13. Đơn giản biểu thức sau E = cotx + sinx / 1+cosx
17. Biết sin a= 5/13 , cos b =3/5 ( π/2 <a < π ; 0 < b < π/2). Hãy tính sin(a +b)
18. Cho cot = π/14=a. Tính K = sin 2π /7 + sin 4π/7 + sin 6π/7
K
Khách
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HN
29 tháng 4 2020
\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)
=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)
d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)
\(=\left(\frac{cos^2x+sinx.\left(1+sinx\right)}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx.\left(1+cosx\right)}{sinx.\left(1+cosx\right)}\right)=\left(\frac{cos^2x+sinx+sin^2x}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx+cos^2x}{sinx.\left(1+cosx\right)}\right)\)
=\(\frac{1}{cosx.sinx}=VP\)
e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)
c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)
=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)
\(=\frac{\left(sin^2x-cos^2x\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=\frac{\left(sinx-cosx\right).\left(sinx+cosx\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=sinx-cosx=VP\)
Đây nha bạn
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 4 2019
\(P=\frac{1-sin^2x.cos^2x}{cos^2x}-cos^2x=\frac{1}{cos^2x}-sin^2x-cos^2x\)
\(=1+tan^2x-\left(sin^2x+cos^2x\right)=1+tan^2x-1=tan^2x\)
\(M=\frac{2cos^2x-1}{sinx+cosx}=\frac{2cos^2x-\left(sin^2x+cos^2x\right)}{sinx+cosx}=\frac{cos^2x-sin^2x}{sinx+cosx}\)
\(\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx+cosx}=cosx-sinx\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 4 2019
\(A=\frac{sin^2x+cos^2x+2sinx.cosx-1}{\frac{cosx}{sinx}-sinx.cosx}=\frac{2sinx^2x.cosx}{cosx-sin^2x.cosx}=\frac{2sin^2x.cosx}{cosx\left(1-sin^2x\right)}\)
\(=\frac{2sin^2x}{1-sin^2x}=\frac{2sin^2x}{cos^2x}=2tan^2x\)
\(N=\left(\frac{sinx+\frac{sinx}{cosx}}{cosx+1}\right)^2+1=\left(\frac{sinx.cosx+sinx}{cosx\left(cosx+1\right)}\right)^2+1\)
\(=\left(\frac{sinx\left(cosx+1\right)}{cosx\left(cosx+1\right)}\right)^2+1=tan^2x+1=\frac{1}{cos^2x}\)
5 tháng 11 2019
đề bài đầy đủ: rút gọn các biểu thức lượng giác sau trên điều kiện xác định của chúng:
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 11 2019
\(\frac{sin^2x}{cosx+cosx.\frac{sinx}{cosx}}-\frac{cos^2x}{sinx+sinx.\frac{cosx}{sinx}}=\frac{sin^2x}{sinx+cosx}-\frac{cos^2x}{sinx+cosx}=\frac{sin^2x-cos^2x}{sinx+cosx}\)
\(=\frac{\left(sinx+cosx\right)\left(sinx-cosx\right)}{sinx+cosx}=sinx-cosx\)
\(\left(\frac{sinx}{cosx}+\frac{cosx}{1+sinx}\right)\left(\frac{cosx}{sinx}+\frac{sinx}{1+cosx}\right)=\left(\frac{sinx+sin^2x+cos^2x}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+cos^2x+sin^2x}{sinx\left(1+cosx\right)}\right)\)
\(=\left(\frac{sinx+1}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+1}{sinx\left(1+cosx\right)}\right)=\frac{1}{sinx.cosx}\)
26 tháng 11 2018
1.
a) \(\left(1-cos_x\right)\left(1+cos_x\right)-sin^2_x=1-cos^2_x-sin^2_x=1-\left(cos^2_x+sin^2_x\right)=1-1=0\)
b) \(tan^2_x\left(2.cos^2_x+sin^2_x-1\right)+cos^2_x=tan^2_x\left(cos^2_x+sin^2_x+cos^2_x-1\right)+cos^2_x=tan^2_x\left(1-1+cos^2_x\right)+cos^2_x=tan^2_x.cos^2_x+cos^2_x=\left(tan_x.cos_x\right)^2+cos^2_x=sin^2_x+cos^2_x=1\)2. Ta có \(9>5\Leftrightarrow\sqrt{9}>\sqrt{5}\Leftrightarrow3>\sqrt{5}\Leftrightarrow3-\sqrt{5}>0\)
Vậy \(3-\sqrt{5}>0\)
CM
4 tháng 3 2018
a) √2 cos(x - π/4)
= √2.(cosx.cos π/4 + sinx.sin π/4)
= √2.(√2/2.cosx + √2/2.sinx)
= √2.√2/2.cosx + √2.√2/2.sinx
= cosx + sinx (đpcm)
b) √2.sin(x - π/4)
= √2.(sinx.cos π/4 - sin π/4.cosx )
= √2.(√2/2.sinx - √2/2.cosx )
= √2.√2/2.sinx - √2.√2/2.cosx
= sinx – cosx (đpcm).
