A8DB关于三角函数有如下的公式: sin(a B)sin a cos B cos a sin Bi cos(a B)cos a cos B-sin a sin B2 tan a tan Btan(a B)31-tana·tan3 利用这些公式可将某些不是特殊角的三角函数转化为特殊角的三角函数来求值,如: tan45°+tan60°1+√3(1+√③)(1+√3)tan105°=tan(45°+60...
{1 + \tan \alpha \tan \beta }(1+ \tan \alpha \tan \beta \neq 0)利用这些公式可以将一些不是特殊角的三角函数转化为特殊角的三角函数来求值.如:\tan 105^{{\circ} }= \tan (45^{{\circ} }+ 60^{{\circ} })= \dfrac{\tan 45^{{\circ} } + \tan 60^{{\circ} }}{1 - \...
sin2Betacos2BetaBaBarCp-asymmetriesThe Standard Model (SM) of particle physics describes charge conjugation-parity (CP) violation as a consequence of a complex phase in the three-generation Cabibbo-Kobayashi-Maskawa (CKM) quark-mixing matrix. In this framework, measurements of CP asymmetries in ...
题目:若\( \cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta \),那么\( \sin(\alpha + \beta) = ___ 。相关知识点: 试题来源: 解析 答案:\( \sin\alpha \cos\beta + \cos\alpha \sin\beta \)
|AB|=\sqrt{(\cos\alpha -\cos\beta )^2+(\sin\alpha -\sin\beta )^2} =\sqrt{\cos^2\alpha +\sin^2\alpha+\cos^2\beta +\sin^2\beta -2\cos\alpha \cos\beta -2\sin\alpha \sin\beta } =\sqrt{2-2(\cos\alpha \cos\beta +\sin\alpha \sin\beta )} =\sqrt{2-2\cos(\...
\alpha 、\beta 均为锐角,\sin \alpha = \dfrac{5}{13},\cos \beta = \dfrac{4}{5},则\sin (\alph
15.阅读材料:关于三角函数还有如下的公式:{\sin}({\alpha}-{\beta})={\sin}{\alpha}{\cos}{\beta}-{\cos}{\al
关于三角函数有如下的公式:\cos (\alpha - \beta )= \cos \alpha \cos \beta \sin \alpha \sin \beta,由该公式可求得\cos 15^ \circ的值是() A. { \sqrt 6+ \sqrt 2}\div 4\ \ B. { \sqrt 6- \sqrt 2}\div 4\ \ C. { \sqrt 3- \sqrt 2}\div 4\ \ D. { \sqrt 3\...
1. 利用正弦函数的加减公式,原式可化为 sin(α). 答案为 **B**. 2. 将 sin 83° 变为 cos 7°,再利用正弦函数的加减公式,原式可化为 -sin 30° = -1/2. 答案为 **B**. 3. 将 cos α + √3 sin α 变为 2(1/2cos α + (√3)/2sin α), 利用正弦函数的加减公式,原...
\sin \alpha +\sin \beta =\frac{21}{65},可得2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=\frac{21}{65}①,\cos \alpha +\cos \beta =\frac{27}{65},2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=\frac{27}{65}②,\frac{①}{...