A= sin a x cos a (tan s + cot a)
a) Biết sin a =\(\dfrac{2}{3}\).Tính cos a,tan a,cot a
b)Biết cos a =\(\dfrac{1}{5}\).Tính sin a, tan a,cot a
c)Biết tan a = 2.Tính sin a,cos a ,cot a.
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
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
Rút gọn các biểu thức sau :
a) (1- sin^2 x) cot^2 x + 1- cot^2 x
b) ( tan x + cot x ) ^2 - ( tan x - cot x ) ^2
c) ( x. Sin a - y. Cos a )^2 + ( x. Cos a + y. Sin a )^2
a, \(\left(1-sin^2x\right)cot^2x+1-cot^2x\)
\(=cot^2x-sin^2x.cot^2x+1-cot^2x\)
\(=1-sin^2x.\frac{\text{cos}^2x}{sin^2x}=1-\text{cos}^2x=sin^2x\)
b,\(\left(tanx+cotx\right)^2-\left(tanx-cotx\right)2\)
\(=tan^2x2.tanx.cotx+cot^2x-tan^2x+2tanx.cotx-cot^2x\)
\(=4tanxcotx=4\)
c,\(\left(xsina-y\text{cos}a\right)^2+\left(x\text{cos}a+ysina\right)^2\)
\(=x^2sin^2a=2xysina\text{cos}a+y^2\text{cos}^2a+2xysina\text{cos}a+y^2sin^2a\)
\(=x^2\left(sin^2a+\text{cos}^2a\right)+y^2\left(sin^2a+\text{cos}^2a\right)\)
\(=x^2+y^2\)
Chứng minh đẳng thức:
a, \(\dfrac{\sin x+\cos x-1}{1-\cos x}=\dfrac{2\cos x}{\sin x-\cos x+1}\)
b, \(\tan a.\tan b=\dfrac{\tan a+\tan b}{\cot a+\cot b}\)
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)
\(\RightarrowĐPCM\)
bài 1: a)biết sin α=√3/2.tính cos α,tan α,cot α
b)cho tan α=2.tính sin α,cos α,cot α
c)biết sin α=5/13.tính cos,tan,cot α
bài 2
biết sin α x cos α=12/25.tính sin,cos α
1:
a: sin a=căn 3/2
\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)
\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)
cot a=1/tan a=1/căn 3
b: \(tana=2\)
=>cot a=1/tan a=1/2
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=5\)
=>cos^2a=1/5
=>cosa=1/căn 5
\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)
c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)
tan a=5/13:12/13=5/12
cot a=1:5/12=12/5
chứng tỏ:
tan a = sin a/ cos a;
cot a = cos a/ sin a;
tan a . cot a =1;
sin^2 a+ cos^2 a =1
chứng minh rằng
a) tanx(cot\(^2\)x - 1) = cotx(1 - tan\(^2\)x)
b) tan\(^2\)x - sin\(^2\)x = tan\(^2\)x.sin\(^2\)x
c) \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}\) - cos\(^2\)x = - cos\(^4\)x
a: tan x(cot^2x-1)
\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)
=cotx-tanx/cotx=cotx(1-tan^2x)
b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)
\(=sin^2x\left(\dfrac{1}{cos^2x}-1\right)=sin^2x\cdot\dfrac{sin^2x}{cos^2x}=sin^2x\cdot tan^2x\)
c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)
\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)
\(=\dfrac{sin^2x\cdot cos^2x}{1}=sin^2x\cdot cos^2x\)
=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)
=-cos^2x*cos^2x=-cos^4x
=>ĐPCM
A= sin a x cos a (tan a + cot a)
CMR:
a, \(\frac{\cot^2x-\sin^2x}{\cot^2x-tan^2x}=sin^2x.\cos^2x\)
b, \(\frac{\tan x}{1-\tan^2x}.\frac{\cot^2-1}{\cot x}=1\)
c, \(\frac{1+\sin x.\cos x}{\sin^2x-\cos^2x}=\frac{\tan x+1}{\cot x+1}\)
d, \(\frac{\sin x+\cos x-1}{\sin x-cosx+1}=\frac{\cos x}{1+sinx}\)
1.tính cos a, tan a, cot a nếu biết a nhọn và sin a =3/5
2. tính sin x, cos x nếu biết x nhọn và tan x=12/35
3. cho góc a nhọn và cos a =5/13.tính sin a, tan a và cot a
giúp mình với gấp lắm rồi mình sẽ tick cho bạn nào giải được. cảm ơn trước nhé