đơn giản biểu thức
a,\(tan^2\)alpha.(2\(cos^2alpha+sin^2alpha-1\))
cho cos alpha=2/3.Tính giá trị của biểu thức A = 1- 2Sin^2 alpha+-5cos^2alpha
\(A=1-2sin^2\alpha-5cos^2\alpha=1-2\left(sin^2\alpha+cos^2\alpha\right)-3cos^2\alpha\)
\(=1-2-3.\left(\dfrac{2}{3}\right)^2=-1-3.\dfrac{4}{9}=-1-\dfrac{4}{3}=-\dfrac{7}{3}\)
cho cos alpha=2/3.Tính giá trị của biểu thức A = 1- 2Sin^2 alpha+-5cos^2alpha
Đề là \(A=1-2sin^2a+5cos^2a\) hay \(A=1-2sin^2a-5cos^2a\) vậy nhỉ?
A= Cot anpha + \(\frac{SinAlpha}{1+CosAlpha}\)
B=\(\frac{1}{1-SinAlpha}+\frac{1}{1+SinAlpha}\)
C=\(\left(1-CosAlpha\right)\left(1+CosAlpha\right)\)
D=\(tan^2Alpha-Sin^2AlphaTan^2Alpha\)
E=\(Cos^2Alpha+tan^2AlphaCos^2Alpha\)
G=\(SinAlpha-SinAlphaCos^2Alpha\)
H=\(Sin^4Alpha\left(1+2cos^2Alpha\right)+Cos^4Alpha\left(1+2Sin^2Alpha\right)\)
K=\(tan^2Alpha\left(2cos^2Alpha+Sin^2Alpha-1\right)\)
Rút gọn các biểu thức sau
Giúp mình vs mn oi!!
Cho sin alpha =2/3 với 90°<alpha<180°. Tính cos alpha; sin (alpha+30)=?; