Hãy xác định kết quả sai: sin 7pi/12 = (căn 6 + căn 2)/4

* Hướng dẫn giải

Đáp án B

$$\begin{gathered} \sin \frac{7\pi }{12}=\sin \left(\frac{\pi }{3}+\frac{\pi }{4}\right) \\ =\sin \frac{\pi }{3}\cos \frac{\pi }{4}+\sin \frac{\pi }{4}\cos \frac{\pi }{3} \\ =\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}+\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4} \\ \cos {285}^0=\cos \left({360}^0-{75}^0\right)=\cos {75}^0=\cos \left({30}^0+{45}^0\right) \\ =\cos {30}^0\cos {45}^0-\sin {30}^0\sin {45}^0 \\ =\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4} \\ \sin \frac{\pi }{12}=\sin \left(\frac{\pi }{3}-\frac{\pi }{4}\right) \\ =\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\sin \frac{\pi }{4}\cos \frac{\pi }{3} \\ =\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4} \\ \sin \frac{103\pi }{12}=\sin \left(8\pi +\frac{7\pi }{12}\right)=\sin \frac{7\pi }{12}=\frac{\sqrt{6}+\sqrt{2}}{4} \end{gathered}$$