《2025年成才之路高中新课程学习指导高中数学必修第二册北师大版》

3.(多选)在$\triangle ABC$中,给出下列四个式子:其中为常数的是(BC)

A.$\sin(A+B)+\sin C$
B.$\cos(A+B)+\cos C$
C.$\sin(2A+2B)+\sin2C$
D.$\cos(2A+2B)+\cos2C$

答案： 3.BC A.$\sin(A + B) + \sin C = 2 \sin C$;
B.$\cos(A + B) + \cos C = -\cos C + \cos C = 0$;
C.$\sin(2A + 2B) + \sin 2C = \sin[2(A + B)] + \sin 2C = \sin[2(\pi - C)] + \sin 2C = \sin(2\pi - 2C) + \sin 2C = -\sin 2C + \sin 2C = 0$;
D.$\cos(2A + 2B) + \cos 2C = \cos[2(A + B)] + \cos 2C = \cos[2(\pi - C)] + \cos 2C = \cos(2\pi - 2C) + \cos 2C = \cos 2C + \cos 2C = 2 \cos 2C$.
故选 BC.

4.(多选)下列三角函数中与$\sin\frac{\pi}{3}$的值相同的是(BD)

A.$\sin(n\pi+\frac{4\pi}{3})$
B.$\cos(2n\pi+\frac{\pi}{6})$
C.$\cos[(2n+1)\pi-\frac{\pi}{6}]$
D.$\sin[(2n+1)\pi-\frac{\pi}{3}](n\in\mathbf{Z})$

答案： 4.BD A.$\sin\left(n\pi + \frac{4}{3}\pi\right) = (-1)^n \sin\frac{4}{3}\pi = (-1)^{n+1} · \sin\frac{\pi}{3}$;
B.$\cos\left(2n\pi + \frac{\pi}{6}\right) = \cos\frac{\pi}{6} = \sin\frac{\pi}{3}$;C.$\cos\left[(2n+1)\pi - \frac{\pi}{6}\right] = \cos\left(\pi - \frac{\pi}{6}\right) = -\cos\frac{\pi}{6} = -\sin\frac{\pi}{3}$;D.$\sin\left[(2n+1)\pi - \frac{\pi}{3}\right] = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\frac{\pi}{3}$,故选 BD.

5.$\cos1°+\cos2°+\cos3°+·s+\cos180°=$$-1$.

答案： 5.$-1$ $\because \cos(\pi - \theta) = -\cos \theta, \therefore \cos \theta + \cos(\pi - \theta) = 0$,
即$\cos 1° + \cos 179° = \cos 2° + \cos 178° = ·s = \cos 90° = 0$.
$\therefore$原式$= 0 + 0 + ·s + 0 + \cos 180° = -1$.

6.若$\sin(\pi-\alpha)-\cos(\pi+\alpha)=\frac{\sqrt{2}}{4},0＜\alpha＜\pi$,则$\sin(\pi+\alpha)+\cos(2\pi-\alpha)$等于$-\frac{\sqrt{30}}{4}$.(注:对任意角$\alpha$,有$\sin^2\alpha+\cos^2\alpha=1$)

答案： 6.$-\frac{\sqrt{30}}{4}$ $\sin(\pi - \alpha) - \cos(\pi + \alpha) = \frac{\sqrt{2}}{4}$,则$\sin \alpha + \cos \alpha = \frac{\sqrt{2}}{4}$
两边平方,化简得$\sin \alpha \cos \alpha = -\frac{7}{16} ＜ 0$,由$\alpha \in (0, \pi)$,得$\alpha \in \left(\frac{\pi}{2}, \pi\right)$,又$\sin(\pi + \alpha) + \cos(2\pi - \alpha) = -\sin \alpha + \cos \alpha$,$(\cos \alpha - \sin \alpha)^2 = 1 - 2 \sin \alpha \cos \alpha = \frac{15}{8}$,又$\cos \alpha - \sin \alpha ＜ 0$,所以$\cos \alpha - \sin \alpha = -\frac{\sqrt{30}}{4}$.

7.已知角$\alpha$终边上一点$P(-4,3)$,求$\frac{\sin(3\pi+\alpha)\sin(-\pi-\alpha)}{\sin(-\alpha+2\pi)\cos(-\alpha-4\pi)}$的值.

答案： 7.方法一:点$P$到原点$O$的距离$|OP| = \sqrt{(-4)^2 + 3^2} = 5$.
根据三角函数的定义得$\sin \alpha = \frac{3}{5}, \cos \alpha = -\frac{4}{5}$
$\frac{\sin(3\pi + \alpha) \sin(-\pi - \alpha)}{\sin(-\alpha + 2\pi) \cos(-\alpha - 4\pi)} = \frac{\sin(\pi + \alpha)[-\sin(\pi + \alpha)]}{\sin(-\alpha) \cos(-\alpha)} = \frac{-\sin \alpha \sin \alpha}{-\sin \alpha \cos \alpha} = \frac{\sin \alpha}{\cos \alpha} × \frac{-5}{4} = -\frac{3}{4}$
方法二:据三角函数定义$\tan \alpha = -\frac{3}{4}$
原式$ = \frac{\sin \alpha}{\cos \alpha} = \tan \alpha = -\frac{3}{4}$

8.已知$f(x)=\frac{\cos^2(n\pi+x)\sin^2(n\pi-x)}{\cos^2[(2n+1)\pi-x]}(n\in\mathbf{Z})$.
(1)化简$f(x)$的表达式;
(2)求$f(\frac{2020}{3}\pi)$.

答案： 8.
(1)当$n$为偶数,即$n = 2k (k \in \mathbf{Z})$时,
$f(x) = \frac{\cos^2(2k\pi + x) · \sin^2(2k\pi - x)}{\cos^2[(2 × 2k + 1)\pi - x]} = \frac{\cos^2 x · \sin^2(-x)}{(-\cos x)^2} = \sin^2 x$;
当$n$为奇数时,即$n = 2k + 1 (k \in \mathbf{Z})$时,
$f(x) = \frac{\cos^2[(2k + 1)\pi + x] · \sin^2[(2k + 1)\pi - x]}{\cos^2|[2 × (2k + 1) + 1]\pi - x|} = \frac{\cos^2(\pi + x) · \sin^2(\pi - x)}{(-\cos x)^2} = \frac{(-\cos x)^2 · \sin^2 x}{(-\cos x)^2} = \sin^2 x$,
综上得$f(x) = \sin^2 x$.
(2)由
(1)知$f\left(\frac{2020}{3}\right) = \sin^2 \frac{2020\pi}{3} = \sin^2\left(672\pi + \frac{4\pi}{3}\right) = \sin^2 \frac{4\pi}{3} = \sin^2\left(\pi + \frac{\pi}{3}\right) = \sin^2 \frac{\pi}{3} = \frac{3}{4}$