\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)....
sin^2\alpha +cos^2\alpha =___。 相关知识点: 试题来源: 解析 1设直角△ABC中,∠C =90°, ∠A=α ,a的 对边是a,邻边是b,斜边是c. 则有 a^2+b^2=c^2 , sinα=a/c , cosα=b/c , 所以 sin^2α+cos^2α=(a^2+b^2)/(c^2)=(c^2)/(c^2)=1 故答案为1. 反馈 ...
已知\sin\alpha =\frac{4}{5},且\alpha 是第二象限的角,求\sin2\alpha ,\cos2\alpha 和\tan2\alp
(1)降幂公式:sin^2\alpha =___,cos^2\alpha =___;(2)升幂公式:1+cos\alpha =___,1-cos\alpha
\$\sin 2 \alpha \cos \alpha = 2 \sin \alpha \left( 1 - \sin ^ { 2 } \alpha \right)\$ 证明:. 相关知识点: 试题来源: 解析 证明: 证明: \$\sin 2 \alpha \cos \alpha = 2 \sin \alpha \cos \alpha \cos \alpha\$ \$= 2 \sin \alpha \cdot \cos ^ { 2 } \...
{b}{c}$,所以$\sin ^{2}\alpha +\cos ^{2}\alpha =\dfrac{a^{2}+b^{2}}{c^{2}}=\dfrac{c^{2}}{c^{2}}=1$.故答案为$1$.【平方关系】sin2A+cos2A=1【积的关系(正余弦与正切之间的关系)】一个角的正切值等于这个角的正弦与余弦的比,即sin A tan A cos A或sin A=tan A-c...
已知$\dfrac{\sin 2\alpha -\cos 2\alpha +1}{\sin \alpha +\cos \alpha }=\dfrac{\sqrt{10}}
2sin2α=cos2α+1 來自Web 搜索的類似問題 If sinα+cosα=0.2, find the numerical value of sin2α. https://math.stackexchange.com/questions/451199/if-sin-alpha-cos-alpha-0-2-find-the-numerical-value-of-sin2-alpha You also know that sin2α+cos2α=1, so square what you are given...
sin(α)+sin(β)=2sin(2α+β)cos(2α−β) Solve for α α∈C Solve for β β∈C Quiz Complex Number sinα+sinβ=2sin2α+βcos2α−β Similar Problems from Web Search Why is sinα+sinβ=2sin(...
素下斯了实导然层军已解办手一而油难观原海已知sin\alpha =0.8,且\alpha \in (0,\pi),求cos2\alpha ,sin2\alpha . 素下