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#875a0b51-7a52-4640-898f-1913466aa33e简单解答题同角三角函数关系三角函数与三角恒等变换

21. (2023春•谯城区校级期中)已知f(α)=sin(α3π)cos(2πα)sin(α+3π2)cos(πα)sin(πα)f(\alpha )=\frac{\sin (\alpha -3\pi )\cos (2\pi -\alpha )\sin (-\alpha +\frac{3\pi }{2})}{\cos (-\pi -\alpha )\sin (-\pi -\alpha )}
(1)化简f(α)f(\alpha )
(2)若α\alpha是第三象限角,且cos(α3π2)=15\cos (\alpha -\frac{3\pi }{2})=\frac{1}{5},求f(α+π6)f(\alpha +\frac{\pi }{6})的值;