Bài em cần đâu

sin^2x+sin^2(60-x)+sinx*sin(60 độ-x)

\(=sin^2x+\left[sin60\cdot cosx-sinx\cdot cos60\right]^2+sinx\cdot\left[sin60\cdot cosx-sinx\cdot cos60\right]\)

\(=sin^2x+\left[-\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right]^2+sinx\left[\dfrac{-1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right]\)

\(=sin^2x+\dfrac{1}{4}sin^2x-\dfrac{\sqrt{3}}{2}\cdot sinx\cdot cosx+\dfrac{3}{4}\cdot cos^2x-\dfrac{1}{2}\cdot sin^2x+\dfrac{\sqrt{3}}{2}\cdot sinx\cdot cosx\)

\(=\dfrac{5}{4}sin^2x+\dfrac{3}{4}\cdot cos^2x-\dfrac{1}{2}\cdot sin^2x\)

=3/4*(sin^2x+cos^2x)=3/4

\(x_1=a>2;x_{n+1}=x_n^2-2,\forall n=1,2,...\)

mà \(n\rightarrow+\infty\)

\(\Rightarrow a\rightarrow+\infty\Rightarrow x_n\rightarrow+\infty\)

\(\Rightarrow\lim\limits_{n\rightarrow+\infty}\dfrac{1}{x_n}=0\) \(\Rightarrow\lim\limits_{n\rightarrow+\infty}\left(\dfrac{1}{x_nx_{n+1}}\right)=0\)

\(\)\(\Rightarrow\lim\limits_{n\rightarrow+\infty}\left(\dfrac{1}{x_1}+\dfrac{1}{x_1x_2}+\dfrac{1}{x_1x_2x_3}+...+\dfrac{1}{x_1x_2...x_n}\right)=0\)

...

\(sin\left(x\right)+\left[sin\left(x+\dfrac{2\pi}{5}\right)-sin\left(x+\dfrac{\pi}{5}\right)\right]+\left[sin\left(x+\dfrac{4\pi}{5}\right)-sin\left(x+\dfrac{3\pi}{5}\right)\right]\)

\(=sin\left(x\right)+2cos\left(x+\dfrac{3\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)+2cos\left(x+\dfrac{7\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)\)

\(=sin\left(x\right)+2sin\left(\dfrac{\pi}{10}\right)\left[cos\left(x+\dfrac{3\pi}{10}\right)+cos\left(x+\dfrac{7\pi}{10}\right)\right]\)

\(=sin\left(x\right)+4sin\left(\dfrac{\pi}{10}\right)cos\left(\dfrac{\pi}{5}\right)cos\left(x+\dfrac{\pi}{2}\right)\)

\(=sin\left(x\right)+cos\left(x+\dfrac{\pi}{2}\right)\)

\(=sin\left(x\right)+cos\left(x\right)cos\left(\dfrac{\pi}{2}\right)-sin\left(x\right)sin\left(\dfrac{\pi}{2}\right)\)

\(=sin\left(x\right)-sin\left(x\right)\)

\(=0\)

B=cos^2x-sin^2x+cosx-sinx

=(cosx-sinx)(cosx+sinx)+(cosx-sinx)

=(cosx-sinx)(cosx+sinx+1)

\(u_{n+1}-u_n=\dfrac{4\left(n+1\right)+3}{3\left(n+1\right)-4}-\dfrac{4n+3}{3n+4}\)

\(=\dfrac{4n+7}{3n-1}-\dfrac{4n+3}{3n+4}\)

\(=\dfrac{12n^2+16n+21n+28-12n^2+4n-9n+3}{\left(3n-1\right)\left(3n+4\right)}\)

\(=\dfrac{32n+31}{\left(3n-1\right)\left(3n+4\right)}>0\) với mọi n thuộc N*

=>(Un) là dãy tăng

Dòng đầu phải là Un+1 = 4(n+1)+3/ 3(n+1)+4 chứ ạ, s lại là 3(n+1)- 4 ạ

(sina+cosa)^2

=sin^2a+cos^2a+2*sina*cosa

=1+sin2a

mấy cái dấu sao là gì v bn ???

cos^4a-sin^4a

=(cos^2a-sin^2a)(cos^2a+sin^2a)

=cos^2a-sin^2a

=cos2a