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update mathformula
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Yuzhiy05 committed Nov 7, 2024
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例如 $\sin x$的图像过原点,在$\sin(\frac{\pi}{2})$时到最大值`1`,过$\pi$时为`0`,过$\frac{3\pi}{2}$时为最小值`-1`。当$f(x+\frac{\pi}{2})$时图像左移,因为三角函数为周期函数的性质。
原先$f(0)$现在为$f(0+\frac{\pi}{2})$,原先为$f(\frac{\pi}{2})$现在为$f(\frac{\pi}{2}+\frac{\pi}{2})$,原先$f(\pi)$现在为$f(\pi+\frac{\pi}{2})$对三角函数某些特殊点的推论可以看出函数向左移动了

1.$$
1.
$\huge\sin(\theta+\frac{\pi}{2})=\cos\theta \\
\cos(\theta+\frac{\pi}{2})=-\sin\theta \\
\sin(\frac{\pi}{2}-\theta)=-\sin(\theta-\frac{\pi}{2})=-(-\cos\theta)=\cos\theta\\
\cos(\frac{\pi}{2}-\theta)=\cos(\theta-\frac{\pi}{2})=\sin\theta$

2.和差化积
$\huge\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\
\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}\\
\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha+\beta}{2}\\
\cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$

2.1 正弦相关
$\huge\tan\alpha$

$\mathrm{Rt}\triangle$

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