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 4 2019
a/
\(\left(\frac{sin2x}{cos2x}-\frac{sinx}{cosx}\right)cos2x=\left(\frac{sin2x.cosx-cos2x.sinx}{cos2x.cosx}\right).cos2x\)
\(=\frac{sin\left(2x-x\right)}{cosx}=\frac{sinx}{cosx}=tanx\)
b/
\(2\left(1-sinx\right)\left(1+cosx\right)=2+2cosx-2sinx-2sinxcosx\)
\(=1+sin^2x+cos^2x-2sinx+2cosx-2sinx.cosx\)
\(=\left(1-sinx+cosx\right)^2\)
c/
\(1+cotx+cot^2x+cot^3x=1+cotx+cot^2x\left(1+cotx\right)\)
\(=\left(1+cotx\right)\left(1+cot^2x\right)=\left(1+\frac{cosx}{sinx}\right)\left(1+\frac{cos^2x}{sin^2x}\right)=\frac{sinx+cosx}{sin^3x}\)
d/
\(\frac{cos3x}{sinx}+\frac{sin3x}{cosx}=\frac{cos3x.cosx+sin3x.sinx}{sinx.cosx}=\frac{cos\left(3x-x\right)}{\frac{1}{2}2sinx.cosx}=\frac{2cos2x}{sin2x}=2cot2x\)
AH
Akai Haruma
Giáo viên
4 tháng 5 2020
Lời giải:
a)
\(\cos 2a=\frac{2}{5}\Rightarrow \sin ^22a=1-(\cos 2a)^2=1-(\frac{2}{5})^2=\frac{21}{25}\)
Vì $a\in (0; \frac{\pi}{4})\Rightarrow 2a\in (0; \frac{\pi}{2})$
$\Rightarrow \sin 2a>0\Rightarrow \sin 2a=\frac{\sqrt{21}}{5}$
$\tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{\sqrt{21}}{5.\frac{2}{5}}=\frac{\sqrt{21}}{2}$
$\cot 2a=\frac{1}{\tan 2a}=\frac{2}{\sqrt{21}}$
-------------------------
$\sin 2a=\frac{24}{25}\Rightarrow \cos ^22a=1-(\sin 2a)^2=\frac{49}{625}$
$a\in [\frac{-3}{4}\pi; \frac{-\pi}{2}]\Rightarrow 2a\in [\frac{-3}{2}\pi ; -\pi]\Rightarrow \cos 2a< 0$
$\Rightarrow \cos 2a=\frac{-7}{25}$
$\Rightarrow \tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{24}{25.\frac{-7}{25}}=\frac{-24}{7}$
$\Rightarrow \cot 2a=\frac{-7}{24}$
\(E=\frac{cosx}{sinx}+\frac{sinx}{1+cosx}=\frac{cosx+cos^2x+sin^2x}{sinx\left(1+cosx\right)}=\frac{cosx+1}{sinx\left(1+cosx\right)}=\frac{1}{sinx}\)
17.
\(\frac{\pi}{2}< a< \pi\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{12}{13}\)
\(0< b< \frac{\pi}{2}\Rightarrow sinb>0\Rightarrow sinb=\sqrt{1-cos^2b}=\frac{4}{5}\)
\(sin\left(a+b\right)=sina.cosb+cosa.sinb=\frac{5}{13}.\frac{3}{5}-\frac{12}{13}.\frac{4}{5}=-\frac{33}{65}\)
18.
\(K=sin\frac{2\pi}{7}+sin\frac{6\pi}{7}+sin\frac{4\pi}{7}\)
\(\Leftrightarrow K.sin\frac{\pi}{7}=sin\frac{\pi}{7}.sin\frac{2\pi}{7}+sin\frac{\pi}{7}.sin\frac{4\pi}{7}+sin\frac{\pi}{7}.sin\frac{6\pi}{7}\)
\(=\frac{1}{2}\left(cos\frac{\pi}{7}-cos\frac{3\pi}{7}+cos\frac{\pi}{7}-cos\frac{5\pi}{7}+cos\frac{5\pi}{7}-cos\frac{7\pi}{7}\right)\)
\(=\frac{1}{2}\left(cos\frac{\pi}{7}-cos\pi\right)=\frac{1}{2}\left(cos\frac{\pi}{7}+1\right)=\frac{1}{2}\left(2cos^2\frac{\pi}{14}-1+1\right)=cos^2\frac{\pi}{14}\)
\(\Leftrightarrow K.2.sin\frac{\pi}{14}.cos\frac{\pi}{14}=cos^2\frac{\pi}{14}\)
\(\Leftrightarrow2K=\frac{cos\frac{\pi}{14}}{sin\frac{\pi}{14}}=cot\frac{\pi}{14}=a\Rightarrow K=\frac{a}{2}\)