【学以致用1】(1)化简$\cos 2θ\cos(2π-θ)-\sin(π+2θ)\sinθ$所得的结果是 ()

A.$\cosθ$
B.$-\cosθ$
C.$\cos3θ$
D.$-\cos3θ$

cos 2θcos(2π−θ)−sin(π+2θ)sin θ = cos 2θcos(−θ)−(−sin 2θ)sin θ = cos 2θcos θ+sin 2θsin θ=cos(2θ−θ)=cos θ。

(2)已知$\cos(α+β)=\dfrac{1}{5}$，$\tanα\tanβ=2$，则$\cos(α-β)=$（）

A.$-\dfrac{3}{5}$
B.$-\dfrac{1}{5}$
C.$\dfrac{1}{5}$
D.$\dfrac{3}{5}$

因为$\cos(α+β)=\frac{1}{5}$，所以$\cos α\cos β-\sin α\sin β=\frac{1}{5}$。
因为$\tan α\tan β=2$，所以$\sin α\sin β=2\cos α\cos β$，
所以$\cos α\cos β=-\frac{1}{5}$，$\sin α\sin β=-\frac{2}{5}$，
所以$\cos(α-β)=\cos α\cos β+\sin α\sin β=-\frac{3}{5}$。

【学以致用2】(1) $\sin 16° \cos 46° - \cos 16° · \sin 46° =$ ()

A.$-\dfrac{\sqrt{3}}{2}$
B.$-\dfrac{1}{2}$
C.$\dfrac{1}{2}$
D.$\dfrac{\sqrt{3}}{2}$

$\sin 16° \cos 46° - \cos 16° \sin 46° = -(\sin 46° \cos 16° - \cos 46° \sin 16°) = -\sin(46° - 16°) = -\sin 30° = -\frac{1}{2}$。

(2)已知角α的顶点在坐标原点上，始边与x轴的非负半轴重合，终边经过点(4,－3)，则$\sin(α + \frac{π}{4})=$（）

A.$-\frac{7\sqrt{2}}{10}$
B.$\frac{7\sqrt{2}}{10}$
C.$-\frac{\sqrt{2}}{10}$
D.$\frac{\sqrt{2}}{10}$

由题意，知角$α$的终边经过点$(4,-3)$，则$\sin α = \frac{-3}{\sqrt{4^2+(-3)^2}} = -\frac{3}{5}$，$\cos α = \frac{4}{\sqrt{4^2+(-3)^2}} = \frac{4}{5}$，
故$\sin(α + \frac{π}{4}) = \sin α \cos \frac{π}{4} + \cos α \sin \frac{π}{4} = -\frac{3}{5}×\frac{\sqrt{2}}{2} + \frac{4}{5}×\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{10}$。