この節について
角の範囲を一般角へ広げ、弧度法、三角関数のグラフ、三角方程式・不等式を扱う節です。度数法と弧度法の対応（\(180^\circ = \pi\)）に慣れることが最初の関門で、以降の微分積分でも弧度法が前提になります。
グラフの問題は \(y = a\sin(b\theta+c)+d\) の形で整理すると扱いやすくなります。振幅が \(|a|\)、周期が \(\dfrac{2\pi}{|b|}\)、\(c\) が横方向の平行移動、\(d\) が縦方向の平行移動に対応します。\(b\) で括ってから平行移動量を読む点を間違えやすいので注意が必要です。
三角方程式・不等式は単位円を描いて考えるのが確実です。指定された範囲の中に解が何個あるかを円周上で数えてから式に落とすと、解の書き落としを防げます。
5章 三角関数
2 三角関数
BASIC
2 三角関数
BASIC
285(1)
\begin{align*}
\quad 630^{\circ} &= 360^{\circ} + 270^{\circ}
\end{align*}
285(2)
\begin{align*}
\quad -310^{\circ} &= -360^{\circ} + 50^{\circ}
\end{align*}
285(3)
\begin{align*}
\quad 490^{\circ} &= 360^{\circ} + 130^{\circ}
\end{align*}
285(4)
\begin{align*}
\quad -1120^{\circ} &= -1080^{\circ} - 40^{\circ}
\end{align*}
285(5)
\begin{align*}
\quad 2000^{\circ} &= 1800^{\circ} + 200^{\circ}
\end{align*}
286(1)
\begin{align*}
\quad &\text{第3象限}
\end{align*}
286(2)
\begin{align*}
\quad 400^{\circ} &= 360^{\circ} + 40^{\circ} \quad \text{第1象限}
\end{align*}
286(3)
\begin{align*}
\quad -740^{\circ} &= -720^{\circ} - 20^{\circ} \quad \text{第4象限}
\end{align*}
286(4)
\begin{align*}
\quad 820^{\circ} &= 720^{\circ} + 100^{\circ} \quad \text{第2象限}
\end{align*}
286(5)
\begin{align*}
\quad -635^{\circ} &= -720^{\circ} + 85^{\circ} \quad \text{第1象限}
\end{align*}
287(1)
\begin{align*}
\sin 210^{\circ} &= -\frac{1}{2}
\end{align*}
287(2)
\begin{align*}
\cos 570^{\circ} &= \cos 210^{\circ} = -\frac{\sqrt{3}}{2}
\end{align*}
287(3)
\begin{align*}
\tan 390^{\circ} &= \tan 30^{\circ} = \frac{1}{\sqrt{3}}
\end{align*}
287(4)
\begin{align*}
\sin 630^{\circ} &= \sin 270^{\circ} = -1
\end{align*}
287(5)
\begin{align*}
\cos \left(-225^{\circ}\right)=\cos 135^{\circ}=-\frac{1}{\sqrt{2}}
\end{align*}
287(6)
\begin{align*}
\tan \left(-480^{\circ}\right)=\tan \left(-120^{\circ}\right)=\sqrt{3}
\end{align*}
288 (1)
\begin{align*}
135^{\circ} &= \frac{3}{4} \pi
\end{align*}
288 (2)
\begin{align*}
36^{\circ} &= \frac{\pi}{5}
\end{align*}
288 (3)
\begin{align*}
-10^{\circ} &= -\frac{1}{18} \pi
\end{align*}
288 (4)
\begin{align*}
240^{\circ} &= \frac{4}{3} \pi
\end{align*}
288 (5)
\begin{align*}
-190^{\circ} &= -\frac{19}{18} \pi
\end{align*}
289 (1)
\begin{align*}
\frac{\pi}{3} &= 60^{\circ}
\end{align*}
289 (2)
\begin{align*}
\frac{5}{4} \pi &= 225^{\circ}
\end{align*}
289 (3)
\begin{align*}
-\frac{2}{5} \pi &= -72^{\circ}
\end{align*}
289 (4)
\begin{align*}
\frac{7}{3} \pi &= 420^{\circ}
\end{align*}
289 (5)
\begin{align*}
-\frac{\pi}{9} &= -20^{\circ}
\end{align*}
290 (1)
\begin{align*}
