解得:cos\beta =\dfrac{-1\pm \root \of {7} }{4},sin\beta =\dfrac{1\pm \root \of {7} }{4}所以:sin\alpha =\dfrac{-1\pm \root \of {7} }{4},cos\alpha =\dfrac{1\pm \root \of {7} }{4}所以sin(\alpha +\beta )=sin\alpha cos\beta +sin\beta cos\alpha 代入上...
百度试题 结果1 题目\( \cos ( \alpha - \beta )= \cos \alpha - \cos \beta \).___ 相关知识点: 试题来源: 解析 ×
{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 - \...
【COS试妆】Alp..#COS试妆# 这次的主题是 #Alpha,Beta&Omega# ,即传说中的ABO设定【Alpha=赤,Beta=金,Omega=蓝】老早就觉得少狼的族群中无论是Alpha还是Beta还是
百度试题 结果1 题目\(\begin{vmatrix} 0&\cos \alpha&\cos \beta \\ -\cos \alpha&0&\cos \gamma\\-\cos \beta&-\cos \gamma& 0 \end{vmatrix}=\)___.相关知识点: 试题来源: 解析 \(0\)
{1 - \tan \alpha \ast \tan \beta }③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:\tan 105^{{\circ} }=\tan (45^{{\circ} }+ 60^{{\circ} }) = \dfrac{\tan 45^{{\circ} } + \tan 60^{{\circ} }}{1 - \tan 45^{{\circ} }\ast \tan 60...
Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.
少经基率劳非何条所选立交\(\begin{vmatrix} 0&\cos \alpha&\cos \beta \\ -\cos \alpha&0&\cos \gamma\
\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}{...
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\ \...