Mathematics
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三角函数

  1. 和差角

    sin(α±β)=sinαcosβ±sinβcosαsin(\alpha\pm\beta)=sin\alpha cos\beta\pm sin\beta cos\alpha

    cos(α±β)=cosαcosβsinαcosβcos(\alpha \pm \beta)=cos\alpha\cos\beta\mp sin\alpha cos\beta

    tan(α±β)=tanα+tanβ1tanαtanβtan(\alpha\pm\beta)=\frac{tan\alpha+tan\beta}{1\mp tan\alpha tan\beta}

  2. 二倍角

    sin2α=2sinαcosαsin2\alpha=2sin\alpha cos\alpha

    cos2αcos2\alpha=2cos2α1=cos2αsin2α=2cos^2\alpha-1=cos^2\alpha-sin^2\alpha=1sin2α=1-sin^2\alpha

    tan2α=2tanα1tan2αtan2\alpha=\frac{2tan\alpha}{1-tan^2\alpha}

  3. 辅助角

    asinα±bcosαasin\alpha\pm bcos\alpha=a2+b2(aa2+b2sinα±ba2+b2cosα)=\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}sin\alpha\pm \frac{b}{\sqrt{a^2+b^2}}cos\alpha)

    a2+b2(sinαcosφ±cosαsinφ)\Rightarrow\sqrt{a^2+b^2}(sin\alpha cos\varphi\pm cos\alpha sin\varphi)

    a2+b2sin(α±φ),tanφ=ba\Rightarrow \sqrt{a^2+b^2}sin(\alpha\pm \varphi),tan\varphi=\frac{b}{a}

  4. 降幂

    sin2α=1cos2α2sin^2\alpha=\frac{1-cos2\alpha}{2}

    cos2α=1+cosα2cos^2\alpha=\frac{1+cos\alpha}{2}

三角函数

  1. 振幅、周期

    振幅:AA,周期:T=2πωT=\frac{2\pi}{\omega}

  2. 函数变换:设y=A0sin(w0x+φ0)y=A_0sin(w_0x+\varphi_0)

    1. 沿xx轴平移向左或向右平移x0x_0个单位

      y=A0sin(w0(x±x0)+φ0)y=A_0sin(w_0(x\pm x_0)+\varphi_0)或者y=A0sin(w0x+(φ0±ω0x0))y=A_0sin(w_0x+(\varphi_0\pm \omega_0x_0))

    2. 横坐标伸长或缩短为原来的ω1\omega_1

      y=A0sin(w0w1x+φ0)y=A_0sin(\frac{w_0}{w_1}x+\varphi_0)

  3. 正弦函数性质

    1. 对称轴:x=π2+kπ,(kZ)x=\frac{\pi}{2}+k\pi,(k\in Z)
    2. 对称中心(根):x=kπ,(kZ)x=k\pi,(k\in Z)
    3. 增区间:[π2+2kπ,π2+2kπ](kZ)[-\frac{\pi}{2}+2k\pi,\frac{\pi}{2}+2k\pi](k\in Z)
    4. 减区间:[π2+2kπ,3π2+2kπ](kZ)[\frac{\pi}{2}+2k\pi,\frac{3\pi}{2}+2k\pi](k\in Z)
  4. cosx=sin(x+π2)cosx=sin(x+\frac{\pi}{2})

三角诱导公式

对于kπ2±α(kZ)\frac{k\pi}{2}\pm \alpha(k \in Z)有“奇变偶不变,符号看象限”。

  1. α\alpha一律看成锐角
  2. 符号取决于原函数名的符号,例如:sin(α+3π2)=cosαsin(\alpha+\frac{3\pi}{2})=-cos\alpha
  3. 类似于sin(π2α)sin(\frac{\pi}{2}-\alpha)可以利用sinsin函数奇偶性变换成sin(απ2)=cosα-sin(\alpha-\frac{\pi}{2})=cos\alpha
  4. sin(α+2π)=sinα,sin(\alpha+2\pi)=sin\alpha,cos(α+2π)=cos(α)cos(\alpha+2\pi)=cos(\alpha)

三角函数探究

  1. 已知函数f(x)=sin(ωx+π3)(ωZ)f(x)=sin(\omega x+\frac{\pi}{3})(\omega\in Z),x(0,π3]x\in(0,\frac{\pi}{3}]时,f(x)=32f(x)=\frac{\sqrt{3}}{2}有唯一解,则满足条件的ω\omega的个数是多少。

    ω>0\omega>0时,2π3π3ω+π3<π3+2π\frac{2\pi}{3}\le\frac{\pi}{3}\omega +\frac{\pi}{3} < \frac{\pi}{3}+2\pi

    1ω<6\Rightarrow1\le\omega<6

    ω<0\omega<0时,5π3<π3ω+π34π3-\frac{5\pi}{3}<\frac{\pi}{3}\omega +\frac{\pi}{3}\le-\frac{4\pi}{3}

    6<ω5\Rightarrow-6<\omega\le-5

    ω={1,2,3,4,5,5}\omega = \{1,2,3,4,5,-5\}

    6\therefore 6

  2. 已知函数f(x)f(x)=sin(ωx+π6)sin(ωx+2π3)=sin(\omega x + \frac{\pi}{6})sin(\omega x + \frac{2\pi}{3})(ω>0,xR)(\omega>0,x\in R),若f(x)f(x)在区间(π2,π)(\frac{\pi}{2},\pi)内没有零点,则ω\omega的取值范围是?

    解答:f(x)=sin(ωx+π6)sin(ωx+π6+π2)f(x)=sin(\omega x+\frac{\pi}{6})sin(\omega x+\frac{\pi}{6}+\frac{\pi}{2})

    f(x)=sin(ωx+π6)cos(ωx+π6)\Rightarrow f(x)=sin(\omega x+\frac{\pi}{6})cos(\omega x+\frac{\pi}{6})

    f(x)=12sin(2ωx+π3)\Rightarrow f(x)=\frac{1}{2}sin(2\omega x+\frac{\pi}{3})

    2ωx+π32\omega x+\frac{\pi}{3}=kπx=π6ω+kπ2ω,=k\pi\Rightarrow x= -\frac{\pi}{6\omega}+\frac{k\pi}{2\omega},kZk \in Z

    π2<π6ω+kπ2ω<π\frac{\pi}{2}<-\frac{\pi}{6\omega}+\frac{k\pi}{2\omega}<\piω+13<k<2ω+13\rightarrow\omega+\frac{1}{3}<k<2\omega+\frac{1}{3}

    \because函数f(x)f(x)在区间(π2,π)(\frac{\pi}{2},\pi)内没有零点,

    \therefore区间(ω+13,2ω+13)(\omega+\frac{1}{3},2\omega+\frac{1}{3})内不存在整数

    T2ππ2ω1\because \frac{T}{2}\ge\pi-\frac{\pi}{2},\therefore\omega\le1

    ω>0,(ω+13,2ω+13)(0,1a)\because \omega>0,\therefore(\omega+\frac{1}{3},2\omega+\frac{1}{3})\subset(0,1a)(ω+13,2ω+13)(1,2)(\omega+\frac{1}{3},2\omega+\frac{1}{3})\subset(1,2)

    0<ω13\therefore0<\omega\le\frac{1}{3}23ω56\frac{2}{3}\le\omega\le\frac{5}{6}

    ω(0,13][23,56]\therefore \omega\in(0,\frac{1}{3}]\cup[\frac{2}{3},\frac{5}{6}]


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