三角函数 / trigonometry
由于打完一些竞赛，整理记录下有用的三角函数公式。
基本知识点
primary: $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$
secondary: $\csc\theta=\frac{1}{\sin\theta}=\frac{\text{hypotenuse}}{\text{opposite}},\sec\theta=\frac{1}{\cos\theta}=\frac{\text{hypotenuse}}{\text{adjacent}},\cot=\frac{1}{\tan\theta}=\frac{\text{adjacent}}{\text{opposite}}$
基本变换
$\tan\theta=\frac{\sin\theta}{\cos\theta},\cot\theta=\frac{\cos\theta}{\sin\theta}$ $\csc\theta=\frac{1}{\sin\theta},\sec\theta=\frac{1}{\cos\theta},\cot=\frac{1}{\tan\theta}$
$1+\tan^2\theta=\sec^2\theta$, $1+\cot^2\theta=\csc^2\theta$
和差公式
$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$
$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$
$\tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}$
二倍角公式
$\sin2\theta=2\sin\theta\cos\theta$
$\cos2\theta=2\cos^2\theta-1=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$
降幂公式
$\sin^2\theta=\frac{1-\cos2\theta}{2}$
$\cos^2\theta=\frac{1+\cos2\theta}{2}$
$\tan^2\theta=\frac{1-\cos2\theta}{1+\cos2\theta}$
$\sin\theta=\pm\sqrt{\frac{1-\cos2\theta}{2}}$
$\cos\theta=\pm\sqrt{\frac{1+\cos2\theta}{2}}$
$\tan\theta=\frac{1-\cos2\theta}{\sin2\theta}=\frac{\sin2\theta}{1+\cos2\theta}$
三倍角公式
$\begin{aligned}
\sin3\theta&=\sin\theta\cos2\theta+\cos\theta\sin2\theta=\sin\theta(\cos^2\theta-\sin^2\theta)+2\sin\theta\cos^2\theta=3\sin\theta\cos^2\theta-\sin^3\theta\\
&=3\sin\theta(1-\sin^2\theta)-\sin^3\theta=\sin\theta(3-4\sin^2\theta)=3\sin\theta-4\sin^3\theta\\
\sin\theta\sin(\frac{\pi}{3}-\theta)\sin(\frac{\pi}{3}+\theta)&=\sin\theta(\frac{\sqrt{3}}{2}\cos\theta-\frac{1}{2}\sin\theta)(\frac{\sqrt{3}}{2}\cos\theta+\frac{1}{2}\sin\theta)=\sin\theta(\frac{3}{4}\cos^2\theta-\frac{1}{4}\sin^2\theta)\\
&=\frac{1}{4}(3\sin\theta\cos^2\theta-\sin^3\theta)=\frac{1}{4}\sin3\theta\Rightarrow\boxed{\sin3\theta=4\sin\theta\sin(\frac{\pi}{3}-\theta)\sin(\frac{\pi}{3}+\theta)}
\end{aligned}$
$\begin{aligned}
\cos3\theta&=\cos\theta\cos2\theta-\sin\theta\sin2\theta=\cos\theta(\cos^2\theta-\sin^2\theta)-2\sin^2\theta\cos\theta=\cos^3\theta-3\sin^2\theta\cos\theta\\
&=\cos^3\theta-3(1-\cos^2\theta)\cos\theta=\cos\theta(4\cos^2\theta-3)=4\cos^3\theta-3\cos\theta\\
