Đề bài yêu cầu điều gì em nhỉ?
6.
\(sinx.cosx.cos2x=\dfrac{1}{2}\left(2sinx.cosx\right).cos2x\)
\(=\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}\left(2sin2x.cos2x\right)\)
\(=\dfrac{1}{4}sin4x\)
7.
\(4sinx.sin\left(x+\dfrac{\pi}{2}\right)sin\left(2x+\dfrac{\pi}{2}\right)\)
\(=4sinx.cos\left(-x\right).cos\left(-2x\right)\)
\(=4sinx.cosx.cos2x\)
\(=2sin2x.cos2x\)
\(=sin4x\)
8.
\(sin^2\left(\dfrac{\pi}{8}+\dfrac{x}{2}\right)-sin^2\left(\dfrac{\pi}{8}-\dfrac{x}{2}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}cos\left(\dfrac{\pi}{4}+x\right)-\left[\dfrac{1}{2}-\dfrac{1}{2}cos\left(\dfrac{\pi}{4}-x\right)\right]\)
\(=\dfrac{1}{2}cos\left(\dfrac{\pi}{4}-x\right)-\dfrac{1}{2}cos\left(\dfrac{\pi}{4}+x\right)\)
\(=\dfrac{1}{2}.\left(-2\right).sin\left(\dfrac{\dfrac{\pi}{4}-x+\dfrac{\pi}{4}+x}{2}\right)sin\left(\dfrac{\dfrac{\pi}{4}-x-\dfrac{\pi}{4}-x}{2}\right)\)
\(=-sin\left(\dfrac{\pi}{4}\right).sin\left(-x\right)\)
\(=\dfrac{\sqrt{2}}{2}sinx\)
9.
\(sin8x+2cos^2\left(45^0+4x\right)\)
\(=sin8x+cos\left(90^0+8x\right)+1\)
\(=sin8x+sin\left(-8x\right)+1\)
\(=sin8x-sin8x+1=1\)
10.
\(sin2x+cos2x-2cosx\left(sinx+cosx\right)+1\)
\(=sin2x+2cos^2x-1-2cosx.sinx-2cos^2x+1\)
\(=sin2x-2sinx.cosx=sin2x-sin2x\)
\(=0\)
ai giúp em với ạ :<< em đang cần gấp lắm, giải hộ em 2 bài thì càng tốt ạ, em xin chân thành cám ơn