\sin \frac{5}{3} \pi &= -\frac{\sqrt{3}}{2}
\end{align*}
290 (2)
\begin{align*}
\cos \frac{5}{4} \pi &= -\frac{1}{\sqrt{2}}
\end{align*}
290 (3)
\begin{align*}
\tan \frac{\pi}{6} &= \frac{1}{\sqrt{3}}
\end{align*}
291 (1)
\begin{align*}
\text{弧の長さ： } & 8\pi \times \frac{\frac{\pi}{6}}{2\pi} = \frac{2}{3}\pi \nr
\text{面積： } & 16\pi \times \frac{\frac{\pi}{6}}{2\pi} = \frac{4}{3}\pi
\end{align*}
291 (2)
中心角を \(\theta\) とすると
\begin{align*}
6\pi \times \frac{\theta}{2\pi} &= 2\pi \implies \theta = \frac{2}{3}\pi \nr
\text{面積： } & 9\pi \times \frac{\frac{2}{3}\pi}{2\pi} = 3\pi
\end{align*}
292 (1)
\begin{align*}
(\text{左辺}) &= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \nr
&= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \nr
&= \frac{1}{\sin \theta \cos \theta} = (\text{右辺})
\end{align*}
292 (2)
\begin{align*}
(\text{左辺}) &= \frac{1}{1-\cos \theta} + \frac{1}{1+\cos \theta} \nr
&= \frac{(1+\cos \theta) + (1-\cos \theta)}{1-\cos^2 \theta} \nr
&= \frac{2}{\sin^2 \theta} = (\text{右辺})
\end{align*}
293 (1)
\(180^{\circ} < \theta < 270^{\circ}\) なので \(\cos \theta < 0\)
\begin{align*}
\cos \theta &= -\sqrt{1 - \left( -\frac{4}{5} \right)^2} = -\frac{3}{5} \nr
\tan \theta &= \frac{\sin \theta}{\cos \theta} = \frac{4}{3}
\end{align*}
293 (2)
\(270^{\circ} < \theta < 360^{\circ}\) なので \(\sin \theta < 0\)
\begin{align*}
\sin \theta &= -\sqrt{1 - \left( \frac{1}{3} \right)^2} = -\frac{2\sqrt{2}}{3} \nr
\tan \theta &= \frac{\sin \theta}{\cos \theta} = -2\sqrt{2}
\end{align*}
293 (3)
\begin{align*}
1 + \tan^2 \theta &= \frac{1}{\cos^2 \theta} \nr
1 + 9 &= \frac{1}{\cos^2 \theta} \implies \cos^2 \theta = \frac{1}{10}
\end{align*}
\(180^{\circ} < \theta < 270^{\circ}\) なので \(\cos \theta < 0\)
\begin{align*}
\cos \theta &= -\frac{1}{\sqrt{10}} \nr
\sin \theta &= \tan \theta \cdot \cos \theta = -\frac{3}{\sqrt{10}}
\end{align*}
294 (1)
\begin{align*}
&\cos \theta \sin \left(\frac{\pi}{2}-\theta\right) + \sin \theta \cos \left(\frac{\pi}{2}-\theta\right) \nr
&= \cos \theta \cdot \cos \theta + \sin \theta \cdot \sin \theta \nr
&= \cos^2 \theta + \sin^2 \theta \nr
&= 1
\end{align*}
294 (2)
\begin{align*}
&\sin \left(\frac{\pi}{2}-\theta\right) - \sin (\pi+\theta) + \\
&\cos \left(\frac{\pi}{2}+\theta\right) + \cos (\pi-\theta) \nr
&= \cos \theta - (-\sin \theta) + (-\sin \theta) + (-\cos \theta) \nr
&= \cos \theta + \sin \theta - \sin \theta - \cos \theta \nr
&= 0
\end{align*}
294 (3)
\begin{align*}
&\tan (\pi+\theta) \sin \left(\frac{\pi}{2}+\theta\right) + \\