\cos\theta\cos(\frac{\pi}{3}-\theta)\cos(\frac{\pi}{3}+\theta)&=\cos\theta(\frac{1}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta)(\frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta)=\cos\theta(\frac{1}{4}\cos^2\theta-\frac{3}{4}\sin^2\theta)\\
&=\frac{1}{4}(\cos^3\theta-3\sin^2\theta\cos\theta)=\frac{1}{4}\cos3\theta\Rightarrow\boxed{\cos3\theta=4\cos\theta\cos(\frac{\pi}{3}-\theta)\cos(\frac{\pi}{3}+\theta)}
\end{aligned}$
$\tan3\theta=\frac{\sin3\theta}{\cos3\theta}=\frac{4\sin\theta\sin(\frac{\pi}{3}-\theta)\sin(\frac{\pi}{3}+\theta)}{4\cos\theta\cos(\frac{\pi}{3}-\theta)\cos(\frac{\pi}{3}+\theta)}=\tan\theta\tan(\frac{\pi}{3}-\theta)\tan(\frac{\pi}{3}+\theta)$
万能(置换)公式
- $\sin\theta=\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$
- $\cos\theta=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$
- $\tan\theta=\frac{2\tan\frac{\theta}{2}}{1-\tan^{2}\frac{\theta}{2}}$
辅助角公式
$p\sin\alpha+q\cos\alpha=\sqrt{p^2+q^2}\sin\beta$, $\beta=\alpha+\theta$
$\theta\in[0,2\pi)$, $\theta=\tan^{-1}(\frac{q}{p})=\arctan(\frac{q}{p})$
$\sin\theta=\frac{q}{\sqrt{p^2+q^2}},\cos\theta=\frac{p}{\sqrt{p^2+q^2}}, \tan\theta=\frac{q}{p}$
和差化积
$\begin{aligned}
\sinα+\sinβ&=2\sin\frac{(α+β)}{2}\cos\frac{(α-β)}{2}=2\sin\frac{(α+β)}{2}\cos\frac{(β-α)}{2},\\
\sinα-\sinβ&=2\cos\frac{(α+β)}{2}\sin\frac{(α-β)}{2}=2\sin\frac{(α-β)}{2}\cos\frac{(α+β)}{2}=-2\sin\frac{(β-α)}{2}\cos\frac{(α+β)}{2},\\
\cosα+\cosβ&=2\cos\frac{(α+β)}{2}\cos\frac{(α-β)}{2}=2\cos\frac{(α+β)}{2}\cos\frac{(β-α)}{2},\\
\cosα-\cosβ&=-2\sin\frac{(α+β)}{2}\sin\frac{(α-β)}{2}=2\sin\frac{(α+β)}{2}\sin\frac{(β-α)}{2}.
\end{aligned}$
积化和差
$\sinα\sinβ=\frac{(\cos\alpha\cos\beta+\sin\alpha\sin\beta)-(\cos\alpha\cos\beta-\sin\alpha\sin\beta)}{2}=\frac{cos(α-β)-cos(α+β)}{2}=-\frac{cos(α+β)-cos(α-β)}{2}=-\frac{cos(α+β)-cos(β-α)}{2}$,
$\sin\alpha\cos\beta=\frac{(\sinα\cosβ+\cosα\sinβ)+(\sinα\cosβ-\cosα\sinβ)}{2}=\frac{\sin{(α+β)}+\sin(α-β)}{2}=\frac{\sin{(α+β)}-\sin(β-α)}{2}$,
$\cos\alpha\sin\beta=\frac{(\sinα\cosβ+\cosα\sinβ)-(\sinα\cosβ-\cosα\sinβ)}{2}=\frac{\sin{(α+β)}-\sin(α-β)}{2}=\frac{\sin{(α+β)}+\sin(β-α)}{2}$,
$\cosα\cosβ=\frac{(\cos\alpha\cos\beta+\sin\alpha\sin\beta)+(\cos\alpha\cos\beta-\sin\alpha\sin\beta)}{2}=\frac{cos(α-β)+cos(α+β)}{2}=\frac{cos(β-α)+cos(α+β)}{2}$.
三角形恒等式
Let $A,B,C$ represent the three interior angles of the triangle $\triangle ABC$.
