由\sin \alpha =2\cos \alpha ,得\tan \alpha =2, 所以;\sin \alpha \cos \alpha =\frac{1}{2}\sin 2\alpha =\frac{1}{2}\times \frac{2\tan \alpha }{1+{{\tan }^{2}}\alpha }=\frac{1}{2}\times \frac{2\times 2}{1+4}=\frac{2}{5}. 故答案为:\frac{2}{5}.反馈...
\cos 2\alpha =\cos^2\alpha -\sin^2\alpha ,\tan 2\alpha =\frac{2\tan \alpha }{1-\tan^2\alpha },同理,我们可以证得如下三倍角公式:\sin 3\alpha =3\sin\alpha -4\sin^3\alpha ,\cos 3\alpha =___(要求用\cos\alpha 来表示),\tan 3\alpha =___(要求用\tan\alpha 来表示)....
Answer Step by step video & image solution for cos(alpha/2)+sin(alpha/2)=sqrt(2)(cos3 6^(@)-sin1 8^(@)) by Maths experts to help you in doubts & scoring excellent marks in Class 11 exams. Updated on:21/07/2023 Doubtnut is No.1 Study App and Learning App with Instant Video...
【解析】\because \left\{\begin{array}{l}\sin \alpha -2\cos \alpha =1\\ {\sin }^{2}\alpha +{\cos }^{2}\alpha =1\end{array}\right.\Rightarrow \left\{\begin{array}{l}\cos \alpha =-\dfrac{4}{5}\\ \sin \alpha =-\dfrac{3}{5}\end{array}\right.或\left\{\begi...
因为锐角\alpha ,所以;\cos \alpha \ne 0, 则\sin \alpha =\frac{1}{2},\cos \alpha =\frac{\sqrt{3}}{2}, 因为0<{}\alpha < {}\frac{ \pi }{2},0<{}\beta < {}\frac{ \pi }{2}, 所以;0<{}\alpha +\beta < {} \pi ,又\cos \left( \alpha +\beta \right)=\frac{1...
Hint: 1+tan2α1−tan2α=1+tan4πtan2αtan4π−tan2α=tan(4π−2α) Why is sinα−sinβ=2cos(2α+β)sin(2α−β)? https://www.quora.com/Why-is-sin-alpha-sin-beta-2-cos-left-frac-alph...
已知:关于x的方程的两根为sin\alpha 和cos\alpha .求:(1)\frac{sin^2\alpha }{sin\alpha -cos\alpha }+\f
-\dfrac{2}{5}/ -0.4 【分析】由任意角三角函数的定义可得\sin \alpha 、\cos \alpha ,即可得解.【详解】由三角函数定义可得\sin \alpha =\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}=\dfrac{4}{5},\cos \alpha =\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}=-\dfrac{3}{5},...
यदि a = cos alpha + sin alpha तथा b = cos alpha - sin alpha, तो (a^(2) - b^(2))/(2ab) का मान होगा
解析 ,且,,sin\alpha =\root \of {1-cos^2\alpha }=\frac{24}{25} 由已知中,且,由二倍角公式,可得cos\alpha 的值,再由同角三角函数关系公式,可得sin\alpha 的值.本题考查的知识点是二倍角公式,同角三角函数关系公式,难度中档.反馈 收藏