&\cos (\pi-\theta) \tan (\pi-\theta) \nr
&= \tan \theta \cdot \cos \theta + (-\cos \theta) \cdot (-\tan \theta) \nr
&= \sin \theta + \sin \theta \nr
&= 2 \sin \theta
\end{align*}
295 (1)
\begin{align*}
y &= \sin \left(x-\frac{\pi}{2}\right) \nr
&= -\cos x \nr
\text{周期：} & 2\pi
\end{align*}
295 (2)
\begin{align*}
y &= \cos \left(x+\frac{\pi}{4}\right) \nr
\text{周期：} & 2\pi
\end{align*}
296 (1)
\begin{align*}
y &= 2 \cos x \nr
\text{周期：} & 2\pi
\end{align*}
296 (2)
\begin{align*}
y &= -\frac{1}{2} \sin x \nr
\text{周期：} & 2\pi
\end{align*}
297 (1)
\begin{align*}
y &= \cos 2x \nr
\text{周期：} & \pi
\end{align*}
297 (2)
\begin{align*}
y &= \sin \frac{x}{3} \nr
\text{周期：} & 6\pi
\end{align*}
298 (1)
\begin{align*}
\sin x &= \frac{\sqrt{2}}{2}
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
x &= \frac{\pi}{4}, \frac{3}{4}\pi
\end{align*}
298 (2)
\begin{align*}
\cos x &= \frac{1}{2}
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
x &= \frac{\pi}{3}, \frac{5}{3}\pi
\end{align*}
298 (3)
\begin{align*}
\sin x &\geqq \frac{\sqrt{3}}{2}
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
\frac{\pi}{3} &\leqq x \leqq \frac{2}{3}\pi
\end{align*}
298 (4)
\begin{align*}
\cos x &< -\frac{1}{\sqrt{2}}
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
\frac{3}{4}\pi &< x < \frac{5}{4}\pi
\end{align*}
299 (1)
\begin{align*}
\tan x &= 0
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
x &= 0, \pi
\end{align*}
299 (2)
\begin{align*}
\tan x &= -1
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
x &= \frac{3}{4}\pi, \frac{7}{4}\pi
\end{align*}
CHECK
300 (1)
\begin{align*}
40^{\circ} &= \frac{40}{180}\pi = \frac{2}{9}\pi
\end{align*}
300 (2)
\begin{align*}
50^{\circ} &= \frac{50}{180}\pi = \frac{5}{18}\pi
\end{align*}
300 (3)
\begin{align*}
-18^{\circ} &= -\frac{18}{180}\pi = -\frac{1}{10}\pi
\end{align*}
300 (4)
\begin{align*}
-210^{\circ} &= -\frac{210}{180}\pi = -\frac{7}{6}\pi
\end{align*}
301 (1)
\begin{align*}
-\frac{\pi}{4} &= -45^{\circ}
\end{align*}
301 (2)
\begin{align*}
\frac{2}{3}\pi &= 120^{\circ}
\end{align*}
301 (3)
\begin{align*}
-\frac{11}{6}\pi &= -330^{\circ}
\end{align*}
301 (4)
\begin{align*}
\frac{7}{5}\pi &= \frac{7}{5} \times 180^{\circ} = 252^{\circ}
\end{align*}
302 (1)
\begin{align*}
\sin (-90^{\circ}) &= -1
\end{align*}
302 (2)
\begin{align*}
\cos 225^{\circ} &= -\frac{1}{\sqrt{2}}
\end{align*}
302 (3)
\begin{align*}
\tan (-780^{\circ}) &= \tan (-60^{\circ}) = -\sqrt{3}
\end{align*}
302 (4)
\begin{align*}
\sin \left(-\frac{17}{3}\pi\right) &= \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}
