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Cho sin a= 
1) \(sin\left(A+2B+C\right)=sin\left(\pi-B+2B\right)\)
=\(sin\left(\pi+B\right)=sin\left(-B\right)=-sinB\)
2) \(sinBsinC-cosBcosC=-cos\left(B+C\right)\)
\(=-cos\left(\pi-A\right)=cosA\)
4) bạn ơi +2 vào vế phải mới đúng nhé
2+ \(2cosAcosBcosC=\left[cos\left(A+B\right)+cos\left(A-B\right)\right]cosC+2\)
\(=cos\left(\pi-C\right)cosC+cos\left(A-B\right)cos\left(\pi-\left(A+B\right)\right)+2\)
=\(-cos^2C-cos\left(A-B\right)cos\left(A+B\right)+2\)
\(=-cos^2C-\frac{1}{2}\left(cos2A+cos2B\right)+2\)
\(=-cos^2C-\frac{1}{2}\left(2cos^2A-1\right)-\frac{1}{2}\left(2cos^2B-1\right)+2\)
\(=-cos^2C-cos^2A+\frac{1}{2}-cos^2C+\frac{1}{2}+2\)
= sin2C - 1 + sin2A - 1 + sin2C - 1 + 3
= sin2A + sin2B + sin2C
Bài 1:
Đk:\(1\le x\le2\)
\(pt\Leftrightarrow x^2-3x-10=-\sqrt{x^2-3x+2}\)
Đặt \(\sqrt{x^2-3x+2}=t\left(t\ge0\right)\) ta có:
\(t^2-12=-t\Leftrightarrow t^2+t-12=0\)\(\Leftrightarrow\left(t+4\right)\left(t-3\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}t=-4\left(loai\right)\\t=3\end{array}\right.\)
Xét \(t=3\Leftrightarrow x^2-3x+2=3\)
\(\Leftrightarrow x^2-3x-1=0\)
\(\Delta=\left(-3\right)^2-\left[\left(-4\right).\left(1.1\right)\right]=13\)\(\Leftrightarrow x_{1,2}=\frac{3\pm\sqrt{13}}{2}\) (thỏa mãn)
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(A=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{4}+x\right)=0\)
a: \(A=\sqrt{3}\left(\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx\right)+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)
\(=\dfrac{\sqrt{3}}{2}sinx-\dfrac{3}{2}cosx+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)
\(=\sqrt{3}sinx-cosx\)
c: \(=2\left[\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right]+4sinx+1\)
\(=\sqrt{3}sin2x-cos2x+4sinx+1\)
d: \(D=\sqrt{3}cos2x+sin2x+2\cdot\left(\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right)\)
\(=\sqrt{3}\cdot cos2x+sin2x+\sqrt{3}\cdot sin2x-cos2x\)
\(=cos2x\left(\sqrt{3}-1\right)+sin2x\left(1+\sqrt{3}\right)\)