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3. Cho tam giác ABC vuông tại A . Vẽ hình và thiết lập các hệ thúc tính TSLG của góc B từ đó suy ra các hệ thức tính TSLG góc C
Bài 2:
\(=\left(sin^2a+cos^2a\right)^3-3sin^2a\cdot cos^2a\left(sin^2a+cos^2a\right)+3sin^2a\cdot cos^2a\)
\(=1-3\cdot sin^2a\cdot cos^2a+3\cdot sin^2a\cdot cos^2a\)
=1
a: sin a=2/3
=>cos^2a=1-(2/3)^2=5/9
=>\(cosa=\dfrac{\sqrt{5}}{3}\)
\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)
\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
b: cos a=1/5
=>sin^2a=1-(1/5)^2=24/25
=>\(sina=\dfrac{2\sqrt{6}}{5}\)
\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)
\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)
c: cot a=1/tana=1/2
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>1/cos^2a=1+4=5
=>cos^2a=1/5
=>cosa=1/căn 5
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
bài 1 : ta có : \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\left(0,6\right)^2=\dfrac{16}{25}\)
\(\Rightarrow cosa=\pm\dfrac{4}{5}\)
\(\Rightarrow tanx=\dfrac{sinx}{cosx}=\pm\dfrac{3}{4}\) \(\Rightarrow cotx=\dfrac{1}{tanx}=\pm\dfrac{4}{3}\)
bài 2)
ý 1 : a) ta có : \(\dfrac{1}{cos^2a}=\dfrac{sin^2a+cos^2a}{cos^2a}=tan^2a+1\left(đpcm\right)\)
b) ta có : \(\dfrac{1}{sin^2a}=\dfrac{sin^2a+cos^2a}{sin^2a}=1+cot^2a\left(đpcm\right)\)
c) \(cos^4a-sin^4a=\left(sin^2a+cos^2a\right)\left(cos^2a-sin^2a\right)\)
\(=cos^2a-sin^2a=2cos^2a-cos^2a-sin^2a=2cos^2a-1\left(đpcm\right)\)
ý 2 :
ta có : \(tana=2\Rightarrow cota=\dfrac{1}{2}\)
ta có : \(tan^2a+1=\dfrac{1}{cos^2a}\Leftrightarrow cos^2a=\dfrac{1}{tan^2a+1}=\dfrac{1}{5}\)
\(\Rightarrow cosa=\pm\dfrac{1}{\sqrt{5}}\Rightarrow sin^2a=1-cos^2a=\dfrac{4}{5}\) \(\Rightarrow sina=\pm\dfrac{2}{\sqrt{5}}\)
vậy ............................................................................
bài 3 bạn tự luyện tập như bài 2 cho quen nha :)
a) 1 + tan22 a =1 +(\(\dfrac{sina}{cosa}\))2 =\(\dfrac{sina+cosa}{cos^2a}\)=\(\dfrac{1}{cos^2a}\)
b) 1 + cot2 a= 1 +(\(\dfrac{cosa}{sina}\))2 = \(\dfrac{cosa+sina}{sin^2a}\)=\(\dfrac{1}{sin^2a}\)
c) tan2 a (2 sin2a + 3 cos2 a - 2)
=tan2 a[cos2 a +2 (\(sina^2+cos^2a\))-2 ]
=\(\dfrac{sin^2a}{cos^2a}\)×\(cos^2a=sin^2a\)
b: \(1+cot^2a=1+\left(\dfrac{cosa}{sina}\right)^2=\dfrac{1}{sin^2a}\)
c: \(=tan^2a\left[2\left(1-cos^2a\right)+3cos^2a-2\right]\)
\(=tan^2a\left[cos^2a\right]\)
\(=\dfrac{sin^2a}{cos^2a}\cdot cos^2a=sin^2a\)
a) Cần chứng minh \(\dfrac{1-cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1+cos\alpha}\)
\(\Rightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Rightarrow sin^2\alpha=1-cos^2\alpha\)
\(\Rightarrow sin^2\alpha+cos^2\alpha=1\)
Giả sử tam giác ABC vuông tại A
Ta có: \(\left\{{}\begin{matrix}sin^2B=\dfrac{AC^2}{BC^2}\\cos^2B=\dfrac{AB^2}{BC^2}\end{matrix}\right.\Rightarrow sin^2B+cos^2B=\dfrac{AC^2+AB^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\)
a)\(\dfrac{1-cosa}{sina}=\dfrac{sina}{1+cosa}\)
<=>\(\left(1-cosa\right)\left(1+cosa\right)=sin^2a\)
<=>\(1-cos^2a=sin^2a\) (lđ)
b)Ta có VT=\(\dfrac{cosa}{1+sina}+tga=\dfrac{cosa}{1+sina}+\dfrac{sina}{cosa}=\dfrac{cos^2a+sin^2a+sina}{\left(1+sina\right)cosa}=\dfrac{1+sina}{\left(1+sina\right)cosa}=\dfrac{1}{cosa}=vp\left(dpcm\right)\)
Đặt \(f\left(A,B,C\right)=cosA+cosB+cosC+\dfrac{1}{sinA}+\dfrac{1}{sinB}+\dfrac{1}{sinC}-2\sqrt{3}-\dfrac{3}{2}\)
Ta có: \(f\left(A,B,C\right)-f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)\)
\(=\left(cosB+cosC-2cos\left(\dfrac{B+C}{2}\right)\right)+\left(\dfrac{1}{sinB}+\dfrac{1}{sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}\right)\)
\(=2cos\left(\dfrac{B+C}{2}\right)\left(cos\left(\dfrac{B-C}{2}\right)-1\right)+\left(\dfrac{1}{sinB}+\dfrac{1}{sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}\right)\left(1\right)\)
Bên cạnh đó ta có:
\(\dfrac{1}{sinB}+\dfrac{1}{sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}\ge\dfrac{4}{sinB+sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}=\dfrac{4\left(1-cos\left(\dfrac{B-C}{2}\right)\right)}{sinB+sinC}\)
Do đó \(\left(1\right)\ge2\left(1-cos\left(\dfrac{B-C}{2}\right)\right)\left(\dfrac{2}{sinB+sinC}-cos\left(\dfrac{B+C}{2}\right)\right)\)
\(=\left(1-cos\left(\dfrac{B-C}{2}\right)\right)\left(\dfrac{1-sin\left(\dfrac{B+C}{2}\right)cos\left(\dfrac{B+C}{2}\right)cos\left(\dfrac{B-C}{2}\right)}{sinB+sinC}\right)\ge0\)
\(\Rightarrow f\left(A,B,C\right)\ge f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)\)
Giờ ta chỉ cần chứng minh bất đẳng thức đúng trong trường hợp tam giác cân.
