加法定理を用いて、以下の値を求めます。 (1) $\sin(\frac{7}{12}\pi)$ (2) $\cos(\frac{3}{4}\pi)$解析学三角関数加法定理sincos2025/6/41. 問題の内容加法定理を用いて、以下の値を求めます。(1) sin(712π)\sin(\frac{7}{12}\pi)sin(127π)(2) cos(34π)\cos(\frac{3}{4}\pi)cos(43π)2. 解き方の手順(1) sin(712π)\sin(\frac{7}{12}\pi)sin(127π)を求めます。712π\frac{7}{12}\pi127πをπ3\frac{\pi}{3}3πとπ4\frac{\pi}{4}4πの和として表します。sin(712π)=sin(π3+π4)\sin(\frac{7}{12}\pi) = \sin(\frac{\pi}{3} + \frac{\pi}{4})sin(127π)=sin(3π+4π)加法定理 sin(A+B)=sinAcosB+cosAsinB\sin(A+B) = \sin A \cos B + \cos A \sin Bsin(A+B)=sinAcosB+cosAsinB を用いると、sin(π3+π4)=sinπ3cosπ4+cosπ3sinπ4\sin(\frac{\pi}{3} + \frac{\pi}{4}) = \sin \frac{\pi}{3} \cos \frac{\pi}{4} + \cos \frac{\pi}{3} \sin \frac{\pi}{4}sin(3π+4π)=sin3πcos4π+cos3πsin4πそれぞれの三角関数の値を代入します。sinπ3=32\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}sin3π=23, cosπ4=22\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}cos4π=22, cosπ3=12\cos \frac{\pi}{3} = \frac{1}{2}cos3π=21, sinπ4=22\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}sin4π=22したがって、sin(712π)=32⋅22+12⋅22\sin(\frac{7}{12}\pi) = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2}sin(127π)=23⋅22+21⋅22sin(712π)=64+24\sin(\frac{7}{12}\pi) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}sin(127π)=46+42sin(712π)=6+24\sin(\frac{7}{12}\pi) = \frac{\sqrt{6} + \sqrt{2}}{4}sin(127π)=46+2(2) cos(34π)\cos(\frac{3}{4}\pi)cos(43π)を求めます。34π\frac{3}{4}\pi43πはπ−π4\pi - \frac{\pi}{4}π−4πと表せます。cos(34π)=cos(π−π4)\cos(\frac{3}{4}\pi) = \cos(\pi - \frac{\pi}{4})cos(43π)=cos(π−4π)cos(π−x)=−cosx\cos(\pi - x) = - \cos xcos(π−x)=−cosxの関係を利用すると、cos(π−π4)=−cosπ4\cos(\pi - \frac{\pi}{4}) = - \cos \frac{\pi}{4}cos(π−4π)=−cos4πcosπ4=22\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}cos4π=22 であるから、cos(34π)=−22\cos(\frac{3}{4}\pi) = - \frac{\sqrt{2}}{2}cos(43π)=−223. 最終的な答え(1) 6+24\frac{\sqrt{6} + \sqrt{2}}{4}46+2(2) −22- \frac{\sqrt{2}}{2}−22