{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...
{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 - \ta...
https://www.quora.com/Why-is-sin-alpha-sin-beta-frac-1-2-cos-alpha-beta-cos-alpha+-beta To show why, our goal will be to manipulate the right side of the identity: sin α sin β = (1/2)[cos(α ...
解:\sin\alpha+\sin\beta=\sin\alpha\sin\beta可化为\sin(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2})+\sin(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2})=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}=\sin\alpha\sin\beta, 则(\cos\frac{\alpha-\beta}2-...
यदि sin (alpha + beta ) =1, sin (alpha - beta) = 1/2 , तब tan (alpha + 2 beta) tan (2alpha + beta) बराबर है
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...
\because \sin\alpha<\sin\beta \therefore \alpha<\beta \therefore -\frac{\pi}{2}<\alpha-\beta<0 \therefore\sin\left(\alpha-\beta\right)=\sin\alpha\cos\beta-\cos\alpha\sin\beta=\frac{\sqrt{5}}{5}\times\frac{\sqrt{10}}{10}-\frac{2\sqrt{5}}{5}\times\frac{3\sqrt{10...
【题目】【题目】 \$\sin \alpha - \sin \beta = 1 - \frac { \sqrt { 3 } } { 2 }\$ , _ 的 _ 值为【题目】 \$\sin \alpha - \sin \beta = 1 - \frac { \sqrt { 3 } } { 2 }\$ , _ 的 _ 值为() 相关知识点: ...
已知\( \sin(\alpha + \beta) = \frac{3}{5} \),\( \cos(\alpha + \beta) = -\frac{4}{5} \),且 \( \alpha < \alpha + \beta < \pi \),那么 \( \sin\alpha \) 的值是: A. \( \frac{4}{5} \) B. \( -\frac{3}{5} \) C. \( \frac{3}{5} \) D...
(1) \$\sin ( \alpha + \beta ) = \sin \alpha + \sin \beta\$ 相关知识点: 试题来源: 解析 (1) (1) (1) 结果一 题目 (1)sin(α+β)=sinα+sinβ. × 答案 答案见上相关推荐 1(1)sin(α+β)=sinα+sinβ. × 反馈 收藏 ...