sinθ+cosθ=aの両辺の2乗

数研出版｜数学Ⅱ[709] p.131 練習10

\({\small (1)}~\)\(\sin\theta-\cos\theta=a\) の両辺を2乗すると、

---

\(\begin{eqnarray}~~~(\sin\theta-\cos\theta)^2&=&a^2
\\[3pt]~~~\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta&=&a^2
\\[3pt]~~~(\sin^2\theta+\cos^2\theta)-2\sin\theta\cos\theta&=&a^2\end{eqnarray}\)

---

ここで、\(\sin^2\theta+\cos^2\theta=1\) より、

---

\(\begin{eqnarray}~~~1-2\sin\theta\cos\theta&=&a^2
\\[3pt]~~~2\sin\theta\cos\theta&=&1-a^2
\\[3pt]~~~\sin\theta\cos\theta&=&\displaystyle\frac{\,1-a^2\,}{\,2\,}\end{eqnarray}\)
 
\({\small (2)}~\)因数分解の公式

---

　\(a^3-b^3=(a-b)(a^2+ab+b^2)\)

---

これを用いて、

---

---

\(\sin\theta-\cos\theta=a~,~\)\(\sin^2\theta+\cos^2\theta=1~,~\)

---

\(\sin\theta\cos\theta=\displaystyle\frac{\,1-a^2\,}{\,2\,}\) を代入すると、

---

\(\begin{eqnarray}~~~&=&a\left\{1+\displaystyle\frac{\,1-a^2\,}{\,2\,}\right\}
\\[5pt]~~~&=&a\cdot\displaystyle\frac{\,2+1-a^2\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle\frac{\,a(3-a^2)\,}{\,2\,}\end{eqnarray}\)

数研出版｜高等学校数学Ⅱ[710] p.161 演習問題A 1

\({\small (1)}~\)\(\sin\theta-\cos\theta=\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,}\) の両辺を2乗すると、

---

\(\begin{eqnarray}~~~(\sin\theta-\cos\theta)^2&=&\left(\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,}\right)^2
\\[3pt]~~~\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta&=&\displaystyle\frac{\,2\,}{\,9\,}
\\[3pt]~~~(\sin^2\theta+\cos^2\theta)-2\sin\theta\cos\theta&=&\displaystyle\frac{\,2\,}{\,9\,}\end{eqnarray}\)

---

ここで、\(\sin^2\theta+\cos^2\theta=1\) より、

---

\(\begin{eqnarray}~~~1-2\sin\theta\cos\theta&=&\displaystyle\frac{\,2\,}{\,9\,}
\\[3pt]~~~2\sin\theta\cos\theta&=&1-\displaystyle\frac{\,2\,}{\,9\,}
\\[3pt]~~~2\sin\theta\cos\theta&=&\displaystyle\frac{\,7\,}{\,9\,}
\\[3pt]~~~\sin\theta\cos\theta&=&\displaystyle\frac{\,7\,}{\,18\,}\end{eqnarray}\)
 
\({\small (2)}~\)\(\sin\theta+\cos\theta\) を2乗すると、

---

\(\begin{eqnarray}~~~&&(\sin\theta+\cos\theta)^2
\\[3pt]~~~&=&\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta
\\[3pt]~~~&=&(\sin^2\theta+\cos^2\theta)+2\sin\theta\cos\theta\end{eqnarray}\)

---

\(\sin^2\theta+\cos^2\theta=1~,~\)\(\sin\theta\cos\theta=\displaystyle\frac{\,7\,}{\,18\,}\) を代入すると、

---

\(\begin{eqnarray}~~~(\sin\theta+\cos\theta)^2&=&1+2\cdot\displaystyle\frac{\,7\,}{\,18\,}
\\[5pt]~~~&=&1+\displaystyle\frac{\,7\,}{\,9\,}
\\[5pt]~~~&=&\displaystyle\frac{\,16\,}{\,9\,}\end{eqnarray}\)

---

\(0{\small ~≦~}\theta{\small ~≦~}\displaystyle\frac{\,\pi\,}{\,2\,}\) のとき、\(\sin\theta{\small ~≧~}0~,~\cos\theta{\small ~≧~}0\) より、

