\(B=\frac{-\sin\left(\frac{\pi}{2}+144^0\right)-\cos126^0}{\sin144^0-\cos126^0}.\tan\left(\pi-144^0\right)\)

\(B=\frac{-\cos144^0-\cos126^0}{\sin144^0-\cos126^0}.\left(-\tan144^0\right)\)

\(B=\frac{\sin144^0.\cos144^0+\sin144^0.\cos126^0}{\sin144^0.\cos144^0-\cos144^0.\cos126^0}\)

\(B=\frac{\sin\left(\pi+\frac{\pi}{2}-126^0\right)[\cos\left(\pi+\frac{\pi}{2}-126^0\right)+\cos126^0]}{\cos\left(\pi+\frac{\pi}{2}-126^0\right)[\sin\left(\pi+\frac{\pi}{2}-126^0\right)-\cos126^0]}\)

\(\sin\left(\pi+\frac{\pi}{2}-126^0\right)=-\sin\left(\frac{\pi}{2}-126^0\right)=-\cos126^0\)

\(\cos\left(\pi+\frac{\pi}{2}-126^0\right)=-\cos\left(\frac{\pi}{2}-126^0\right)=-\sin126^0\)

\(\Rightarrow B=\frac{-\cos126^0\left(-\sin126^0+\cos126^0\right)}{-\sin126^0\left(-\cos126^0-\cos126^0\right)}\)

\(=\cot126^0.\frac{\sin126^0-\cos126^0}{2\cos126^0}\)

\(=\cot126^0\left(\frac{1}{2}.\tan126^0-\frac{1}{2}\right)\)

\(=\frac{1}{\tan126^0}.\frac{1}{2}.\tan126^0-\frac{1}{2}.\cot126^0=\frac{1}{2}\left(1-\cot126^0\right)\)

Thế này là gọn nhất rồi đấy :<

\(B=\frac{sin126^0-cos144^0}{sin144^0-cos126^0}.tan36^0=\frac{cos36^0+sin54^0}{cos54^0+sin36^0}.tan36^0\)

\(=\frac{cos36^0+cos36^0}{sin36^0+sin36^0}.tan36^0=cot36^0.tan36^0=1\)

\(A=\dfrac{sin\left(a-b\right)}{cosa.cosb}+\dfrac{sin\left(b-c\right)}{cosb.cosc}+\dfrac{sin\left(c-a\right)}{cosc.cosa}\)

\(=\dfrac{sina.cosb-cosa.sinb}{cosa.cosb}+\dfrac{sinb.cosc-cosb.sinc}{cosb.cosc}+\dfrac{sinc.cosa-cosc.sina}{cosc.cosa}\)

\(=\dfrac{sina}{cosa}-\dfrac{sinb}{cosb}+\dfrac{sinb}{cosb}-\dfrac{sinc}{cosc}+\dfrac{sinc}{cosc}-\dfrac{sina}{cosa}\)

\(=0\)

\(=\frac{cos36-sin\left(180+54\right)}{sin\left(180-36\right)-cos\left(90+36\right)}.cos54=\frac{cos36+sin54}{sin36+sin36}.cos54\)

\(=\frac{cos36+cos36}{2sin36}.sin36=cos36\)

Đây nè bn:

rút gọn biểu thức:

E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))

tìm cả đk giúp mik vs

ĐKXĐ: \(x>0;x\ne1\)

\(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)

\(=\left(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\right)\)

\(=\dfrac{\left(x+2\sqrt{x}\right).x.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)}=\dfrac{x}{\sqrt{x}-1}\)

b.

\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)

\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)

c.

Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)

Ta có:

\(\sqrt{A}=\sqrt{\dfrac{x}{\sqrt{x}-1}}=\sqrt{\dfrac{x}{\sqrt{x}-1}-4+4}=\sqrt{\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4}\ge\sqrt{4}=2\)

Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)

\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)

\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)

\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)

\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)

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

\(=cosx-cosx+sin^2x+cos^2x+sinx\)

\(=1+sinx\)

\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)

\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)

\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)

\(=1+cosx\)