\end{align*}
302 (5)
\begin{align*}
\cos \frac{17}{6}\pi &= \cos \frac{5}{6}\pi = -\frac{\sqrt{3}}{2}
\end{align*}
302 (6)
\begin{align*}
\tan \left(\frac{9}{4}\pi\right) &= \tan \left(\frac{\pi}{4}\right) = 1
\end{align*}
303
\begin{align*}
\text{弧の長さ： } & 10\pi \times \frac{\frac{\pi}{4}}{2\pi} = \frac{5}{4}\pi \nr
\text{面積： } & 25\pi \times \frac{\frac{\pi}{4}}{2\pi} = \frac{25}{8}\pi
\end{align*}
304
\begin{align*}
(\text{左辺}) &= \frac{\sin \theta \{ (1+\cos \theta) - (1-\cos \theta) \}}{(1-\cos \theta)(1+\cos \theta)} \nr
&= \frac{2 \sin \theta \cos \theta}{1-\cos^2 \theta} \nr
&= \frac{2 \sin \theta \cos \theta}{\sin^2 \theta} \nr
&= \frac{2}{\tan \theta} = (\text{右辺})
\end{align*}
305
\(180^{\circ} < \theta < 270^{\circ}\) なので \(\cos \theta < 0\)
\begin{align*}
\cos \theta &= -\sqrt{1 - \left( -\frac{1}{4} \right)^2} \nr
&= -\frac{\sqrt{15}}{4} \nr
\tan \theta &= \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{15}}
\end{align*}
306 (1)
\begin{align*}
y &= \cos \left(x-\frac{\pi}{6}\right) \nr
\text{周期：} & 2\pi
\end{align*}
306 (2)
\begin{align*}
y &= 2 \sin 3x \nr
\text{周期：} & \frac{2}{3}\pi
\end{align*}
307 (1)
\begin{align*}
\sin x &= -\frac{\sqrt{2}}{2}
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
x &= \frac{5}{4}\pi, \ \frac{7}{4}\pi
\end{align*}
307 (2)
\begin{align*}
\tan x &= -\sqrt{3}
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
x &= \frac{2}{3}\pi, \ \frac{5}{3}\pi
\end{align*}
307 (3)
\begin{align*}
2 \sin x - 1 &< 0 \nr
\sin x &< \frac{1}{2}
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
0 \leqq x &< \frac{\pi}{6}, \ \frac{5}{6}\pi < x < 2\pi
\end{align*}
307 (4)
\begin{align*}
\cos x &\leqq \frac{\sqrt{2}}{2}
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
\frac{\pi}{4} \leqq x &\leqq \frac{7}{4}\pi
\end{align*}
308
\begin{align*}
y &= \sin x
\end{align*}
(1) \(a\) の値を求める：
\begin{align*}
a &= \sin \frac{5}{4}\pi = -\frac{1}{\sqrt{2}}
\end{align*}
(2) \(b\) の値を求める（\(0 < b < \frac{5}{4}\pi\)）：
\begin{align*}
\frac{1}{2} &= \sin b \implies b = \frac{5}{6}\pi
\end{align*}
(3) \(c\) の値を求める（\(-\pi < c < 0\)）：
\begin{align*}
-1 &= \sin c \implies c = -\frac{\pi}{2}
\end{align*}
STEP UP
309
\begin{align*}
&OA : OB = r_1 : r_2 \nr
&r_2 \cdot OA = r_1 \cdot (OA + l) \nr
&OA \cdot (r_2 - r_1) = r_1 l \nr
&OA = \frac{r_1}{r_2 - r_1} l, \quad OB = OA + l = \frac{r_2}{r_2 - r_1} l
\end{align*}
側面の展開図のおうぎ形の中心角を \(\theta\) とすると
\begin{align*}
&2\pi r_1 = 2 OA \cdot \pi \cdot \frac{\theta}{2\pi} \nr
