知识点二 二倍角的正弦、余弦、正切公式$$ \sin 2 \alpha = 2 \sin \alpha \cos \alpha ; $$$ \cos 2 \alpha = \cos ^ { 2 } \alpha - \sin ^ { 2 } \alpha = 2 \cos ^ { 2 } \alpha - 1 = 1 - 2 \sin ^ { 2 } \alpha $$;$$ \tan 2 \alpha = \frac { 2...
(sin(α)+cos(α))2 Expand sin(2α)+1 Evaluate (sin(α)+cos(α))2 Share Copy Copied to clipboard
5. 解:因为$$ ( \sin \alpha + \cos \alpha ) ^ { 2 } = \sin ^ { 2 } \alpha + \cos ^ { 2 } \alpha + 2 \sin $$ $$ a \cos \alpha = 1 + 2 \sin \alpha \cos \alpha = 2 , $$ , 所以$$ \sin a \cos a = \frac { 1 } { 2 } $$ ,因此$$ ( \...
If ∫sinxsin(x−α)dx=Ax+Blogsin(x−α)+C, then value of (A,B) is (A) (−sinα,cosα) (B) (−cosα,sinα) (C) (sinα,cosα) (D) (cosα,sinα) View Solution If tanθ=sinα−cosαsinα+cosα, then (A) sinα−cosα=±√2sinθ (B) sinα+cosα=±√...
Hint: 1+tan2α1−tan2α=1+tan4πtan2αtan4π−tan2α=tan(4π−2α) Why is sinα−sinβ=2cos(2α+β)sin(2α−β)? https://www.quora.com/Why-is-sin-alpha-sin-beta-2-cos-left-frac-alpha...
\tan \alpha = \frac { 1 } { 2 } , \sin \alpha = \pm \frac { \sqrt { 5 } } { 5 } , \cos \alpha = \pm \frac { 2 \sqrt { 5 } } { 5 } , $$ 即$$ \sin \alpha = \frac { \sqrt { 5 } } { 5 } \cos \alpha = \frac { 2 \sqrt { 5 } } { 5 ...
দেখাও যে, cos theta (sin theta pm sqrt (sin^2 theta + sin^2 alpha)) -এর মান সর্বদাই pm sqrt (1 + sin^2 alpha) -র মধ্যে থাকে|
【解析】 解:由题意,得$$\left\{ \begin{matrix} \frac { \sin \alpha } { \cos \alpha } = 2 , \\ \sin ^ { 2 } \alpha + \cos h \alpha = 1 \end{matrix} \right.$$解方程组得 或 $$\left\{ \begin{matrix} \sin \alpha = \frac { 2 \sqrt { 5 } } { 5 } \\ ...
যদি alpha+gamma=2beta হয় তবে দেখাও যে,cotbeta=(sinalpha-singamma)/(cosgamma-cosalpha)
知识点二 二倍角公式1.基本公式(1)$$ \sin 2 \alpha = 2 \cos \alpha \cos \alpha $$(2)$$ \cos 2 \alpha = \cos ^ { 2 } \alpha - \sin ^ { 2 } \alpha = 2 \cos ^ { 2 } \alpha - 1 = $$$ 1 - 2 \sin ^ { 2 } \alpha $$(3)$$ \tan 2 \alpha = \frac...