---

　\(\sin\theta+\cos\theta{\small ~≧~}0\)

---

したがって、\(\sin\theta+\cos\theta=\displaystyle\frac{\,4\,}{\,3\,}\)
 
\({\small (3)}~\)\(\sin\theta-\cos\theta=\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,}~,~\)\(\sin\theta+\cos\theta=\displaystyle\frac{\,4\,}{\,3\,}\) を連立すると、

---

\(\begin{eqnarray}~~~ \left\{~\begin{array}{l}
\sin\theta-\cos\theta=\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,} \\[3pt] \sin\theta+\cos\theta=\displaystyle\frac{\,4\,}{\,3\,}
\end{array}\right.\end{eqnarray}\)

---

2式を足すと、

---

\(\begin{eqnarray}~~~2\sin\theta&=&\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,}+\displaystyle\frac{\,4\,}{\,3\,}
\\[5pt]~~~\sin\theta&=&\displaystyle\frac{\,\sqrt{\,2\,}+4\,}{\,6\,}\end{eqnarray}\)

---

2式を引くと、

---

\(\begin{eqnarray}~~~2\cos\theta&=&\displaystyle\frac{\,4\,}{\,3\,}-\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,}
\\[5pt]~~~\cos\theta&=&\displaystyle\frac{\,4-\sqrt{\,2\,}\,}{\,6\,}\end{eqnarray}\)

数研出版｜高等学校数学Ⅱ[710] p.121 練習10

\({\small (1)}~\)\(\sin\theta-\cos\theta=\displaystyle\frac{\,1\,}{\,3\,}\) の両辺を2乗すると、

---

\(\begin{eqnarray}~~~(\sin\theta-\cos\theta)^2&=&\left(\displaystyle\frac{\,1\,}{\,3\,}\right)^2
\\[3pt]~~~\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta&=&\displaystyle\frac{\,1\,}{\,9\,}
\\[3pt]~~~(\sin^2\theta+\cos^2\theta)-2\sin\theta\cos\theta&=&\displaystyle\frac{\,1\,}{\,9\,}\end{eqnarray}\)

---

ここで、\(\sin^2\theta+\cos^2\theta=1\) より、

---

\(\begin{eqnarray}~~~1-2\sin\theta\cos\theta&=&\displaystyle\frac{\,1\,}{\,9\,}
\\[3pt]~~~2\sin\theta\cos\theta&=&1-\displaystyle\frac{\,1\,}{\,9\,}
\\[3pt]~~~2\sin\theta\cos\theta&=&\displaystyle\frac{\,8\,}{\,9\,}
\\[3pt]~~~\sin\theta\cos\theta&=&\displaystyle\frac{\,4\,}{\,9\,}\end{eqnarray}\)
 
\({\small (2)}~\)因数分解の公式

---

　\(a^3-b^3=(a-b)(a^2+ab+b^2)\)

---

これを用いて、

---

---

\(\sin\theta-\cos\theta=\displaystyle\frac{\,1\,}{\,3\,}~,~\)\(\sin^2\theta+\cos^2\theta=1~,~\)

---

\(\sin\theta\cos\theta=\displaystyle\frac{\,4\,}{\,9\,}\) を代入すると、

---

\(\begin{eqnarray}~~~&=&\displaystyle\frac{\,1\,}{\,3\,}\left(1+\displaystyle\frac{\,4\,}{\,9\,}\right)
\\[5pt]~~~&=&\displaystyle\frac{\,1\,}{\,3\,}\cdot\displaystyle\frac{\,9+4\,}{\,9\,}
\\[5pt]~~~&=&\displaystyle\frac{\,13\,}{\,27\,}\end{eqnarray}\)

数研出版｜高等学校数学Ⅱ[710] p.150 章末問題A 2
数研出版｜新編数学Ⅱ[711] p.146 章末問題A 3

\({\small (1)}~\)\(\theta\) の動径は第4象限より、

---

　\(\sin\theta\lt0~,~\cos\theta\gt0\)