- $\sin A+\sin B+\sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$
$\begin{align}
&\sin A+\sin B+\sin C\\
=&2\sin\frac{A+B}{2}\cos\frac{A-B}{2}+2\sin\frac{C}{2}\cos\frac{C}{2}\\
=&2\cos\frac{C}{2}\cos\frac{A-B}{2}+2\cos\frac{A+B}{2}\cos\frac{C}{2}\\
=&2\cos\frac{C}{2}\left(\cos\frac{A-B}{2}+\cos\frac{A+B}{2}\right)\\
=&2\cos\frac{C}{2}\cdot2\cos\frac{A}{2}\cos\frac{B}{2}\\
=&4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}
\end{align}$
- $\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C$
$\begin{align}&\sin2A+\sin2B+\sin2C\\
=&2\sin(A+B)\cos(A-B)+2\sin C\cos C\\
=&2\sin C\cos(A-B)-2\sin C\cos(A+B)\\
=&2\sin C\left[\cos(A-B)-\cos(A+B)\right]\\
=&2\sin C\cdot2\sin A\sin B\\
=&4\sin A\sin B\sin C
\end{align}$
- $\sin^{2}A+\sin^{2}B+\sin^{2}C=2+2\cos A\cos B\cos C$
$\begin{aligned}
&\sin^{2}A+\sin^{2}B+\sin^{2}C\\
=&\frac{1-\cos2A}{2}+\frac{1-\cos2B}{2}+\frac{1-\cos2C}{2}\\
=&\frac{3}{2}-\frac{1}{2}(\cos2A+\cos2B+\cos2C)\\
=& \frac{3}{2}-\frac{1}{2}(-1-4\cos A\cos B\cos C)\\
=&2+2\cos A\cos B\cos C
\end{aligned}$
- $\cos A+\cos B+\cos C=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
$\begin{align}
&\cos A+\cos B+\cos C=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\\
&\left(\cos A-1\right)+\left[\cos B-\cos(A+B)\right]=4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\\
&-2\sin^2\frac{A}{2}+2\sin\frac{A}{2}\sin\frac{A+2B}{2}=4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\\
&2\sin\frac{A}{2}\left(\sin\frac{A+2B}{2}-\sin\frac{A}{2}\right)4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\\
&2\cos\frac{A+B}{2}\sin\frac{B}{2}=2\sin\frac{B}{2}\sin\frac{C}{2}\\
&\cos\frac{A+B}{2}=\sin\left(\frac{\pi}{2}-\frac{A+B}{2}\right)=\sin\frac{\pi-A-B}{2}=\sin\frac{C}{2}
&\end{align}$
- $\cos2A+\cos2B+\cos2C=-1-4\cos A\cos B\cos C$
$\begin{align}
&\cos2A+\cos2B+\cos2C=-1-4\cos A\cos B\cos C\\
&(\cos2A+1)+(\cos2B+\cos2C)=-4\cos A\cos B\cos C\\
&2\cos^2A+2\cos(B+C)\cos(B-C)=-4\cos A\cos B\cos C\\
&2\cos^2A+2\cos A\cos(A+2B)=-4\cos A\cos B\cos C\\
&2\cos A\big(\cos A+\cos(A+2B)\big)=-4\cos A\cos B\cos C\\
&2\cos A\cdot2\cos(A+B)\cos B=-4\cos A\cos B\cos C\\
&-4\cos A\cos B\cos C=-4\cos A\cos B\cos C
\end{align}$
- $\cos^{2}A+\cos^{2}B+\cos^{2}C=1-2\cos A\cos B\cos C$
$\begin{aligned}
&\cos^{2}A+\cos^{2}B+\cos^{2}C=1-2\cos A\cos B\cos C\\
=&3-\sin^{2}A+\sin^{2}B+\sin^{2}C\\
=&3-(2+2\cos A\cos B\cos C)\\
=&1-2\cos A\cos B\cos C
\end{aligned}$
- $\tan A+\tan B+\tan C=\tan A\tan B\tan C$
$\begin{align}
&\tan A+\tan B+\tan C=\tan A\tan B\tan C\\
