos^3 \alpha -3\cos \alpha$\\
Õ[RQ܂

$\cos 3\alpha=4\cos^3 \alpha -3\cos \alpha$




$\sin \alpha \cos \beta$

process
\begin{eqnarray*}
 \sin (\alpha+\beta) &=& \sin\alpha \cos\beta + \cos\alpha \sin\beta  \\
+\sin (\alpha-\beta) &=& \sin\alpha \cos\beta - \cos\alpha \sin\beta  \\
\hline
\sin (\alpha+\beta)+\sin (\alpha-\beta)&=&2\sin\alpha \cos\beta
\end{eqnarray*}\\
ӂQŊ

$\sin \alpha \cos \beta=\dfrac12 \{ \sin(\alpha+\beta)+\sin(\alpha-\beta) \}$




$\cos \alpha \sin \beta$

process
\begin{eqnarray*}
 \sin (\alpha+\beta) &=& \sin\alpha \cos\beta + \cos\alpha \sin\beta  \\
-\sin (\alpha-\beta) &=& \sin\alpha \cos\beta - \cos\alpha \sin\beta  \\
\hline
\sin (\alpha+\beta)-\sin (\alpha-\beta)&=&2\cos\alpha \sin\beta
\end{eqnarray*}\\
ӂQŊ

$\cos \alpha \sin \beta=\dfrac12 \{ \sin(\alpha+\beta)-\sin(\alpha-\beta) \}$




$\cos \alpha \cos \beta$

process
\begin{eqnarray*}
 \cos (\alpha+\beta) &=& \cos\alpha \cos\beta - \sin\alpha \sin\beta  \\
+\cos (\alpha-\beta) &=& \cos\alpha \cos\beta + \sin\alpha \sin\beta  \\
\hline
\cos (\alpha+\beta)+\cos (\alpha-\beta)&=&2\cos\alpha \cos\beta
\end{eqnarray*}\\
ӂQŊ

$\cos \alpha \cos \beta=\dfrac12 \{ \cos (\alpha+\beta)+\cos (\alpha-\beta) \}$





$\sin\alpha \sin\beta$

process
\begin{eqnarray*}
 \cos (\alpha+\beta) &=& \cos\alpha \cos\beta - \sin\alpha \sin\beta  \\
-\cos (\alpha-\beta) &=& \cos\alpha \cos\beta + \sin\alpha \sin\beta  \\
\hline
\cos (\alpha+\beta)-\cos (\alpha-\beta)&=&-2\sin\alpha \sin\beta
\end{eqnarray*}\\
ӂQŊ

$\sin\alpha \sin\beta=-\dfrac12 \{ \cos (\alpha+\beta)-\cos (\alpha-\beta) \}$





$\sin A + \sin B$

process
$\alpha+\beta=A,\ \alpha-\beta=B$Ƃ\\
$\alpha=\dfrac{A+B}{2},\ \beta=\dfrac{A-B}{2}$ƂȂB\\
L̎ɑB\\
$\sin (\alpha+\beta)+\sin (\alpha-\beta)=2\sin\alpha \cos\beta$\\
$\sin A+ \sin B= 2\sin \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$

$\sin A+ \sin B= 2\sin \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$





$\sin A - \sin B$

process
$\alpha+\beta=A,\ \alpha-\beta=B$Ƃ\\
$\alpha=\dfrac{A+B}{2},\ \beta=\dfrac{A-B}{2}$ƂȂB\\
L̎ɑB\\
$\sin (\alpha+\beta)-\sin (\alpha-\beta)=2\cos\alpha \sin\beta$\\
$\sin A- \sin B= 2\cos \dfrac{A+B}{2} \sin \dfrac{A-B}{2}$

$\sin A- \sin B= 2\cos \dfrac{A+B}{2} \sin \dfrac{A-B}{2}$





$\cos A + \cos B$

process
$\alpha+\beta=A,\ \alpha-\beta=B$Ƃ\\
$\alpha=\dfrac{A+B}{2},\ \beta=\dfrac{A-B}{2}$ƂȂB\\
L̎ɑB\\
$\cos (\alpha+\beta)+\cos (\alpha-\beta)=2\cos\alpha \cos\beta$\\
$\cos A+\cos B=2\cos \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$

$\cos A+\cos B=2\cos \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$




$\cos A - \cos B$

process
$\alpha+\beta=A,\ \alpha-\beta=B$Ƃ\\
$\alpha=\dfrac{A+B}{2},\ \beta=\dfrac{A-B}{2}$ƂȂB\\
L̎ɑB\\
$\cos (\alpha+\beta)-\cos (\alpha-\beta)=-2\sin\alpha \sin\beta$\\
$\cos A-\cos B=-2\sin \dfrac{A+B}{2} \sin \dfrac{A-B}{2}$

$\cos A-\cos B=-2\sin \dfrac{A+B}{2} \sin \dfrac{A-B}{2}$





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