与えられた式を簡略化します。式は $sin(\theta + \frac{4}{3}\pi) + \frac{1}{\sqrt{2}}cos(\theta + \frac{\pi}{4})$ です。解析学三角関数加法定理三角関数の合成2025/6/161. 問題の内容与えられた式を簡略化します。式はsin(θ+43π)+12cos(θ+π4)sin(\theta + \frac{4}{3}\pi) + \frac{1}{\sqrt{2}}cos(\theta + \frac{\pi}{4})sin(θ+34π)+21cos(θ+4π)です。2. 解き方の手順まず、sin(θ+43π)sin(\theta + \frac{4}{3}\pi)sin(θ+34π) と cos(θ+π4)cos(\theta + \frac{\pi}{4})cos(θ+4π) を加法定理を用いて展開します。sin(θ+43π)=sin(θ)cos(43π)+cos(θ)sin(43π)sin(\theta + \frac{4}{3}\pi) = sin(\theta)cos(\frac{4}{3}\pi) + cos(\theta)sin(\frac{4}{3}\pi)sin(θ+34π)=sin(θ)cos(34π)+cos(θ)sin(34π)cos(θ+π4)=cos(θ)cos(π4)−sin(θ)sin(π4)cos(\theta + \frac{\pi}{4}) = cos(\theta)cos(\frac{\pi}{4}) - sin(\theta)sin(\frac{\pi}{4})cos(θ+4π)=cos(θ)cos(4π)−sin(θ)sin(4π)ここで、cos(43π)=−12cos(\frac{4}{3}\pi) = -\frac{1}{2}cos(34π)=−21, sin(43π)=−32sin(\frac{4}{3}\pi) = -\frac{\sqrt{3}}{2}sin(34π)=−23, cos(π4)=12cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}cos(4π)=21, sin(π4)=12sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}sin(4π)=21 です。したがって、sin(θ+43π)=−12sin(θ)−32cos(θ)sin(\theta + \frac{4}{3}\pi) = -\frac{1}{2}sin(\theta) - \frac{\sqrt{3}}{2}cos(\theta)sin(θ+34π)=−21sin(θ)−23cos(θ)cos(θ+π4)=12cos(θ)−12sin(θ)cos(\theta + \frac{\pi}{4}) = \frac{1}{\sqrt{2}}cos(\theta) - \frac{1}{\sqrt{2}}sin(\theta)cos(θ+4π)=21cos(θ)−21sin(θ)与えられた式に代入します。sin(θ+43π)+12cos(θ+π4)=(−12sin(θ)−32cos(θ))+12(12cos(θ)−12sin(θ))sin(\theta + \frac{4}{3}\pi) + \frac{1}{\sqrt{2}}cos(\theta + \frac{\pi}{4}) = (-\frac{1}{2}sin(\theta) - \frac{\sqrt{3}}{2}cos(\theta)) + \frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}}cos(\theta) - \frac{1}{\sqrt{2}}sin(\theta))sin(θ+34π)+21cos(θ+4π)=(−21sin(θ)−23cos(θ))+21(21cos(θ)−21sin(θ))=−12sin(θ)−32cos(θ)+12cos(θ)−12sin(θ)= -\frac{1}{2}sin(\theta) - \frac{\sqrt{3}}{2}cos(\theta) + \frac{1}{2}cos(\theta) - \frac{1}{2}sin(\theta)=−21sin(θ)−23cos(θ)+21cos(θ)−21sin(θ)=−sin(θ)+(12−32)cos(θ)= -sin(\theta) + (\frac{1}{2} - \frac{\sqrt{3}}{2})cos(\theta)=−sin(θ)+(21−23)cos(θ)=−sin(θ)+1−32cos(θ)= -sin(\theta) + \frac{1 - \sqrt{3}}{2}cos(\theta)=−sin(θ)+21−3cos(θ)3. 最終的な答え−sin(θ)+1−32cos(θ)-sin(\theta) + \frac{1 - \sqrt{3}}{2}cos(\theta)−sin(θ)+21−3cos(θ)