Ta có: \(\left\{{}\begin{matrix}B=\dfrac{\pi}{2}-\dfrac{A}{2}\\cosB=cosC=\dfrac{sinA}{2}\\sinB=sinC=\dfrac{cosA}{2}\end{matrix}\right.\)
\(f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)=\left(cosA+2sin\left(\dfrac{A}{2}\right)-\dfrac{3}{2}\right)+\left(\dfrac{1}{sinA}+\dfrac{2}{cos\left(\dfrac{A}{2}\right)}-2\sqrt{3}\right)\)
\(=\dfrac{-2\left(sin\left(\dfrac{A}{2}\right)-1\right)^2}{2}+\dfrac{1+4sin\left(\dfrac{A}{2}\right)-2\sqrt{3}sinA}{sinA}\)
Mà ta có: \(1\ge sin\left(\dfrac{A}{2}+\dfrac{\pi}{3}\right)\)
\(\Rightarrow8sin\left(\dfrac{A}{2}\right)\ge2\sqrt{3}sinA+4sin^2\left(\dfrac{A}{2}\right)\)
\(\Rightarrow1+4sin\left(\dfrac{A}{2}\right)-2\sqrt{3}sinA\ge4sin^2\left(\dfrac{A}{2}\right)-4sin\left(\dfrac{A}{2}\right)+1=\left(2sin\left(\dfrac{A}{2}-1\right)\right)^2\)
Từ đó ta suy ra:
\(f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)\ge\left(2sin-1\right)^2\left(\dfrac{1}{sinA}-\dfrac{1}{2}\right)\ge0\)
Vậy bài toán đã được chứng minh. Dấu = xảy ra khi \(A=B=C=\dfrac{\pi}{3}\)
Hàm số \(f\left(x\right)=\cos\left(x\right)+\dfrac{1}{\sin\left(x\right)}\) là hàm lồi trên \(\left(0,\pi\right)\)
Do đó theo BĐT Jensen ( trường hợp của Karamata) có:
\(f\left(A\right)+f\left(B\right)+f\left(c\right)\ge3f\left(\dfrac{A+B+C}{3}\right)=3f\left(\dfrac{\pi}{3}\right)=2\sqrt{3}+\dfrac{3}{2}\)
P/s:Tính độ "lầy" của hàm số:
\(f''(x)=-\cos(x)-\frac{1}{\sin(x)}+\frac{2}{(\sin(x))^3}\)
Và cho \(x\in (0,\pi);f''(x)>0\) nếu \(2>(\sin(x))^2(\sin(x)\cos(x)+1)\) là xài dc Jensen :D
b,ta có :\(\frac{sin^2a-cos^2a\left(1-cos^2a\right)}{cos^2a-sin^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)
=>\(\frac{sin^2a-sin^2a.cos^2a}{cos^2a-sin^2a.cos^2a}=\frac{sin^4a}{cos^4a}\)
=>\(\frac{sin^2a\left(1-cos^2a\right)}{cos^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)
=>\(\frac{sin^4a}{cos^4a}=\frac{sin^4a}{cos^4a}\)luon dung => dpcm
a) \(1+tan^2B=1+\dfrac{AC^2}{AB^2}=\dfrac{AB^2+AC^2}{AB^2}=\dfrac{BC^2}{AB^2}=\dfrac{1}{\left(\dfrac{AB}{BC}\right)^2}=\dfrac{1}{cos^2B}\)
b) Ta có: \(a.sinB.cosB=BC.\dfrac{AC}{BC}.\dfrac{AB}{BC}=\dfrac{AC.AB}{BC}=\dfrac{AH.BC}{BC}=AH\)
\(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=BC.\left(\dfrac{AB}{BC}\right)^2=BC.cos^2B\)
Tương tự \(\Rightarrow CH=BC.sin^2B\)
1+\(^{ }\tan^{2^{ }}\alpha\)= \(1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2\)=\(\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}\)=\(\dfrac{1}{cos^2\alpha}\)
b: \(1+tan^2a=\dfrac{1}{cos^2a}=1+\dfrac{9}{25}=\dfrac{34}{25}\)
=>cos^2a=25/34
=>\(cosa=\dfrac{5}{\sqrt{34}}\)
\(sina=\sqrt{1-\dfrac{25}{34}}=\dfrac{3}{\sqrt{34}}\)