特殊的三角函数等式
收集一些特殊的三角函数等式，特殊之处在于，等式左边是两个三角函数的线性和，右边是一个简单的代数表达式：1.$tan((6pi)/7)+4sin((2pi)/7)=sqrt(7)$
2.$tan((4pi)/11)+4sin(pi/11)=sqrt(11)$
3.$tan((2pi)/13)+4sin((6pi)/13)=sqrt(13+2sqrt(13)$
分母再大就很难有类似的等式了。 本帖最后由 lsr314 于 2018-2-2 15:10 编辑
三个正切函数的乘积：
4.$\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}$ 5.$\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$ 都不是很显然的呢，这都能整理出来，赞 1.$a\tan((6pi)/7)+b\sin((2pi)/7)=x$
$x$为下列方程的第二个正根(按大小排列第二个正根)
$-64x^6+(112b^2-448ab+1344a^2)x^4+(-2912a^2b^2-56b^4-2240a^4+560ab^3+4480ba^3)x^2+7(-b^3+12ab^2-20ba^2+8a^3)^2=0$ \(当\D\tan(A+B)=\frac{3}{4}\ \ \ \ \ \ \frac{\tan(A)}{\tan(B)}=\frac{1}{2}时\ \ \ \ \ 5\sin(A)\cos(B)=1\)
\(当\D\tan(A+B)=\frac{9}{40}\ \ \ \ \ \frac{\tan(A)}{\tan(B)}=\frac{1}{2}时\ \ \ \ 41\sin(A)\cos(B)=3\)
\(当\D\tan(A+B)=\frac{15}{112}\ \ \ \frac{\tan(A)}{\tan(B)}=\frac{1}{2}时\ \ \ \ 113\sin(A)\cos(B)=5\)
\(当\D\tan(A+B)=\frac{21}{220}\ \ \ \frac{\tan(A)}{\tan(B)}=\frac{1}{2}时\ \ \ \ 221\sin(A)\cos(B)=7\)
\(当\D\tan(A+B)=\frac{27}{364}\ \ \ \frac{\tan(A)}{\tan(B)}=\frac{1}{2}时\ \ \ \ 365\sin(A)\cos(B)=9\)
\(当\D\tan(A+B)=\frac{33}{544}\ \ \ \frac{\tan(A)}{\tan(B)}=\frac{1}{2}时\ \ \ \ 545\sin(A)\cos(B)=11\)
\(当\D\tan(A+B)=\frac{39}{760}\ \ \ \frac{\tan(A)}{\tan(B)}=\frac{1}{2}时\ \ \ \ 761\sin(A)\cos(B)=13\)
\(\cosh\left(0×\ln(2+\sqrt{3})\right)=1\)
\(\cosh\left(1×\ln(2+\sqrt{3})\right)=2\)
\(\cosh\left(2×\ln(2+\sqrt{3})\right)=7\)
\(\cosh\left(3×\ln(2+\sqrt{3})\right)=26\)
\(\cosh\left(4×\ln(2+\sqrt{3})\right)=97\)
\(\cosh\left(5×\ln(2+\sqrt{3})\right)=362\)
\(\cosh\left(6×\ln(2+\sqrt{3})\right)=1351\)
\(\cosh\left(7×\ln(2+\sqrt{3})\right)=5042\)
\(\cosh\left(8×\ln(2+\sqrt{3})\right)=18817\)
\(\cosh\left(9×\ln(2+\sqrt{3})\right)=70226\)
本帖最后由 chyanog 于 2018-6-7 11:56 编辑
一点补充
$\tan \left(\frac{6 \pi }{7}\right)+4 \sin \left(\frac{2 \pi }{7}\right)=\tan \left(\frac{5 \pi }{7}\right)+4 \sin \left(\frac{3 \pi }{7}\right)=\tan \left(\frac{\pi }{7}\right) \tan \left(\frac{2 \pi }{7}\right) \tan \left(\frac{3 \pi }{7}\right)=\tan \left(\frac{3 \pi }{7}\right)+\tan \left(\frac{5 \pi }{7}\right)+\tan \left(\frac{6 \pi }{7}\right)=\sqrt{7}$
$\tan \left(\frac{4 \pi }{11}\right)+4 \sin \left(\frac{\pi }{11}\right)=\tan \left(\frac{3 \pi }{11}\right)+4 \sin \left(\frac{2 \pi }{11}\right)=\sqrt{11}$
$\tan \left(\frac{2 \pi }{13}\right)+4 \sin \left(\frac{6 \pi }{13}\right)=\tan \left(\frac{5 \pi }{13}\right)+4 \sin \left(\frac{2 \pi }{13}\right)=\sqrt{13+2 \sqrt{13}}$
$\tan \left(\frac{4 \pi }{13}\right)+4 \sin \left(\frac{\pi }{13}\right)=\tan \left(\frac{10 \pi }{13}\right)+4 \sin \left(\frac{4 \pi }{13}\right)=\tan \left(\frac{12 \pi }{13}\right)+4 \sin \left(\frac{3 \pi }{13}\right)=\sqrt{13-2 \sqrt{13}}$
$\tan \left(\frac{\pi }{13}\right) \tan \left(\frac{3 \pi }{13}\right) \tan \left(\frac{4 \pi }{13}\right)=\sqrt{65-18 \sqrt{13}}$
$\tan \left(\frac{2 \pi }{13}\right) \tan \left(\frac{5 \pi }{13}\right) \tan \left(\frac{6 \pi }{13}\right)=\tan \left(\frac{2 \pi }{13}\right)+\tan \left(\frac{5 \pi }{13}\right)+\tan \left(\frac{6 \pi }{13}\right)=\sqrt{65+18 \sqrt{13}}$
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