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a: \(sin^2x+cos^2x=1\)
=>\(sin^2x=1-cos^2x\)
=>\(sinx=\sqrt{1-cos^2x}\)
b: \(sin^2x+cos^2x=1\)
=>\(cos^2x=1-sin^2x\)
=>\(cosx=\sqrt{1-sin^2x}\)
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.
Thực hiện phép biến đổi tương đương:
\(\dfrac{sinx+cosx-1}{1-cosx}=\dfrac{2cosx}{sinx-cosx+1}\)
\(\Leftrightarrow\left(sinx+cosx-1\right)\left(sinx-cosx+1\right)=2cosx\left(1-cosx\right)\)
\(\Leftrightarrow sin^2x-\left(cosx-1\right)^2=2cosx-2cos^2x\)
\(\Leftrightarrow sin^2x-cos^2x+2cosx-1=2cosx-2cos^2x\)
\(\Leftrightarrow1-cos^2x-cos^2x-1=-2cos^2x\)
\(\Leftrightarrow-2cos^2x=-2cos^2x\) (luôn đúng)
Vậy đẳng thức đã cho được chứng minh
b.
\(cot^2x-cos^2x=\dfrac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\dfrac{1}{sin^2x}-1\right)=\dfrac{cos^2x\left(1-sin^2x\right)}{sin^2x}=cot^2x.cos^2x\)
Cho biết \(cosx=-\dfrac{1}{2}\)
\(sin^2x+cos^2x=1\Rightarrow sin^2x=1-cos^2x\)
\(\Rightarrow sin^2x=1-\dfrac{1}{4}=\dfrac{3}{4}\)
\(S=4sin^2x+8tan^2x\)
\(\Rightarrow S=4\left(sin^2x+2\dfrac{sin^2x}{cos^2x}\right)\)
\(\Rightarrow S=4\left(\dfrac{3}{4}+2\dfrac{\dfrac{3}{4}}{\dfrac{1}{4}}\right)\)
\(\Rightarrow S=4\left(\dfrac{3}{4}+6\right)\)
\(\Rightarrow S=4.\dfrac{27}{4}=27\)
a, \(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)
\(=\dfrac{1+cos2x+cosx+cos3x}{2cos^2x+cosx-1}\)
\(=\dfrac{2cos^2x+2cos2x.cosx}{cos2x+cosx}\)
\(=\dfrac{2cosx\left(cos2x+cosx\right)}{cos2x+cosx}=2cosx\)
b) \(cos\dfrac{5x}{2}.cos\dfrac{3x}{2}+sin\dfrac{7x}{2}.sin\dfrac{x}{2}\)
\(=cos\dfrac{4x+x}{2}.cos\dfrac{4x-x}{2}+sin\dfrac{4x+3x}{2}.sin\dfrac{4x-3x}{2}\)
\(=\dfrac{1}{2}\left(cos4x+cosx\right)-\dfrac{1}{2}\left(cos4x-cos3x\right)\)
\(=\dfrac{1}{2}\left(cosx+cos3x\right)=\dfrac{1}{2}.2cos2x.cos\left(-x\right)\)\(=cosx.cos2x\)
a) \(\left(sinx+cosx\right)^2=sin^2x+2sinxcosx+cos^2x\)\(=1+2sinxcosx\).
b) \(\left(sinx-cosx\right)^2=sin^2x-2sinxcosx+cos^2x\)\(=1-2sinxcosx\).
c) \(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\)
\(=1-2sin^2xcos^2x\).
với (cosx khác 0)
VT: \(\dfrac{cosx+sinx}{cosx^3}=\dfrac{\dfrac{cosx}{cosx}+\dfrac{sinx}{cosx}}{\dfrac{cosx^3}{cosx}}=\dfrac{1+tanx}{cosx^2}\)
VP:
\(tanx^3+tanx^2+tanx+1=\left(tanx+1\right)\left(tanx^2+1\right)\\ =\left(tanx+1\right).\dfrac{1}{cosx^2+1}\)
Vậy VT=VP
1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
\(=3\left(sin^4x+cos^4x\right)-2\left(sin^4x-sin^2x.cos^4x+cos^4x\right)\)
\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)
Vậy...
2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)
\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)
\(=-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)
\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)
Vậy...
3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)
\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)
\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)
Vậy...
4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)
\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)
\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)
Vậy...
5, Xem lại đề
6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)
\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)
Vậy...
\(A=\dfrac{4sin^4x-cos^2x\left(1-cos^2x\right)+sin^2x.cos^2x-2cos^2x}{sin^2x}+\dfrac{2}{tan^2x}\)
\(=\dfrac{4sin^4x-sin^2x.cos^2x+sin^2x.cos^2x-2cos^2x}{sin^2x}+2cot^2x\)
\(=4sin^2x-2cot^2x+2cot^2x=4sin^2x\)
\(\Rightarrow\left\{{}\begin{matrix}a=4\\b=2\end{matrix}\right.\)
a: Ta sẽ có hình vẽ sau:
Đặt \(x=\widehat{B}\)
sin x=sin B=AC/BC
cosx=cosB=AB/BC
\(tanx=tanB=\dfrac{AC}{AB}=\dfrac{sinx}{cosx}\)
=>\(tan^2x=\dfrac{sin^2x}{cos^2x}\)
b: \(cot^2x=\dfrac{1}{tan^2x}=1:\dfrac{sin^2x}{cos^2x}=\dfrac{cos^2x}{sin^2x}\)