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【学以致用1】(1)在单位圆中,已知角α是第二象限角,它的终边与单位圆交于点$P(-\dfrac{3}{5},y)$,则$\sinα=$（
隐含条件是横纵坐标的平方和等于1.
A.$-\dfrac{4}{5}$ B.$-\dfrac{3}{5}$ C.$\dfrac{3}{5}$ D.$\dfrac{4}{5}$
(2)已知角θ的顶点为坐标原点,始边与x轴的非负半轴重合,终边过点$P(4,-3)$,则$\cosθ-\sinθ=$（
A.$\dfrac{1}{5}$ B.$-\dfrac{1}{5}$ C.$\dfrac{7}{5}$ D.$-\dfrac{7}{5}$
D
）隐含条件是横纵坐标的平方和等于1.
A.$-\dfrac{4}{5}$ B.$-\dfrac{3}{5}$ C.$\dfrac{3}{5}$ D.$\dfrac{4}{5}$
(2)已知角θ的顶点为坐标原点,始边与x轴的非负半轴重合,终边过点$P(4,-3)$,则$\cosθ-\sinθ=$（
C
）A.$\dfrac{1}{5}$ B.$-\dfrac{1}{5}$ C.$\dfrac{7}{5}$ D.$-\dfrac{7}{5}$
答案
(1)由题意,得$(-\dfrac{3}{5})^2 + y^2 = 1,y＞0$,解得$y=\dfrac{4}{5}$,所以$\sin α = y=\dfrac{4}{5}$。
(2)因为角$θ$的终边过点$P(4,-3)$,所以$\cos θ = \dfrac{4}{\sqrt{4^2+(-3)^2}}=\dfrac{4}{5}$,$\sin θ = \dfrac{-3}{\sqrt{4^2+(-3)^2}}=-\dfrac{3}{5}$,所以$\cos θ - \sin θ = \dfrac{4}{5}-(-\dfrac{3}{5})=\dfrac{7}{5}$。
(2)因为角$θ$的终边过点$P(4,-3)$,所以$\cos θ = \dfrac{4}{\sqrt{4^2+(-3)^2}}=\dfrac{4}{5}$,$\sin θ = \dfrac{-3}{\sqrt{4^2+(-3)^2}}=-\dfrac{3}{5}$,所以$\cos θ - \sin θ = \dfrac{4}{5}-(-\dfrac{3}{5})=\dfrac{7}{5}$。