&\tan C-\tan A\tan B\tan C=-(\tan A+\tan B)\\
&\tan C(1-\tan A\tan B)=-(\tan A+\tan B)\\
&\tan C=-\frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan(A+B)=\tan(\pi-A-B)
\end{align}$
- $\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1$
$\begin{align}
&\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}\\
=&\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{C}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}\right)\\
=&\tan\frac{A}{2}\tan\frac{B}{2}+\frac{1}{\tan\frac{A+B}{2}}\cdot\left(\tan\frac{A}{2}+\tan\frac{B}{2}\right)\\
=&\tan\frac{A}{2}\tan\frac{B}{2}+\frac{1-\tan\frac{A}{2}\tan\frac{B}{2}}{\tan\frac{A}{2}+\tan\frac{B}{2}}\cdot\left(\tan\frac{A}{2}+\tan\frac{B}{2}\right)\\
=&\tan\frac{A}{2}\tan\frac{B}{2}+1-\tan\frac{A}{2}\tan\frac{B}{2}\\
=&1
\end{align}$
- $\cot A\cot B+\cot B\cot C+\cot C\cot A=1$
$\begin{align}
&\cot A\cot B+\cot B\cot C+\cot C\cot A\\
=&\tan(\frac{\pi}{2}-A)\tan(\frac{\pi}{2}-B)+\tan(\frac{\pi}{2}-B)\tan(\frac{\pi}{2}-C)+\tan(\frac{\pi}{2}-C)\tan(\frac{\pi}{2}-A)\\
=&\tan\frac{\pi-2A}{2}\tan\frac{\pi-2B}{2}+\tan\frac{\pi-2B}{2}\tan\frac{\pi-2C}{2}+\tan\frac{\pi-2C}{2}\tan\frac{\pi-2A}{2}\\
=&1
\end{align}$
because $\frac{\pi-2A}{2}+\frac{\pi-2B}{2}+\frac{\pi-2A}{2}=\frac{3\pi}{2}-(A+B+C)=\frac{3\pi}{2}-\pi=\frac{\pi}{2}$
交错公式
本质上是想办法让东西互相抵消，然后就可以把很长的式子压缩到只有几项。
$\begin{align}
&\sin x+\sin2x+...+\sin nx\quad (x\ne2k\pi,k\in\mathbb{Z})\\
=&\frac{1}{\sin x}\left(\sin x\sin x+\sin2x\sin x+...+\sin nx\sin x\right)\\
=&\frac{1}{\sin x}\left(\sum_{k=1}^{n}\frac{\cos(k-1)x-\cos(k+1)x}{2}\right)\\
=&\frac{1}{\sin x}\frac{1-\cos(n+1)x}{2}\\
=&\frac{1+\cos x-\cos nx-\cos(n+1)x}{2\sin x}\\
=&\frac{1+\cos^2\frac{x}{2}-\sin^2\frac{x}{2}-\cos nx-\cos(n+1)x}{4\sin\frac{x}{2}\cos\frac{x}{2}}\\
=&\frac{2\cos^2\frac{x}{2}-2\cos\frac{x}{2}\cos\frac{(2n+1)x}{2}}{4\sin\frac{x}{2}\cos\frac{x}{2}}\\
=&\frac{\cos\frac{x}{2}-\cos\frac{(2n+1)x}{2}}{2\sin\frac{x}{2}}
\end{align}$
$\begin{align}
&\sin x+\sin2x+...+\sin nx\quad (x\ne2k\pi,k\in\mathbb{Z})\\
=&\frac{1}{\sin\frac{x}{2}}\left(\sin x\sin\frac{x}{2}+\sin2x\sin\frac{x}{2}+...+\sin nx\sin\frac{x}{2}\right)\\
=&\frac{1}{\sin\frac{x}{2}}\left(\sum_{k=1}^{n}\frac{\cos\frac{2n-1}{2}x-\cos\frac{2n+1}{2}x}{2}\right)\\
=&\frac{\cos\frac{x}{2}-\cos\frac{(2n+1)x}{2}}{2\sin\frac{x}{2}}
\end{align}$
$\begin{align}
&\cos x+\cos2x+...+\cos nx\\
=&\frac{1}{\sin\frac{x}{2}}\left(\cos x\sin\frac{x}{2}+\cos2x\sin\frac{x}{2}+...+\cos nx\sin\frac{x}{2}\right)\\
=&\frac{1}{\sin\frac{x}{2}}\left(\sum_{i=1}^{n}\frac{\sin\frac{2n+1}{2}x-\sin\frac{2n-1}{2}x}{2}\right)\\
