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\(a)sin^4x+cos^4x=1-2sin^2x\cdot cos^2x\)
\(\Leftrightarrow sin^4x+2sin^2x\cdot cos^2x+cos^4x=1\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2=1\)(luôn đúng)
Lời giải:
a)
\(\frac{1-\cos x}{\sin x}=\frac{(1-\cos x)(1+\cos x)}{\sin x(1+\cos x)}=\frac{1-\cos ^2x}{\sin x(1+\cos x)}=\frac{\sin ^2x}{\sin x(1+\cos x)}=\frac{\sin x}{1+\cos x}\)
b)
\((\sin x+\cos x-1)(\sin x+\cos x+1)=(\sin x+\cos x)^2-1^2\)
\(=\sin ^2x+\cos ^2x+2\sin x\cos x-1=1+2\sin x\cos x-1=2\sin x\cos x\)
c)
\(\frac{\sin ^2x+2\cos x-1}{2+\cos x-\cos ^2x}=\frac{1-\cos ^2x+2\cos x-1}{2+\cos x-\cos ^2x}=\frac{-\cos ^2x+2\cos x}{2+\cos x-\cos ^2x}\)
\(=\frac{\cos x(2-\cos x)}{(2-\cos x)(\cos x+1)}=\frac{\cos x}{\cos x+1}\)
d)
\(\frac{\cos ^2x-\sin ^2x}{\cot ^2x-\tan ^2x}=\frac{\cos ^2x-\sin ^2x}{\frac{\cos ^2x}{\sin ^2x}-\frac{\sin ^2x}{\cos ^2x}}=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{\cos ^4x-\sin ^4x}\)
\(=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{(\cos ^2x-\sin ^2x)(\cos ^2x+\sin ^2x)}=\frac{\sin ^2x\cos ^2x}{\sin ^2x+\cos ^2x}=\sin ^2x\cos ^2x\)
e)
\(1-\cot ^4x=1-\frac{\cos ^4x}{\sin ^4x}=\frac{\sin ^4x-\cos ^4x}{\sin ^4x}=\frac{(\sin ^2x-\cos ^2x)(\sin ^2x+\cos ^2x)}{\sin ^4x}\)
\(=\frac{\sin ^2x-\cos ^2x}{\sin ^4x}=\frac{\sin ^2x-(1-\sin ^2x)}{\sin ^4x}=\frac{2\sin ^2x-1}{\sin ^4x}=\frac{2}{\sin ^2x}-\frac{1}{\sin ^4x}\)
Ta có ddpcm.
a) Ta có: \(1-\frac{\sin^2x}{1+\cot x}-\frac{\cos^2x}{1+\tan x}=1-\frac{\sin^2x}{1+\frac{\cos x}{\sin x}}-\frac{\cos^2x}{1+\frac{\sin x}{\cos x}}\) (Đk: sinx và cosx khác 0)
\(=1-\frac{\sin^3x}{\sin x+\cos x}-\frac{\cos^3x}{\cos x+\sin x}\)
\(=1-\frac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x.\cos x+\cos^2x\right)}{\sin x+\cos x}\)
\(=1-\left(\sin^2x+\cos^2x-\sin x.\cos x\right)\) (do sinx + cosx luôn khác 0)
\(=\sin x.\cos x\) ( do \(\sin^2x+\cos^2x=1\))
b) Ta có: \(\frac{\sin^2x+2\cos x-1}{2+\cos x-\cos^2x}=\frac{\left(\sin^2x-1\right)+2\cos x}{-\left(\cos x+1\right)\left(\cos x-2\right)}\) (Đk: cosx khác -1 và 2)
\(=\frac{-\cos x\left(\cos x-2\right)}{-\left(\cos x+1\right)\left(\cos x-2\right)}\)
\(=\frac{\cos x}{1+\cos x}\)
a) Ta có: 1−sin2x1+cotx −cos2x1+tanx =1−sin2x1+cosxsinx −cos2x1+sinxcosx (Đk: sinx và cosx khác 0)
=1−sin3xsinx+cosx −cos3xcosx+sinx
=1−(sinx+cosx)(sin2x−sinx.cosx+cos2x)sinx+cosx
=1−(sin2x+cos2x−sinx.cosx) (do sinx + cosx luôn khác 0)
=sinx.cosx ( do sin2x+cos2x=1)
b) Ta có: sin2x+2cosx−12+cosx−cos2x =(sin2x−1)+2cosx−(cosx+1)(cosx−2) (Đk: cosx khác -1 và 2)
=−cosx(cosx−2)−(cosx+1)(cosx−2)
=cosx1+cosx
a/ \(\frac{\pi}{6}< x< \frac{\pi}{3}\Rightarrow cosx>0\)
\(cos^2x=\frac{1}{1+tan^2x}=\frac{1}{10}\)
\(cotx=\frac{1}{tanx}=\frac{1}{3}\)
Thay số và bấm máy
b/ \(\frac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\tana< 0\end{matrix}\right.\)
\(sina=\sqrt{1-cos^2a}=\frac{3}{5}\)
\(tana=\frac{sina}{cosa}=-\frac{3}{4}\)
\(A=\frac{6sina.cosa-\frac{2tana}{1-tan^2a}}{cosa-\left(2cos^2a-1\right)}\)
Thay số và bấm máy
c/ \(\frac{3\pi}{2}< x< 2\pi\Rightarrow\left\{{}\begin{matrix}cosx>0\\sinx< 0\end{matrix}\right.\)