---

よって　\(\sin\theta-\cos\theta\lt0\)

---

\(\sin\theta-\cos\theta\) を2乗すると、

---

\(\begin{eqnarray}~~~&&(\sin\theta-\cos\theta)^2
\\[3pt]~~~&=&\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta
\\[3pt]~~~&=&(\sin^2\theta+\cos^2\theta)-2\sin\theta\cos\theta\end{eqnarray}\)

---

\(\sin^2\theta+\cos^2\theta=1~,~\)\(\sin\theta\cos\theta=-\displaystyle\frac{\,1\,}{\,4\,}\) を代入すると、

---

\(\begin{eqnarray}~~~(\sin\theta-\cos\theta)^2&=&1-2\cdot\left(-\displaystyle\frac{\,1\,}{\,4\,}\right)
\\[5pt]~~~(\sin\theta-\cos\theta)^2&=&1+\displaystyle\frac{\,1\,}{\,2\,}
\\[5pt]~~~(\sin\theta-\cos\theta)^2&=&\displaystyle\frac{\,3\,}{\,2\,}\end{eqnarray}\)

---

\(\sin\theta-\cos\theta\lt0\) より、

---

\(\begin{eqnarray}~~~\sin\theta-\cos\theta&=&-\sqrt{\,\displaystyle\frac{\,3\,}{\,2\,}\,}
\\[5pt]~~~&=&-\displaystyle\frac{\,\sqrt{\,3\,}\,}{\,\sqrt{\,2\,}\,}
\\[5pt]~~~&=&-\displaystyle\frac{\,\sqrt{\,6\,}\,}{\,2\,}\end{eqnarray}\)
 
\({\small (2)}~\)因数分解の公式

---

　\(a^3-b^3=(a-b)(a^2+ab+b^2)\)

---

これを用いて、

---

---

\(\sin\theta-\cos\theta=-\displaystyle\frac{\,\sqrt{\,6\,}\,}{\,2\,}~,~\)\(\sin^2\theta+\cos^2\theta=1~,~\)

---

\(\sin\theta\cos\theta=-\displaystyle\frac{\,1\,}{\,4\,}\) を代入すると、

---

\(\begin{eqnarray}~~~&=&-\displaystyle\frac{\,\sqrt{\,6\,}\,}{\,2\,}\left\{1+\left(-\displaystyle\frac{\,1\,}{\,4\,}\right)\right\}
\\[5pt]~~~&=&-\displaystyle\frac{\,\sqrt{\,6\,}\,}{\,2\,}\cdot\displaystyle\frac{\,3\,}{\,4\,}
\\[5pt]~~~&=&-\displaystyle\frac{\,3\sqrt{\,6\,}\,}{\,8\,}\end{eqnarray}\)

数研出版｜新編数学Ⅱ[711] p.120 練習11

\(\sin\theta+\cos\theta=a\) の両辺を2乗すると、

---

\(\begin{eqnarray}~~~(\sin\theta+\cos\theta)^2&=&a^2
\\[3pt]~~~\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta&=&a^2
\\[3pt]~~~(\sin^2\theta+\cos^2\theta)+2\sin\theta\cos\theta&=&a^2\end{eqnarray}\)

---

ここで、\(\sin^2\theta+\cos^2\theta=1\) より、

---

\(\begin{eqnarray}~~~1+2\sin\theta\cos\theta&=&a^2
\\[3pt]~~~\sin\theta\cos\theta&=&\displaystyle\frac{\,a^2-1\,}{\,2\,}\end{eqnarray}\)

---

ここで、因数分解の公式

---

　\(a^3+b^3=(a+b)(a^2-ab+b^2)\)

---

これを用いて、

---

---

\(\sin\theta+\cos\theta=a~,~\)\(\sin^2\theta+\cos^2\theta=1~,~\)

---

\(\sin\theta\cos\theta=\displaystyle\frac{\,a^2-1\,}{\,2\,}\) を代入すると、

---

\(\begin{eqnarray}~~~&=&a\left\{1-\displaystyle\frac{\,a^2-1\,}{\,2\,}\right\}
\\[5pt]~~~&=&a\cdot\displaystyle\frac{\,2-(a^2-1)\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle\frac{\,a(3-a^2)\,}{\,2\,}\end{eqnarray}\)