=&\frac{\sin\frac{2n+1}{2}x-\sin\frac{x}{2}}{2\sin\frac{x}{2}}
\end{align}$
$\begin{align}
&\frac{1}{\cosθ\cos2θ}+\frac{1}{\cos2θ\cos3θ}+...+\frac{1}{\cos(n-1)θ\cos nθ}\\
=&\frac{1}{\sinθ}\left(\frac{\sinθ}{\cosθ\cos2θ}+\frac{\sinθ}{\cos2θ\cos3θ}+...+\frac{\sinθ}{\cos(n-1)θ\cos nθ}\right)\\
=&\frac{1}{\sinθ}\left(\frac{\sin(2-1)θ}{\cosθ\cos2θ}+\frac{\sin(3-2)θ}{\cos2θ\cos3θ}+...+\frac{\sin[n-(n-1)]θ}{\cos(n-1)θ\cos nθ}\right)\\
=&\frac{1}{\sinθ}\left(\frac{\sin2θ\cosθ-\cos2θ\sinθ}{\cosθ\cos2θ}+\frac{\sin3θ\cos2θ-\cos3θ\sin2θ}{\cos2θ\cos3θ}+...+\frac{\sin nθ\cos(n-1)θ-\cos nθ\sin(n-1)θ}{\cos(n-1)θ\cos nθ}\right)\\
=&\frac{1}{\sinθ}\cdot\Bigg(\left(\frac{\sin2θ}{\cos2θ}-\frac{\sinθ}{\cosθ}\right)+\left(\frac{\sin3θ}{\cos3θ}-\frac{\sin2θ}{\cos2θ}\right)+...+\left(\frac{\sin nθ}{\cos nθ}-\frac{\sin(n-1)θ}{\cos(n-1)θ}\right)\Bigg)\\
=&\frac{1}{\sinθ}\cdot\big((\tan2θ-\tanθ)+(\tan3θ-\tan2θ)+...+(\tan nθ-\tan(n-1)θ)\big)\\
=&\frac{\tan nθ-\tanθ}{\sinθ}
\end{align}$
$\begin{align}
&\frac{1}{\cosθ\cos2θ}+\frac{1}{\cos2θ\cos3θ}+...+\frac{1}{\cos(n-1)θ\cos nθ}=0\\
\Rightarrow\,&\tan nθ-\tanθ=0\\
\Rightarrow\,&θ=\frac{m\pi}{n-1}\quad m\in\mathbb{Z}
\end{align}$
$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\
1-\tan\alpha\tan\beta=\frac{\tan\alpha+\tan\beta}{\tan(\alpha+\beta)}\\
\tan\alpha\tan\beta=1-\frac{\tan\alpha+\tan\beta}{\tan(\alpha+\beta)}$ and $\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\\
1+\tan\alpha\tan\beta=\frac{\tan\alpha-\tan\beta}{\tan(\alpha-\beta)}\\
\tan\alpha\tan\beta=\frac{\tan\alpha-\tan\beta}{\tan(\alpha-\beta)}-1$
一些例子（题目+解析）
$\begin{aligned} &\tan 3^{\circ} \tan 7^{\circ}+\tan 7^{\circ} \tan 11^{\circ}+\cdots+\tan 179^{\circ} \tan 183^{\circ} \\ =&\left[\frac{\tan 3^{\circ}-\tan 7^{\circ}}{\tan \left(3^{\circ}-7^{\circ}\right)}-1\right]+\left[\frac{\tan 7^{\circ}-\tan 11^{\circ}}{\tan \left(7^{\circ}-11^{\circ}\right)}-1\right]+\cdots+\left[\frac{\tan 179^{\circ}-\tan 183^{\circ}}{\tan \left(179^{\circ}-183^{\circ}\right)}-1\right] \\ =&\frac{\tan 3^{\circ}-\tan 7^{\circ}}{\tan \left(-4^{\circ}\right)}+\frac{\tan 7^{\circ}-\tan 11^{\circ}}{\tan \left(-4^{\circ}\right)}+\cdots+\frac{\tan 179^{\circ}-\tan 183^{\circ}}{\tan \left(-4^{\circ}\right)}-45 \\ =&\frac{\tan 3^{\circ}-\tan 183^{\circ}}{\tan \left(-4^{\circ}\right)}-45=\frac{\tan 3^{\circ}-\tan 3^{\circ}}{\tan \left(-4^{\circ}\right)}=0-45=-45 \end{aligned}$
$\begin{aligned}
(\cos1^{\circ}\cos3^{\circ}\cdots\cos89^{\circ})^2&=\sin ^{2} 1^{\circ} \sin ^{2} 3^{\circ} \cdots \sin ^{2} 89^{\circ}\\
