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...
(sin(α)+cos(α))2 Expand sin(2α)+1 Evaluate (sin(α)+cos(α))2 Quiz Trigonometry (sinα+cosα)2 Similar Problems from Web Search If sinα+cosα=1.2, then what is sin3α+cos3α? https://math.stackexchange.com/questions/2001078/if-sin-alpha-cos-alpha-1-2-then-what-is-sin...
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...
\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)....
,求sin\alpha -cos\alpha 的值. 相关知识点: 试题来源: 解析 解:将两边平方得,,因为,所以,则,所以. 将两边平方,利用平方关系化简求出的值,根据三角函数的符号缩小的范围,判断出的符号,利用平方关系求出的值.本题考查同角三角函数的平方关系,以及三角函数的符号,属于中档题....
解:(1)由于关于x的方程的两根为sin\alpha 和cos\alpha , 故有sin\alpha +cos\alpha =\frac{\sqrt{3}+1}{2},\frac{sin^2\alpha }{sin\alpha -cos\alpha }+\frac{cos\alpha }{1-tan\alpha }=\frac{sin^2\alpha }{sin\alpha -cos\alpha }+\frac{cos^2\alpha }{cos\alpha -sin\alpha }=...
解:因为{{\rm \cos }}^{2}α={\rm \sin }α,所以\dfrac{1}{{\rm \sin }α}+{{\rm \cos }}^{4}α= \dfrac{1}{{\rm \sin }α}+{{\rm \sin }}^{2}α= \dfrac{1}{{\rm \sin }α}+1-{{\rm \cos }}^{2}α= \dfrac{1}{{\rm \sin }α}-{\rm \sin...
-\frac{1}{2} C. \frac{1}{2} D. \frac{8}{9} 相关知识点: 试题来源: 解析 A 由题:sin\alpha +cos\alpha =\frac{1}{3}两边平方得: sin^2\alpha +cos^2\alpha +2sin\alpha cos\alpha =1+sin2\alpha =\frac{1}{9} 所以:sin2\alpha =-\frac{8}{9} 故本题选A反馈 收藏 ...
半角公式(1)sin\frac{\alpha }{2}=___.(2)cos\frac{\alpha }{2}=___. (3)tan\frac{\alpha }{
यदि A = [(cos alpha, sin alpha),(-sin alpha,cos alpha)] हो, तो सिद्ध कीजिए कि A' . A = I