1.$\cos 75°$的值为 （）

A.$\frac{\sqrt{6}+\sqrt{2}}{4}$
B.$\frac{\sqrt{2}-\sqrt{6}}{4}$
C.$\frac{\sqrt{6}-\sqrt{2}}{4}$
D.$-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\cos 75° = \cos(45° + 30°) = \cos 45° \cos 30° - \sin 45° \sin 30° = \frac{\sqrt{2}}{2}×\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}×\frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}$。

2. $\sin 165° \cos 75° + \cos 15° \sin 105°$的值为（）

A.0
B.$\frac{1}{2}$
C.$\frac{\sqrt{2}}{2}$
D.1

$\sin 165° \cos 75° + \cos 15° \sin 105° = \sin(180° - 15°)·\cos 75° + \cos 15° \sin(180° - 75°) = \sin 15° \cos 75° + \cos 15° \sin 75° = \sin(15° + 75°) = 1$。

3.若$\cosα+\sinα=\frac{1}{3}$，则$\sin2α=$（）

A.$\frac{1}{9}$
B.$-\frac{8}{9}$
C.$\frac{\sqrt{6}}{3}$
D.$\frac{\sqrt{17}}{9}$

$(\cos α + \sin α)^2 = 1 + 2\sin α \cos α = 1 + \sin 2α = \frac{1}{9}$，得$\sin 2α = -\frac{8}{9}$。

4.若$α \in (0, \dfrac{π}{2})$，$3\sin 2α = 1 - \cos 2α$，则$\tan α =$（）

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

由$3\sin 2α = 1 - \cos 2α$，$α \in (0,\frac{π}{2})$，得$6\sin α \cos α = 1 - (1 - 2\sin^2 α) = 2\sin^2 α$，
即$3\cos α = \sin α$，故$\tan α = 3$。

5.(多选)下列式子中,计算正确的是 ()

A.$\cos \frac{π}{12}\cos \frac{π}{6} - \sin \frac{π}{12}\sin \frac{π}{6} = \frac{\sqrt{2}}{2}$
B.$\sin 47°\cos 17° + \cos 47°\cos 107° = \frac{1}{2}$
C.$\frac{1 - \tan 15°}{1 + \tan 15°} = \sqrt{3}$
D.$\cos^4 \frac{π}{8} - \sin^4 \frac{π}{8} = \frac{1}{2}$

$\cos \frac{π}{12}\cos \frac{π}{6} - \sin \frac{π}{12}\sin \frac{π}{6} = \cos( \frac{π}{12} + \frac{π}{6} ) = \cos \frac{π}{4} = \frac{\sqrt{2}}{2}$，选项A正确。
$\sin 47° \cos 17° + \cos 47° \cos 107° = \sin 47° \cos 17° - \cos 47° \sin 17° = \sin 30° = \frac{1}{2}$，故B正确。
$\frac{1-\tan 15°}{1+\tan 15°} = \frac{\tan 45° - \tan 15°}{1+\tan 45° \tan 15°} = \tan(45° - 15°) = \tan 30° = \frac{\sqrt{3}}{3}$，选项C错误。
$\cos^4 \frac{π}{8} - \sin^4 \frac{π}{8} = ( \cos^2 \frac{π}{8} - \sin^2 \frac{π}{8} )( \cos^2 \frac{π}{8} + \sin^2 \frac{π}{8} ) = \cos \frac{π}{4} × 1 = \frac{\sqrt{2}}{2} ≠ \frac{1}{2}$，选项D错误。

6.已知$\sin(θ + \dfrac{π}{4})=\dfrac{1}{3}(-\dfrac{3π}{4}≤ θ≤ \dfrac{π}{4})$，则$\cosθ=$（）

A.$\dfrac{4-\sqrt{2}}{6}$
B.$\dfrac{4+\sqrt{2}}{6}$
C.$\dfrac{2-\sqrt{2}}{6}$
D.$\dfrac{2+\sqrt{2}}{6}$

由$-\frac{3π}{4}\le θ \le \frac{π}{4}$，得$-\frac{π}{2}\le θ + \frac{π}{4}\le \frac{π}{2}$。由$\sin( θ + \frac{π}{4} ) = \frac{1}{3}$，得$\cos( θ + \frac{π}{4} ) = \sqrt{1 - ( \frac{1}{3} )^2} = \frac{2\sqrt{2}}{3}$，
所以$\cos θ = \cos[ ( θ + \frac{π}{4} ) - \frac{π}{4} ] = \cos( θ + \frac{π}{4} )\cos \frac{π}{4} + \sin( θ + \frac{π}{4} )\sin \frac{π}{4} = \frac{2\sqrt{2}}{3}×\frac{\sqrt{2}}{2} + \frac{1}{3}×\frac{\sqrt{2}}{2} = \frac{4+\sqrt{2}}{6}$。

7.已知$α,β\in(0,\dfrac{π}{2})$，$\cosα=\dfrac{5}{13}$，$\cos(α+β)=-\dfrac{3}{5}$，则$\sinβ=$.

