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- $\sin x^{\circ}+\cos x^{\circ}={\sin x^{\circ}\over\cos x^{\circ}}\implies\cos^2 x^{\circ}=\sin x^{\circ}(1-\cos x^{\circ})$
设$c=\cos x(-\frac1{\sqrt2}\le c\le1)$,平方得$c^4=(1-c^2)(1-c)^2$ - $6 \tan ^{-1} x+4 \tan ^{-1}(3 x)=\pi\implies\tan(6\tan^{-1}x)=-\tan(4\tan^{-1}3x)$$\implies -\frac{2\left(3 x^{5}-10 x^{3}+3 x\right)}{x^{6}-15 x^{4}+15 x^{2}-1}=\frac{12\left(9 x^{3}-x\right)}{81 x^{4}-54 x^{2}+1}$
$x\in\left\{0,{4\pm\sqrt{7}\over3},{-4\pm\sqrt{7}\over3},\pm\sqrt{\frac{15+8 \sqrt{3}}{33} },\pm\sqrt{\frac{15-8 \sqrt{3}}{33} }\right\}$
代回去检验,原方程只有一解$x=\sqrt{\frac{15-8 \sqrt{3}}{33} }$ - $⇒\sin2x\sin3x=\sin x\sin 2 x+\sin x\sin 3 x$
$⇒\cos x-\cos5x=\cos x-\cos3x+\cos2x-\cos4x$
$⇒\cos3x-\cos5x=\cos2x-\cos4x$
$⇒\sin x\sin 4x=\sin x\sin 3x$
$(\because \sin x\ne0)⇒\sin4x=\sin3x$
最小的正数解是$x=2 \tan ^{-1}\left(\sqrt{\frac{5}{3}-\frac{8}{3} \cos \left(\frac{1}{3} \cos ^{-1}\left(-\frac{13}{14}\right)\right)}\right)$ - 由$\cos(nθ)=\cos((17-n)θ)$得
$\cos \theta+\cos 3 \theta+⋯+\cos 15 \theta=\frac12(\cosθ+\cos2θ+⋯+\cos16θ)=\frac{\sin(17θ-\fracθ2)-\sin\fracθ2}{4\sin\fracθ2}=-\frac12$
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