Ответ:
\(\cos\left(\frac{\pi}{4}-\varphi\right)=\cos\frac{\pi}{4}\cos\varphi+\sin\frac{\pi}{4}\sin\varphi\).
Так как \(\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\), получаем:
\(\cos\left(\frac{\pi}{4}-\varphi\right)=\frac{\sqrt{2}}{2}(\cos\varphi+\sin\varphi)\).
\(\cos\left(\frac{\pi}{4}+\varphi\right)=\cos\frac{\pi}{4}\cos\varphi-\sin\frac{\pi}{4}\sin\varphi\).
\(\cos\left(\frac{\pi}{4}+\varphi\right)=\frac{\sqrt{2}}{2}(\cos\varphi-\sin\varphi)\).
\(\sin\left(\varphi+\frac{\pi}{4}\right)=\sin\varphi\cos\frac{\pi}{4}+\cos\varphi\sin\frac{\pi}{4}\).
\(\sin\left(\varphi+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}(\sin\varphi+\cos\varphi)\).
\(\sin\left(\varphi-\frac{\pi}{4}\right)=\sin\varphi\cos\frac{\pi}{4}-\cos\varphi\sin\frac{\pi}{4}\).
\(\sin\left(\varphi-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}(\sin\varphi-\cos\varphi)\).
Ответ: а) \(\frac{\sqrt{2}}{2}(\cos\varphi+\sin\varphi)\); б) \(\frac{\sqrt{2}}{2}(\cos\varphi-\sin\varphi)\); в) \(\frac{\sqrt{2}}{2}(\sin\varphi+\cos\varphi)\); г) \(\frac{\sqrt{2}}{2}(\sin\varphi-\cos\varphi)\).