東京書籍｜Advanced数学Ⅱ[701] p.121 問12
東京書籍｜Standard数学Ⅱ[702] p.128 問11

\(\sin\theta-\cos\theta=\displaystyle\frac{\,1\,}{\,5\,}\) の両辺を2乗すると、

---

\(\begin{eqnarray}~~~(\sin\theta-\cos\theta)^2&=&\left(\displaystyle\frac{\,1\,}{\,5\,}\right)^2
\\[3pt]~~~\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta&=&\displaystyle\frac{\,1\,}{\,25\,}
\\[3pt]~~~(\sin^2\theta+\cos^2\theta)-2\sin\theta\cos\theta&=&\displaystyle\frac{\,1\,}{\,25\,}\end{eqnarray}\)

---

ここで、\(\sin^2\theta+\cos^2\theta=1\) より、

---

\(\begin{eqnarray}~~~1-2\sin\theta\cos\theta&=&\displaystyle\frac{\,1\,}{\,25\,}
\\[3pt]~~~2\sin\theta\cos\theta&=&1-\displaystyle\frac{\,1\,}{\,25\,}
\\[3pt]~~~2\sin\theta\cos\theta&=&\displaystyle\frac{\,24\,}{\,25\,}
\\[3pt]~~~\sin\theta\cos\theta&=&\displaystyle\frac{\,12\,}{\,25\,}\end{eqnarray}\)
 
因数分解の公式

---

　\(a^3-b^3=(a-b)(a^2+ab+b^2)\)

---

これを用いて、

---

---

\(\sin\theta-\cos\theta=\displaystyle\frac{\,1\,}{\,5\,}~,~\)\(\sin^2\theta+\cos^2\theta=1~,~\)

---

\(\sin\theta\cos\theta=\displaystyle\frac{\,12\,}{\,25\,}\) を代入すると、

---

\(\begin{eqnarray}~~~&=&\displaystyle\frac{\,1\,}{\,5\,}\left(1+\displaystyle\frac{\,12\,}{\,25\,}\right)
\\[5pt]~~~&=&\displaystyle\frac{\,1\,}{\,5\,}\cdot\displaystyle\frac{\,25+12\,}{\,25\,}
\\[5pt]~~~&=&\displaystyle\frac{\,1\,}{\,5\,}\cdot\displaystyle\frac{\,37\,}{\,25\,}
\\[5pt]~~~&=&\displaystyle\frac{\,37\,}{\,125\,}\end{eqnarray}\)

東京書籍｜Advanced数学Ⅱ[701] p.135 問題 2
東京書籍｜Standard数学Ⅱ[702] p.156 Level Up 1

\({\small (1)}~\)\(\theta\) は第3象限の角より、

---

　\(\sin\theta\lt0~,~\cos\theta\lt0\)

---

よって　\(\sin\theta+\cos\theta\lt0\)

---

\(\sin\theta+\cos\theta\) を2乗すると、

---

\(\begin{eqnarray}~~~&&(\sin\theta+\cos\theta)^2
\\[3pt]~~~&=&\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta
\\[3pt]~~~&=&(\sin^2\theta+\cos^2\theta)+2\sin\theta\cos\theta\end{eqnarray}\)

---

\(\sin^2\theta+\cos^2\theta=1~,~\)\(\sin\theta\cos\theta=\displaystyle\frac{\,1\,}{\,4\,}\) を代入すると、

---

\(\begin{eqnarray}~~~(\sin\theta+\cos\theta)^2&=&1+2\cdot\displaystyle\frac{\,1\,}{\,4\,}
\\[5pt]~~~(\sin\theta+\cos\theta)^2&=&1+\displaystyle\frac{\,1\,}{\,2\,}
\\[5pt]~~~(\sin\theta+\cos\theta)^2&=&\displaystyle\frac{\,3\,}{\,2\,}\end{eqnarray}\)