因为$α,β \in (0,\frac{π}{2})$，所以$0＜α+β＜π$。
由同角三角函数的基本关系可得，$\sin α = \sqrt{1-\cos^2 α} = \sqrt{1-( \frac{5}{13} )^2} = \frac{12}{13}$，$\sin(α+β) = \sqrt{1-( -\frac{3}{5} )^2} = \frac{4}{5}$。
由两角差的正弦公式，得$\sin β = \sin[ (α+β) - α ] = \sin(α+β)\cos α - \cos(α+β)\sin α = \frac{4}{5}×\frac{5}{13} - ( -\frac{3}{5} )×\frac{12}{13} = \frac{56}{65}$。

8.已知$\tan α=2$，则$\dfrac{2\sin^{2}α + 3\cos 2α}{2\cos^{2}α + 1}=$。

$\frac{2\sin^2 α + 3\cos 2α}{2\cos^2 α + 1} = \frac{2\sin^2 α + 3(\cos^2 α - \sin^2 α)}{2\cos^2 α + \sin^2 α + \cos^2 α} = \frac{3\cos^2 α - \sin^2 α}{3\cos^2 α + \sin^2 α} = \frac{3 - \tan^2 α}{3 + \tan^2 α} = \frac{3 - 2^2}{3 + 2^2} = \frac{3-4}{3+4} = -\frac{1}{7}$。

9.(1)已知$\cos x=\frac{4}{5},x\in(-\frac{π}{2},0)$,求$\sin 2x,\cos 2x,\tan 2x$的值；
(2)已知$α$为锐角,且$\tan α=\frac{1}{2}$,求$\frac{\sin 2α \cos α - \sin α}{\sin 2α \cos 2α}$的值。

解:(1)因为$\cos x = \frac{4}{5}$，$x \in ( -\frac{π}{2},0 )$，
所以$\sin x = -\sqrt{1-\cos^2 x} = -\frac{3}{5}$，
所以$\sin 2x = 2\sin x \cos x = 2×( -\frac{3}{5} )×\frac{4}{5} = -\frac{24}{25}$，
$\cos 2x = 2\cos^2 x - 1 = 2×( \frac{4}{5} )^2 - 1 = \frac{7}{25}$，$\tan 2x = \frac{\sin 2x}{\cos 2x} = -\frac{24}{7}$。
(2)因为$α$为锐角，且$\tan α = \frac{1}{2}$，所以$\tan^2 α = \frac{\sin^2 α}{\cos^2 α} = \frac{1-\cos^2 α}{\cos^2 α} = \frac{1}{4}$，即$\cos^2 α = \frac{4}{5}$，解得$\cos α = \frac{2\sqrt{5}}{5}$，
所以$\frac{\sin 2α \cos α - \sin α}{\sin 2α \cos 2α} = \frac{\sin α(2\cos^2 α - 1)}{2\sin α \cos α(2\cos^2 α - 1)} = \frac{1}{2\cos α} = \frac{\sqrt{5}}{4}$。

10.已知$θ \in (\dfrac{π}{2}, π)$，满足$\cos(θ + \dfrac{π}{4})=\cos 2θ$，则$\tan 2θ=$（）

A.$\dfrac{1}{2}$
B.$\sqrt{3}$
C.$-\dfrac{\sqrt{3}}{3}$
D.$\dfrac{\sqrt{3}}{3}$
(提示:观察$2θ$与$θ + \dfrac{π}{4}$之间的关系，有$\cos 2θ=\cos[2(θ + \dfrac{π}{4})-\dfrac{π}{2}]=\sin[2(θ + \dfrac{π}{4})]$，从而求出$\sin(θ + \dfrac{π}{4})$的值，进而根据角的范围确定$θ$的值，即可得解)

$\cos( θ + \frac{π}{4} ) = \cos 2θ = \sin[ 2( θ + \frac{π}{4} ) ] = 2\sin( θ + \frac{π}{4} )\cos( θ + \frac{π}{4} )$。
因为$θ \in ( \frac{π}{2},π )$，所以$θ + \frac{π}{4} \in ( \frac{3π}{4},\frac{5π}{4} )$，即$\cos( θ + \frac{π}{4} ) ≠ 0$，
所以$\sin( θ + \frac{π}{4} ) = \frac{1}{2}$，所以$θ + \frac{π}{4} = \frac{5π}{6}$，
所以$θ = \frac{7π}{12}$，
所以$\tan 2θ = \tan \frac{7π}{6} = \tan \frac{π}{6} = \frac{\sqrt{3}}{3}$。