\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 यदि A= [{:(cos alpha,sinalpha),(-sin alpha,cos alpha):}] है, तो A^(2) का मान है - by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams.Updated on:21/07/2023Cla...
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-sin3-alpha-cos3-alpha Hint :(si...
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 Solution...
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...
所以;0<{}\alpha +\beta < {} \pi ,又\cos \left( \alpha +\beta \right)=\frac{1}{7}, 所以;\sin \left( \alpha +\beta \right)=\frac{4\sqrt{3}}{7}, 则\cos \beta =\cos \left[ \left( \alpha +\beta \right)-\alpha \right] ~~~=\cos \left( \alpha +\beta \right)...
cos(2α) Differentiate w.r.t. α −2sin(2α) Evaluate cos(2α)
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...
即7tan^2\alpha -20tan\alpha -3=0,解得tan\alpha =-\frac{1}{7}或3(不满足\alpha \in(-\frac{\pi}{2},0),舍去),故\frac{cos^2\alpha }{1+sin2\alpha }=\frac{cos^2\alpha }{sin^2\alpha +cos^2\alpha +2sin\alpha cos\alpha }=\frac{1}{tan^2\alpha +1+2tan\alpha ...
解析 D ∵2sin2\alpha =1-cos2\alpha ,∴4sin\alpha cos\alpha =2sin^2\alpha ∴tan\alpha =2又sin^2\alpha +cos^2\alpha =1 ,\alpha \in (0,\dfrac{π}{2} )∴sin\alpha =\dfrac {2\sqrt {5}}{5}综上所述,本题答案为D。