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ơ bạn :\(\dfrac{cos\left(x+y\right)+cosx}{cos\left(x+y\right)-cosx}=\dfrac{2cos\left(\dfrac{2x+y}{2}\right).cos\left(\dfrac{y}{2}\right)}{-2sin\left(\dfrac{2x+y}{2}\right).sin\left(\dfrac{y}{2}\right)}=-2.cot\left(\dfrac{2x+y}{2}\right).cot\left(\dfrac{y}{2}\right)\) L không thể bẳng 0 được
phần chứng minh biểu thức không phụ thuộc \(x\)
ta có : \(A=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{sinacosa}{cota}=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{cos^2a}{cot^2a}\)
\(=\dfrac{cot^2a-cos^2a+cos^2a}{cot^2a}=\dfrac{cot^2a}{cot^2a}=1\left(đpcm\right)\)
ý còn lại : xem lại đề nha bn
phần chứng minh đẳng thức
ta có : \(\dfrac{sin2a-2sina}{sin2a+2sina}+tan^2\dfrac{a}{2}=\dfrac{2sinacosa-2sina}{2sinacosa+2sina}+tan^2\dfrac{a}{2}\)
\(=\dfrac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}+tan^2\dfrac{a}{2}=\dfrac{cosa-1}{cosa+1}+tan^2\dfrac{a}{2}\)
\(=\dfrac{1-2sin^2\dfrac{a}{2}-1}{2cos^2\dfrac{a}{2}-1+1}+tan^2\dfrac{a}{2}=\dfrac{-2sin^2\dfrac{a}{2}}{2cos^2\dfrac{a}{2}}+tan^2\dfrac{a}{2}\)
\(=-tan^2\dfrac{a}{2}+tan^2\dfrac{a}{2}=0\left(đpcm\right)\)
ta có : \(\dfrac{sina}{1+cosa}+\dfrac{1+cosa}{sina}=\dfrac{sin^2a+\left(1+cosa\right)^2}{sina\left(1+cosa\right)}\)
\(=\dfrac{sin^2a+cos^2a+2cosa+1}{sina\left(1+cosa\right)}=\dfrac{2cosa+2}{sina\left(cosa+1\right)}\)
\(=\dfrac{2\left(cosa+1\right)}{sina\left(cosa+1\right)}=\dfrac{2}{sina}\left(đpcm\right)\)
còn 2 câu kia để chừng nào rảnh mk giải cho nha
mk lm 2 câu còn lại nha
ta có : \(\dfrac{sin^2x}{sinx-cosx}-\dfrac{sinx+cosx}{tan^2x-1}=\dfrac{\left(1-cos^2x\right)\left(tan^2x-1\right)-\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}\)
\(=\dfrac{tan^2x-sin^2x-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2x-1\right)}=\dfrac{\dfrac{sin^4x}{cos^2x}-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2-1\right)}\)
\(=\dfrac{tan^2x\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}=\dfrac{\left(tan^2x-1\right)\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}\)
\(=sinx+cosx\left(đpcm\right)\)
ta có : \(\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-tan^2a.cot^2b}=\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-\dfrac{sin^2a.cos^2b}{cos^2a.sin^2b}}\)
\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{\dfrac{cos^2a.sin^2b-sin^2a.cos^2b}{cos^2a.sin^2b}}=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-\left(sin^2a.cos^2b-cos^2a.sin^2b\right)}\)
\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-\left(\left(sina.cosb-cosa.sinb\right)\left(sina.cosb+cosa.sinb\right)\right)}\)
\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-sin\left(a-b\right)sin\left(a+b\right)}=-cos^2a.sin^2b\left(đpcm\right)\)
mk lm hơi tắc ! do tối rồi , mà mk lại đang ở quán nek nên không tiện làm dài . bạn thông cảm
`A=[sin x + sin 2x + sin 3x]/[cos x + cos 2x + cos 3x]`
`A=[2sin2x.cosx+sin2x]/[2cos2x.cosx+cos2x]`
`A=[sin2x(2cosx+1)]/[cos2x(2cosx+1)]`
`A=tan 2x`
\(A=\dfrac{sinx-sin2x+sin3x}{cosx-cos2x+cos3x}\)
\(ĐK\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\) \(A=\dfrac{sinx+sin3x-sin2x}{cosx+cos3x-cos2x}\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}=\dfrac{2sin2x.cosx-sin2x}{2cos2x.cosx-cos2x}\\=\dfrac{sin2x\left(2cosx-1\right)}{cos2x\left(2cosx-1\right)}\end{matrix}\right.\) \(\Rightarrow\) \(A=tan2x\)
\(A=\dfrac{\sqrt{2}.cosx-2cos\left(\dfrac{\pi}{4}+x\right)}{-\sqrt{2}.sinx+2sin\left(\dfrac{\pi}{4}+x\right)}\)
\(=\dfrac{\sqrt{2}.cosx-2\left(cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\right)}{-\sqrt{2}.sinx+2\left(sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx\right)}\)
\(=\dfrac{\sqrt{2}.cosx-\sqrt{2}.cosx+\sqrt{2}.sinx}{-\sqrt{2}.sinx+\sqrt{2}.cosx+\sqrt{2}.sinx}\)
\(=\dfrac{\sqrt{2}.sinx}{\sqrt{2}.cosx}=tanx\)
\(A=\dfrac{sinx+sin3x+sin2x}{cosx+cos3x+cos2x}=\dfrac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\dfrac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=tan2x\)
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
\(P=\dfrac{sin2x+sinx}{\dfrac{1}{2}\cdot cosx\cdot sin2x+sin2x}=\dfrac{sinx\left(2cosx+1\right)}{sin2x\left(\dfrac{1}{2}cosx+1\right)}\)
\(=\dfrac{2cosx+1}{2\cdot cosx\cdot\left(\dfrac{1}{2}cosx+1\right)}\)