---

\(\sin\theta+\cos\theta\lt0\) より、

---

\(\begin{eqnarray}~~~\sin\theta+\cos\theta&=&-\sqrt{\,\displaystyle\frac{\,3\,}{\,2\,}\,}
\\[5pt]~~~&=&-\displaystyle\frac{\,\sqrt{\,3\,}\,}{\,\sqrt{\,2\,}\,}
\\[5pt]~~~&=&-\displaystyle\frac{\,\sqrt{\,6\,}\,}{\,2\,}\end{eqnarray}\)
 
\({\small (2)}~\)因数分解の公式

---

　\(a^3+b^3=(a+b)(a^2-ab+b^2)\)

---

これを用いて、

---

---

\(\sin\theta+\cos\theta=-\displaystyle\frac{\,\sqrt{\,6\,}\,}{\,2\,}~,~\)\(\sin^2\theta+\cos^2\theta=1~,~\)

---

\(\sin\theta\cos\theta=\displaystyle\frac{\,1\,}{\,4\,}\) を代入すると、

---

\(\begin{eqnarray}~~~&=&-\displaystyle\frac{\,\sqrt{\,6\,}\,}{\,2\,}\left(1-\displaystyle\frac{\,1\,}{\,4\,}\right)
\\[5pt]~~~&=&-\displaystyle\frac{\,\sqrt{\,6\,}\,}{\,2\,}\cdot\displaystyle\frac{\,3\,}{\,4\,}
\\[5pt]~~~&=&-\displaystyle\frac{\,3\sqrt{\,6\,}\,}{\,8\,}\end{eqnarray}\)

東京書籍｜Advanced数学Ⅱ[701] p.154 練習問題A 2

\({\small (1)}~\)\(\sin\theta+\cos\theta=\displaystyle\frac{\,2\,}{\,3\,}\) の両辺を2乗すると、

---

\(\begin{eqnarray}~~~(\sin\theta+\cos\theta)^2&=&\left(\displaystyle\frac{\,2\,}{\,3\,}\right)^2
\\[3pt]~~~\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta&=&\displaystyle\frac{\,4\,}{\,9\,}
\\[3pt]~~~(\sin^2\theta+\cos^2\theta)+2\sin\theta\cos\theta&=&\displaystyle\frac{\,4\,}{\,9\,}\end{eqnarray}\)

---

ここで、\(\sin^2\theta+\cos^2\theta=1\) より、

---

\(\begin{eqnarray}~~~1+2\sin\theta\cos\theta&=&\displaystyle\frac{\,4\,}{\,9\,}
\\[3pt]~~~2\sin\theta\cos\theta&=&\displaystyle\frac{\,4\,}{\,9\,}-1
\\[3pt]~~~2\sin\theta\cos\theta&=&-\displaystyle\frac{\,5\,}{\,9\,}
\\[3pt]~~~\sin\theta\cos\theta&=&-\displaystyle\frac{\,5\,}{\,18\,}\end{eqnarray}\)
 
\({\small (2)}~\)\(-\displaystyle\frac{\,\pi\,}{\,2\,}\lt\theta\lt0\) のとき、\(\sin\theta\lt0~,~\cos\theta\gt0\) より、

---

よって　\(\sin\theta-\cos\theta\lt0\)

---

\(\sin\theta-\cos\theta\) を2乗すると、

---

\(\begin{eqnarray}~~~&&(\sin\theta-\cos\theta)^2
\\[3pt]~~~&=&\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta
\\[3pt]~~~&=&(\sin^2\theta+\cos^2\theta)-2\sin\theta\cos\theta\end{eqnarray}\)

---

\(\sin^2\theta+\cos^2\theta=1~,~\)\(\sin\theta\cos\theta=-\displaystyle\frac{\,5\,}{\,18\,}\) を代入すると、

---

\(\begin{eqnarray}~~~(\sin\theta-\cos\theta)^2&=&1-2\cdot\left(-\displaystyle\frac{\,5\,}{\,18\,}\right)
\\[5pt]~~~(\sin\theta-\cos\theta)^2&=&1+\displaystyle\frac{\,5\,}{\,9\,}
\\[5pt]~~~(\sin\theta-\cos\theta)^2&=&\displaystyle\frac{\,14\,}{\,9\,}\end{eqnarray}\)