&\theta = \frac{r_1}{OA} \cdot 2\pi = \frac{2(r_2 - r_1)}{l} \pi \nr
&S = \pi \cdot OB^2 \cdot \frac{\theta}{2\pi} - \pi \cdot OA^2 \cdot \frac{\theta}{2\pi} \nr
&= \frac{\theta}{2} \left( \frac{r_2 l}{r_2 - r_1} \right)^2 - \frac{\theta}{2} \left( \frac{r_1 l}{r_2 - r_1} \right)^2 \nr
&= \frac{\pi(r_2 - r_1)}{l} \cdot \left( \frac{l}{r_2 - r_1} \right)^2 (r_2^2 - r_1^2) \nr
&= \frac{\pi l}{r_2 - r_1} \cdot (r_2 + r_1)(r_2 - r_1) \nr
&= \pi l (r_1 + r_2)
\end{align*}
310
半径を \(r\) とすると弧の長さは \(12 - 2r \quad (0 < r < 6)\)
中心角を \(\theta\) とおくと
\begin{align*}
&12 - 2r = r \theta \implies \theta = \frac{12 - 2r}{r}
\end{align*}
面積 \(S\) は
\begin{align*}
&S = \frac{1}{2} r^2 \theta \nr
&= \frac{1}{2} r^2 \left( \frac{12 - 2r}{r} \right) \nr
&= 6r - r^2 \nr
&= -(r - 3)^2 + 9
\end{align*}
\(r = 3\) で最大値 \(9\) をとる
311
解と係数の関係より
\begin{align*}
&\begin{cases} \sin \theta + \cos \theta = \frac{2}{3} \\ \sin \theta \cos \theta = \frac{k}{3} \end{cases}
\end{align*}
\((\sin \theta + \cos \theta)^2 = \frac{4}{9}\) より
\begin{align*}
&1 + 2 \sin \theta \cos \theta = \frac{4}{9} \nr
&2 \cdot \frac{k}{3} = -\frac{5}{9} \nr
&k = -\frac{5}{6}
\end{align*}
312 (1)
\begin{align*}
y &= \sin \left( 2x - \frac{\pi}{2} \right) \nr
&= -\cos 2x \nr
\text{周期：} & \pi
\end{align*}
312 (2)
\begin{align*}
y &= \frac{1}{2} \cos \left\{ 3 \left( x - \frac{\pi}{12} \right) \right\}
\end{align*}
\(y = \frac{1}{2} \cos 3x\) のグラフを \(x\) 軸方向に \(\frac{\pi}{12}\) だけ平行移動したもの
\begin{align*}
\text{周期：} & \frac{2}{3}\pi
\end{align*}
312 (3)
\begin{align*}
y &= -\tan \left\{ 2 \left( x - \frac{\pi}{4} \right) \right\}
\end{align*}
\(y = -\tan 2x\) のグラフを \(x\) 軸方向に \(\frac{\pi}{4}\) だけ平行移動したもの
\begin{align*}
\text{周期：} & \frac{\pi}{2}
\end{align*}
313 (1)
\begin{align*}
2 \cos \left(x+\frac{\pi}{3}\right) &= \sqrt{3} \nr
\cos \left(x+\frac{\pi}{3}\right) &= \frac{\sqrt{3}}{2}
\end{align*}
\(0 \leqq x < 2\pi\) より \(\frac{\pi}{3} \leqq x+\frac{\pi}{3} < \frac{7}{3}\pi\) なので
\begin{align*}
x+\frac{\pi}{3} &= \frac{11}{6}\pi, \ \frac{13}{6}\pi \nr
x &= \frac{3}{2}\pi, \ \frac{11}{6}\pi
\end{align*}
313 (2)
\begin{align*}
\sin 2x &= \frac{1}{2}
\end{align*}
\(0 \leqq x < 2\pi\) より \(0 \leqq 2x < 4\pi\) なので
\begin{align*}
2x &= \frac{\pi}{6}, \ \frac{5}{6}\pi, \ \frac{13}{6}\pi, \ \frac{17}{6}\pi \nr
x &= \frac{\pi}{12}, \ \frac{5}{12}\pi, \ \frac{13}{12}\pi, \ \frac{17}{12}\pi
\end{align*}
313 (3)
\begin{align*}
2 \sin \left(2x - \frac{\pi}{6} \right) &= 1 \nr
\sin \left(2x - \frac{\pi}{6} \right) &= \frac{1}{2}
