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#34781c9b-1c1c-4a0f-944d-bf57b86677c6简单解答题同角三角函数关系三角函数与三角恒等变换

19. (2023春•朝阳区校级月考)已知函数f(x)=sin(3πx)cos(x+4π)sin(3π2+x)sin(π2x)sin(7π2x)f(x)=\frac{\sin (3\pi -x)\cos (x+4\pi )\sin (\frac{3\pi }{2}+x)}{\sin (\frac{\pi }{2}-x)\sin (\frac{7\pi }{2}-x)}
(1)化简函数f(x)f(x)的解析式;
(2)若f(x+π5)=23f(x+\frac{\pi }{5})=-\frac{\sqrt{2}}{3}x(0,π)x\in (0,\pi ),求sin(3π10x)\sin (\frac{3\pi }{10}-x)的值.