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/...
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 ...
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 ...
f(α \sin ( \pi - \alpha ) \cos (2 \pi - \alpha ) \sin ( \dfrac{5 \pi }{2}- \alpha ) \cos (- \pi - \alpha ) \cos ( \dfrac{3 \pi }{2}- \alpha ) ,则 f(- \dfrac{25 \pi }{3} 的值为( )A. - \dfrac{1}{2} B. \dfrac{1}{2} C. - \dfrac{...
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) 单位圆:以坐标原点为圆心,单位长(1)为半径的圆。 2)任意角的正弦、余弦和正切,以及物理意义: ...
সরল করো: (cos(2pi+alpha)cosec(pi-alpha)tan(pi/2+alpha))/(sec(pi/2+alpha)sin(3pi/2-alpha)cot(2pi-alpha)
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','...
\$\cos ( 2 \pi - \alpha ) \sin ( 3 \pi + \alpha ) \cos ( \frac { 3 \pi } { 2 } - \alpha )\$ \$\cos ( - \frac { \pi } { 2 } + \alpha ) \cos ( \alpha - 3 \pi ) \sin ( - \pi - \alpha )\$ 相关知识点: 试题来源: 解析 5.-1 反馈...
$$\begin{aligned} \Vert f_{\text {NIG}}^{(j)}\Vert _{\infty }\le \frac{\exp (T\delta \alpha )j!}{(T\delta )^{j+1}\pi },\quad j=0,1,2,\ldots \end{aligned}$$We need Lemma 3.7 to obtain a bound for \(B_{f}(L)\). We use the following abbreviation: the maxim...
[解](1) _ [解](1) _ [解](1) _ [解](1) _ [解](1) _ \$\frac { \sin a \cdot \cos a \cdot \tan \left( - a + \frac { 3 \pi } { 2 } - 2 \pi \right) } { \tan \left( \frac { \pi } { 2 } + a \right) \cdot \sin a }\$ [解](1) ...