\end{align*}
\(0 \leqq x < 2\pi\) より \(-\frac{\pi}{6} \leqq 2x - \frac{\pi}{6} < \frac{23}{6}\pi\) なので
\begin{align*}
2x - \frac{\pi}{6} &= \frac{\pi}{6}, \ \frac{5}{6}\pi, \ \frac{13}{6}\pi, \ \frac{17}{6}\pi \nr
2x &= \frac{\pi}{3}, \ \pi, \ \frac{7}{3}\pi, \ 3\pi \nr
x &= \frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{7}{6}\pi, \ \frac{3}{2}\pi
\end{align*}
314
\begin{align*}
\tan x &\leqq \sqrt{3}
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
&0 \leqq x < \frac{\pi}{3}, \ \frac{\pi}{2} < x \leqq \frac{4}{3}\pi, \ \frac{3}{2}\pi < x < 2\pi
\end{align*}
315 (1)
\begin{align*}
2 \cos^2 x + \sin x - 1 &= 0 \nr
2(1 - \sin^2 x) + \sin x - 1 &= 0 \nr
2 \sin^2 x - \sin x - 1 &= 0 \nr
(2 \sin x + 1)(\sin x - 1) &= 0 \nr
\sin x &= -\frac{1}{2}, \ 1
\end{align*}
\(0 \leqq x < 2\pi\) より
\begin{align*}
x &= \frac{\pi}{2}, \ \frac{7}{6}\pi, \ \frac{11}{6}\pi
\end{align*}
315 (2)
\begin{align*}
2 \sin^2 x + 5 \cos x - 4 &< 0 \nr
2(1 - \cos^2 x) + 5 \cos x - 4 &< 0 \nr
2 \cos^2 x - 5 \cos x + 2 &> 0 \nr
(2 \cos x - 1)(\cos x - 2) &> 0 \nr
\cos x < \frac{1}{2}, \ 2 &< \cos x
\end{align*}
\(0 \leqq x < 2\pi\) かつ \(-1 \leqq \cos x \leqq 1\) より
\begin{align*}
\frac{\pi}{3} < x < \frac{5}{3}\pi
\end{align*}
316 (1)
\begin{align*}
y &= \sin^2 x - \sin x - 1
\end{align*}
\(t = \sin x\) とおくと、\(0 \leqq x < 2\pi\) より \(-1 \leqq t \leqq 1\)
\begin{align*}
y &= t^2 - t - 1 \nr
&= \left( t - \frac{1}{2} \right)^2 - \frac{5}{4}
\end{align*}
316 (2)
\(-1 \leqq t \leqq 1\) において
\begin{align*}
t &= -1 \text{ すなわち } x = \frac{3}{2}\pi \text{ で最大値 } 1 \nr
t &= \frac{1}{2} \text{ すなわち } x = \frac{\pi}{6}, \ \frac{5}{6}\pi \text{ で最小値 } -\frac{5}{4}
\end{align*}
317 (1)
\begin{align*}
&\begin{cases} 2 \sin x - 1 > 0 \nr 2 \cos x - \sqrt{2} \leqq 0 \end{cases} \nr
&\begin{cases} \sin x > \frac{1}{2} \nr \cos x \leqq \frac{1}{\sqrt{2}} \end{cases} \nr
&\begin{cases} \frac{\pi}{6} < x < \frac{5}{6} \pi \nr \frac{\pi}{4} \leqq x \leqq \frac{7}{4} \pi \end{cases} \nr
&\iff \frac{\pi}{4} < x < \frac{5}{6} \pi
\end{align*}
317 (2)
\begin{align*}
&\begin{cases} \tan x + 1 < 0 \nr 2 \cos x < 1 \end{cases} \nr
&\begin{cases} \tan x < -1 \nr \cos x < \frac{1}{2} \end{cases} \nr
&\begin{cases} \frac{\pi}{2} < x < \frac{3}{4} \pi, \ \frac{3}{2} \pi < x < \frac{7}{4} \pi \nr \frac{\pi}{3} < x < \frac{5}{3} \pi \end{cases} \nr
&\iff \frac{\pi}{2} < x < \frac{3}{4} \pi, \ \frac{3}{2} \pi < x < \frac{5}{3} \pi
\end{align*}
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