1 Α α alpha a:lf 阿尔法 2 Β β beta bet 贝塔 3 Γ γ gamma ga:m 伽马 4 Δ δ delta delt 德尔塔 5 Ε ε epsilon ep`silon 伊普西龙 6 Ζ ζ zeta zat 截塔 7 Η η eta eit 艾塔 8 Θ θ theta θit 西塔 9 Ι ι iota aiot 约塔 10 Κ κ kappa kap 卡帕 11 ∧...
-$$ \frac { 7 } { 5 } $$ 解析:因为$$ \tan \alpha = \frac { 4 } { 3 } $$,α为第一象限角,所以$$ \sin \alpha = \frac { 4 } { 5 } , $$ $$ \cos \alpha = \frac { 3 } { 5 } $$,所以$$ \sin ( \pi + \alpha ) + \cos ( \pi - \alpha ) =...
In the case of the Basel problem, it is the hyperbolic 3-manifold SL₂/SL₂. The zeta function also satisfies Riemann's functional equation, which involves π as well as the gamma function: displaystylezeta(s)=2ˢpiˢ⁻¹sinleft(fracpi s2right)Gamma(1-s)zeta(1-s). Further...
在\displaystyle(2k\pi-\frac{\pi}{2},2k\pi+\frac{\pi}{2})(k\in Z)上单调递增; 在\displaystyle(2k\pi+\frac{\pi}{2},2k\pi+\frac{3\pi}{2})(k\in Z)上单调递减。 ⑥最值: 当\displaystyle x=2k\pi+\frac{\pi}{2}时,\displaystyle y_{max}=1; 当\displaystyle x=2k\pi+\frac...
\(\alpha x^{3}+\beta x\)形式的三阶近似,最大近似误差 \(0.005 \ rad = 0.29^{\circ}\), \[\arctan (x) \approx \frac{\pi}{4} x+x\left(0.186982-0.191942 x^{2}\right), \quad-1 \leq x \leq 1 \] \(x(x-1)(\alpha x-\beta)\)形式的三阶近似,最大近似误差 \(0.0015 \...
2\alpha = -\sin \alpha ,\alpha \in (\dfrac{\pi }{2},\, \pi ),∴2\sin \alpha \cos \alpha = -\sin \alpha ,∴\cos \alpha = -\dfrac{1}{2},\alpha \in (\dfrac{\pi }{2},\, \pi )∴\alpha = \dfrac{2\pi }{3},∴\tan \alpha = -\sqrt{3}.故答案为:-\sqr...
正切函数的诱导公式 (1)$$ \tan ( 2 \pi + \alpha ) = \tan \alpha $$ ;(2)$$ \tan ( - \alpha ) = - \tan \alpha $$ (3)$$ \tan ( 2 \pi - \alpha ) = - \tan \alpha $$ (4)$$ \tan ( \pi - \alpha ) = - \tan \alpha $$;(5)$$ \tan ( \pi...
=\cos(\frac{3\pi}{2}-\alpha)/*再加上一个2\pi*/ =-\cos(\frac{\pi}{2}-\alpha)/*减去\pi并加负号*/ =-\sin\alpha 可以看出,按照这个步骤,完全不需要记忆那么多公式,甚至连「奇变偶不变,负号看象限」都不需要,只要按部就班地做就可以得到正确答案。
首先由正切三倍角公式知tan(3×21∘)=tan21∘tan39∘tan81∘，及合分比性质...
1doublealpha =4;//探狼比例因子2intexploring_wolf_num;//探狼数量,(取[n/(α+1),n/α]之间的整数)3//探狼数量,(取[n/(α+1),n/α]之间的整数)4intMIN_VALUE = wolf_num / (alpha +1);5intMAX_VALUE = wolf_num /alpha;6exploring_wolf_num= rand() % (MAX_VALUE - MIN_VALUE +1...