cot α=∠α的邻边 / ∠α的对边 倍角公式 Sin2A=2SinA?CosA Cos2A=CosA^2-SinA^2=1-2SinA^2=2CosA^2-1 tan2A=(2tanA)/(1-tanA^2)(注:SinA^2 是sinA的平方 sin2(A) )三倍角公式 sin3α=4sinα·sin(π/3+α)sin(π/...
百度试题 结果1 题目已知cos(\frac{\pi}{2} \alpha )=2cos(\pi-\alpha ),则tan(\frac{\pi}{4} -\alpha )=( ).A. -4 B. 4 C. -\frac{1}{3} D. \frac{1}{3} 相关知识点: 试题来源: 解析 C. , 综上所述,答案选择:反馈 收藏 ...
扇形面积相当于是这个角度所占圆形角360°的多少:S_{扇}=\pi r^2\frac{\alpha}{2\pi}=\frac{1}{2}\alpha r^2; 弧长相当于是这个弧占圆的周长所对应的多少:l_{弧}=2\pi r\frac{\alpha}{2\pi}=\alpha r 节选自2011年上半年编撰的高一教材 4、单位圆与三角函数值在不同象限正负判断 1) 单位...
Let alpha=(pi)/(5) and A,=[[cos alpha,sin alpha-sin alpha,cos alpha]] then B=A^(4)-A^(3)+A^(2)-A is View Solution If alpha=cos((8 pi)/(11))+i sin((8 pi)/(11)) then Re(alpha+alpha^(2)+alpha^(3)+alpha^(4)+alpha^(5)) is ...
$$\begin{aligned} \Vert f_{\text {NIG}}^{(j)}\Vert _{\infty }\le \frac{\exp (T\delta \alpha )j!}{(T\delta )^{j+1}\pi },\quad j=0,1,2,\ldots \end{aligned}$$We need Lemma 3.7 to obtain a bound for \(B_{f}(L)\). We use the following abbreviation: the maxim...
百度试题 结果1 题目已知Zc< underline>os(x a)= < /underline> \cos (\pi \div 2 \alpha )则\tan a= ()A. 2 B. -2 C. 1\div 2\ \ D. -1\div 2\ \ 相关知识点: 试题来源: 解析 A 反馈 收藏
varlanguages=require("@cospired/i18n-iso-languages");console.log("German => "+languages.getAlpha2Code('German','en'));// German => deconsole.log("German => "+languages.getAlpha3TCode('German','en'));// German => deuconsole.log("German => "+languages.getAlpha3BCode('German','...
(i)y=x2,x=1,x=2एव X - अक्ष (ii)y=x4,x=1,x=5एव X - अक्ष Prove that:∫π0x1+cosαsinxdx=παsinα int_(0)^( pi)(xdx)/(1+cos alpha*sin alpha)(0 ∫π0xdx1+cosα⋅sinx=παsinα,0<α<π ...
所以2\sin\alpha (2-\sin\alpha )=1-2\sin^2\alpha , 所以\sin\alpha ={1\over 4}, 因为\alpha \in (0,{\pi\over 2}), 所以\cos\alpha ={\sqrt{15}\over 4},\tan \alpha ={\sqrt{15}\over 15}。 故本题正确答案为A。 反馈 收藏 ...
结果1 题目 { \tan (\pi - \alpha )\cos (2\pi - \alpha )\sin (- \alpha }\div { \cos (- \alpha - \pi )\sin (- \pi - \alpha )}例4值为 A. -2 B. -1 C. 1 D. 2 相关知识点: 试题来源: 解析 B 反馈 收藏 ...