\(cosx=\frac{1}{\sqrt{1+tan^2x}}=\frac{1}{\sqrt{5}}\)
\(sinx=cosx.tanx=-\frac{2}{\sqrt{5}}\)
\(B=\frac{cos^2x+2sinx.cosx}{\frac{2tanx}{1-tan^2x}-\left(2cos^2x-1\right)}\)
Thay số
1) \(\dfrac{1-cosx+cos2x}{sin2x-sinx}=cotx\)
\(VT=\dfrac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}\)
\(VT=\dfrac{cosx\left(2cos-1\right)}{sinx\left(2cosx-1\right)}\)
\(VT=\dfrac{cosx}{sinx}=cotx=VP\) ( đpcm )
b) \(\dfrac{sinx+sin\dfrac{x}{2}}{1+cosx+cos\dfrac{x}{2}}=tan\dfrac{x}{2}\)
\(VT=\dfrac{sin\left(2.\dfrac{x}{2}\right)+sin\dfrac{x}{2}}{1+cos\left(2.\dfrac{x}{2}\right)+cos\dfrac{x}{2}}\)
\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{1+2cos^2\dfrac{x}{2}-1+cos\dfrac{x}{2}}\)
\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{2cos^2\dfrac{x}{2}+cos\dfrac{x}{2}}\)
\(VT=\dfrac{sin\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}{cos\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}\)
\(VT=\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}=tan\dfrac{x}{2}=VP\) ( đpcm )
c) \(\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=tan^2\left(\dfrac{\pi}{4}-x\right)\)
\(VT=\dfrac{2cos2x-sin\left(2.2x\right)}{2cos2x+sin\left(2.2x\right)}\)
\(VT=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)
\(VT=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}\)
\(VT=\dfrac{1-sin2x}{1+sin2x}\)
\(VP=tan^2\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-cos2\left(\dfrac{\pi}{4}-x\right)}{1+cos2\left(\dfrac{\pi}{4}-x\right)}\)
\(VP=\dfrac{1-cos\left(\dfrac{\pi}{2}-2x\right)}{1+cos\left(\dfrac{\pi}{2}-2x\right)}\)
\(VP=\dfrac{1-sin2x}{1+cos2x}=VT\) ( đpcm )
d) \(tanx-tany=\dfrac{sin\left(x-y\right)}{cosx.cosy}\)
\(VP=\dfrac{sin\left(x-y\right)}{cosx.cosy}=\dfrac{sinx.cosy-cosx.siny}{cosx.cosy}\)
\(VP=\dfrac{sinx.cosy}{cosx.cosy}-\dfrac{cosx.siny}{cosx.cosy}\)
\(VP=\dfrac{sinx}{cosx}-\dfrac{siny}{cosy}=tanx-tany=VT\) ( đpcm )
a/ \(\dfrac{\sin x+\cos x-1}{1-\cos x}=\dfrac{2\cos x}{\sin x-\cos x+1}\)
\(\Leftrightarrow-2\cos^2x+2\cos x-2\cos x+2\cos^2x=0\)
\(\Leftrightarrow0=0\) (đúng)
\(\RightarrowĐPCM\)
b/ \(\tan a.\tan b=\dfrac{\tan a+\tan b}{\cot a+\cot b}\)
\(\Leftrightarrow\tan a.\tan b.\left(\cot a+\cot b\right)=\tan a+\tan b\)
\(\Leftrightarrow\tan a.\tan b.\cot a+\tan a.\tan b.\cot b=\tan a+\tan b\)
\(\Leftrightarrow\tan b+\tan a=\tan a+\tan b\) (đúng)
\(\RightarrowĐPCM\)
Giả sử tất cả các biểu thức đều xác định
a/
\(tan^2x-sin^2x=\frac{sin^2x}{cos^2x}-sin^2x=sin^2x\left(\frac{1}{cos^2x}-1\right)\)
\(=sin^2x\left(\frac{1-cos^2x}{cos^2x}\right)=sin^2x.\frac{sin^2x}{cos^2x}=sin^2x.tan^2x\)
b/
\(tanx+cotx=\frac{sinx}{cosx}+\frac{cosx}{sinx}=\frac{sin^2x+cos^2x}{sinx.cosx}=\frac{1}{sinx.cosx}\)
c/
\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)
d/
\(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{1}{1+\frac{1}{tanx}}=\frac{1}{1+tanx}+\frac{tanx}{1+tanx}=\frac{1+tanx}{1+tanx}=1\)
e/
\(\left(1-\frac{1}{cosx}\right)\left(1+\frac{1}{cosx}\right)+tan^2x=1-\frac{1}{cos^2x}+tan^2x\)
\(=\frac{cos^2x-1}{cos^2x}+tan^2x=\frac{-sin^2x}{cos^2x}+tan^2x=-tan^2x+tan^2x=0\)
Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)
2.