&=\sin 1^{\circ} \sin 3^{\circ} \ldots \sin 179^{\circ}\\
&=\frac{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin 179^{\circ}}{\sin 2^{\circ} \sin 4^{\circ} \cdots \sin 178^{\circ}}\\
&=\frac{\sin 90^{\circ} \sin 91^{\circ} \cdots \sin 179^{\circ}}{2\cos1^{\circ} \cdot 2\cos2^{\circ} \cdots 2\cos89^{\circ}}\\
&=\frac{1 \cdot \cos \left(99^{\circ}-9^{\circ}\right) \cos \left(92^{\circ}-90^{\circ}\right) \cdots \cos \left(179^{\circ}-90^{\circ}\right)}{2^{89} \cdot \cos 1^{\circ} \cos 2^{\circ} \cdots \cos 89^{\circ}}\\
&=\frac{1}{2^{89}}
\end{aligned}$
$\begin{align}
&\sin2^{\circ}\sin4^{\circ}...\sin90^{\circ}\\
=&(\frac{1}{4})^{14}\cdot\sin6^{\circ}\sin12^{\circ} ...\sin84^{\circ}\cdot\sin30^{\circ}\sin60^{\circ}\\
=&(\frac{1}{4})^{18}\cdot\sin18^{\circ}\sin36^{\circ}\sin54^{\circ}\sin72^{\circ}\cdot\sin^230^{\circ}\sin^260^{\circ}\\
=&\frac{3}{4^{20}}\cdot\sin36^{\circ}\cos36^{\circ}\sin72^{\circ}\cos72^{\circ}\\
=&\frac{3}{2^{41}}\cdot\sin^272^{\circ}\cos72^{\circ}
\end{align}$
Because $\sin^272^{\circ}=\frac{5+2\sqrt5}{6+2\sqrt5}$ and $\cos72^{\circ}=\frac{1}{\sqrt5+1}$,
$\begin{align}
&\sin2^{\circ}\sin4^{\circ} ... \sin90^{\circ}\\
=&\frac{3}{2^{41}}\cdot\frac{5+2\sqrt5}{16+8\sqrt5}=\frac{3}{2^{44}}\cdot\frac{5+2\sqrt5}{2+\sqrt5}\\
=&\frac{3\sqrt5}{2^{44}}\\
=&\frac{192\sqrt5}{2^{50}}
\end{align}$
$\begin{align}
&\sin3^{\circ}\sin6^{\circ}...\sin87^{\circ}\\
=&^{-9}\frac{\sqrt{3}}{4}\cdot4^{-9}\cdot\sin9^{\circ}\sin18^{\circ}...\sin81^{\circ}\\
=&\frac{\sqrt{6}}{2^3}\cdot4^{-9}\cdot\sin9^{\circ}\sin18^{\circ}\sin27^{\circ}\sin36^{\circ}\cos9^{\circ}\cos18^{\circ}\cos27^{\circ}\cos36^{\circ}\\
=&\frac{\sqrt{6}}{2^7}\cdot4^{-9}\cdot\sin18^{\circ}\sin36^{\circ}\sin54^{\circ}\sin72^{\circ}\\
=&\frac{\sqrt{6}}{2^9}\cdot2^{-18}\cdot\sin36^{\circ}\sin72^{\circ}\\
=&\frac{\sqrt{6}}{2^{27}}\cdot\sin36^{\circ}\sin72^{\circ}\\
=&\frac{\sqrt{6}}{2^{27}}\cdot2\sin18^{\circ}\cos^218^{\circ}\\
=&\frac{\sqrt{6}}{2^{26}}\sin18^{\circ}\left(1-\sin^218^{\circ}\right)
\end{align}$
Because $\sin18^{\circ}=\frac{1}{\sqrt5+1},\sin^218^{\circ}=\frac{1}{(\sqrt5+1)^2}=\frac{1}{6+2\sqrt5}$,
$\begin{align}
&\sin3^{\circ}\sin6^{\circ}...\sin87^{\circ}\\
=&\frac{\sqrt{6}}{2^{26}}\sin18^{\circ}\left(1-\sin^218^{\circ}\right)\\
=&\frac{\sqrt{6}}{2^{26}}\cdot\frac{1}{\sqrt5+1}\cdot\frac{5+2\sqrt5}{6+2\sqrt5}\\
=&\frac{\sqrt{6}}{2^{26}}\cdot\frac{5+2\sqrt5}{16+8\sqrt5}\\
=&\frac{\sqrt{6}}{2^{26}}\cdot\frac{\sqrt5(\sqrt5+2)}{8(2+\sqrt5)}\\
=&\frac{\sqrt{30}}{2^{29}}
\end{align}$
