Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
rút gọn biểu thức:
E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
\(S=sinx+siny+sin\left(3x+y\right)-sin\left(3x+y\right)-sin\left(x+y\right)\)
\(=sinx+siny-sin\left(x+y\right)\)
\(S^2=\left(sinx+siny-sin\left(x+y\right)\right)^2\le3\left(sin^2x+sin^2y+sin^2\left(x+y\right)\right)\)
\(S^2\le3\left(1-\dfrac{1}{2}\left(cos2x+cos2y\right)+sin^2\left(x+y\right)\right)\)
\(S^2\le3\left[1-cos\left(x+y\right)cos\left(x-y\right)+1-cos^2\left(x-y\right)\right]\)
\(S^2\le3\left[2+\dfrac{1}{4}cos^2\left(x+y\right)-\left[cos\left(x-y\right)-\dfrac{1}{2}cos\left(x+y\right)\right]^2\right]\le3\left[2+\dfrac{1}{4}cos^2\left(x+y\right)\right]\)
\(S^2\le3\left(2+\dfrac{1}{4}\right)=\dfrac{27}{4}\)
\(\Rightarrow S\le\dfrac{3\sqrt{3}}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=3\\c=2\end{matrix}\right.\)
\(A=\dfrac{sin\left(a-b\right)}{cosa.cosb}+\dfrac{sin\left(b-c\right)}{cosb.cosc}+\dfrac{sin\left(c-a\right)}{cosc.cosa}\)
\(=\dfrac{sina.cosb-cosa.sinb}{cosa.cosb}+\dfrac{sinb.cosc-cosb.sinc}{cosb.cosc}+\dfrac{sinc.cosa-cosc.sina}{cosc.cosa}\)
\(=\dfrac{sina}{cosa}-\dfrac{sinb}{cosb}+\dfrac{sinb}{cosb}-\dfrac{sinc}{cosc}+\dfrac{sinc}{cosc}-\dfrac{sina}{cosa}\)
\(=0\)