\(\dfrac{1-sinx}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{\left(1-sinx\right)\left(1+sinx\right)-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{1-sin^2x-cos^2x}{cosx\left(1+sinx\right)}\)
\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)
3.
\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)
4.
\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)
5.
\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)
\(=tan^2x+1+tan^2x=1+2tan^2x\)
a)
\((\sin x+\cos x)^2=\sin ^2x+2\sin x\cos x+\cos ^2x\)
\(=(\sin ^2x+\cos ^2x)+2\sin x\cos x=1+2\sin x\cos x\)
b)
\(\sin ^4x+\cos ^4x=\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x-2\sin ^2\cos ^2x\)
\(=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x\)
\(=1-2\sin ^2x\cos ^2x\)
c)
\(\tan ^2x-\sin ^2x=(\frac{\sin x}{\cos x})^2-\sin ^2x\)
\(=\sin ^2x\left(\frac{1}{\cos ^2x}-1\right)=\sin ^2x. \frac{1-\cos ^2x}{\cos ^2x}=\sin ^2x.\frac{\sin ^2x}{\cos ^2x}\)
\(=\sin ^2x\left(\frac{\sin x}{\cos x}\right)^2=\sin ^2x\tan ^2x\)
d)
\(\sin ^6x+\cos ^6x=(\sin ^2x)^3+(\cos ^2x)^3\)
\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)\)
\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x\)
\(=(\sin ^4x+\cos ^4x)-\sin ^2x\cos ^2x=1-2\sin ^2x\cos ^2x-\sin ^2x\cos ^2x\)
\(=1-3\sin ^2x\cos ^2x\) (theo kq phần b)
e)
\(\sin x\cos x(1+\tan x)(1+\cot x)=\sin x\cos x(1+\frac{\sin x}{\cos x})(1+\frac{\cos x}{\sin x})\)
\(=\sin x\cos x.\frac{\cos x+\sin x}{\cos x}.\frac{\sin x+\cos x}{\sin x}\)
\(=(\sin x+\cos x)^2=\sin ^2x+\cos ^2x+2\sin x\cos x\)
\(=1+2\sin x\cos x\)
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P/s: Nói chung cứ bám vào công thức \(\sin ^2x+\cos ^2x=1\)
Giả sử các biểu thức đều xác định
a/
\(sinx.cotx+cosx.tanx=sinx.\frac{cosx}{sinx}+cosx.\frac{sinx}{cosx}=sinx+cosx\)
b/
\(\left(1+cosx\right)\left(sin^2x+cos^2x-cosx\right)=\left(1+cosx\right)\left(1-cosx\right)=1-cos^2x=sin^2x\)
c/
\(\frac{sinx+cosx}{cos^3x}=\frac{1}{cos^2x}\left(\frac{sinx+cosx}{cosx}\right)=\left(1+tan^2x\right)\left(tanx+1\right)=tan^3x+tan^2x+tanx+1\)
d/
\(tan^2x-sin^2x=\frac{sin^2x}{cos^2x}-sin^2x=sin^2x\left(\frac{1}{cos^2x}-1\right)\)
\(=sin^2x\left(\frac{1-cos^2x}{cos^2x}\right)=sin^2x.\frac{sin^2x}{cos^2x}=sin^2x.tan^2x\)
e/ \(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=cos^2x\left(\frac{1-sin^2x}{sin^2x}\right)\)
\(=cos^2x.\frac{cos^2x}{sin^2x}=cos^2x.cot^2x\)