---

\(\sin\theta-\cos\theta\lt0\) より、

---

\(\begin{eqnarray}~~~\sin\theta-\cos\theta&=&-\sqrt{\,\displaystyle\frac{\,14\,}{\,9\,}\,}
\\[5pt]~~~&=&-\displaystyle\frac{\,\sqrt{\,14\,}\,}{\,3\,}\end{eqnarray}\)

東京書籍｜Standard数学Ⅱ[702] p.144 Training 3

\({\small (1)}~\)\(\sin\theta-\cos\theta=\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,}\) の両辺を2乗すると、

---

\(\begin{eqnarray}~~~(\sin\theta-\cos\theta)^2&=&\left(\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,}\right)^2
\\[3pt]~~~\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta&=&\displaystyle\frac{\,2\,}{\,9\,}
\\[3pt]~~~(\sin^2\theta+\cos^2\theta)-2\sin\theta\cos\theta&=&\displaystyle\frac{\,2\,}{\,9\,}\end{eqnarray}\)

---

ここで、\(\sin^2\theta+\cos^2\theta=1\) より、

---

\(\begin{eqnarray}~~~1-2\sin\theta\cos\theta&=&\displaystyle\frac{\,2\,}{\,9\,}
\\[3pt]~~~2\sin\theta\cos\theta&=&1-\displaystyle\frac{\,2\,}{\,9\,}
\\[3pt]~~~2\sin\theta\cos\theta&=&\displaystyle\frac{\,7\,}{\,9\,}
\\[3pt]~~~\sin\theta\cos\theta&=&\displaystyle\frac{\,7\,}{\,18\,}\end{eqnarray}\)
 
\({\small (2)}~\)\(\theta\) は第1象限の角より、\(\sin\theta\gt0~,~\cos\theta\gt0\) より、

---

よって　\(\sin\theta+\cos\theta\gt0\)

---

\(\sin\theta+\cos\theta\) を2乗すると、

---

\(\begin{eqnarray}~~~&&(\sin\theta+\cos\theta)^2
\\[3pt]~~~&=&\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta
\\[3pt]~~~&=&(\sin^2\theta+\cos^2\theta)+2\sin\theta\cos\theta\end{eqnarray}\)

---

\(\sin^2\theta+\cos^2\theta=1~,~\)\(\sin\theta\cos\theta=\displaystyle\frac{\,7\,}{\,18\,}\) を代入すると、

---

\(\begin{eqnarray}~~~(\sin\theta+\cos\theta)^2&=&1+2\cdot\displaystyle\frac{\,7\,}{\,18\,}
\\[5pt]~~~(\sin\theta+\cos\theta)^2&=&1+\displaystyle\frac{\,7\,}{\,9\,}
\\[5pt]~~~(\sin\theta+\cos\theta)^2&=&\displaystyle\frac{\,16\,}{\,9\,}\end{eqnarray}\)

---

\(\sin\theta+\cos\theta\gt0\) より、

---

　\(\sin\theta+\cos\theta=\displaystyle\frac{\,4\,}{\,3\,}\)
 
\({\small (3)}~\)因数分解の公式

---

　\(a^3-b^3=(a-b)(a^2+ab+b^2)\)

---

これを用いて、

---

---

\(\sin\theta-\cos\theta=\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,}~,~\)\(\sin^2\theta+\cos^2\theta=1~,~\)

---

\(\sin\theta\cos\theta=\displaystyle\frac{\,7\,}{\,18\,}\) を代入すると、

---

\(\begin{eqnarray}~~~&=&\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,}\left(1+\displaystyle\frac{\,7\,}{\,18\,}\right)
\\[5pt]~~~&=&\displaystyle\frac{\,\sqrt{\,2\,}\,}{\,3\,}\cdot\displaystyle\frac{\,25\,}{\,18\,}
\\[5pt]~~~&=&\displaystyle\frac{\,25\sqrt{\,2\,}\,}{\,54\,}\end{eqnarray}\)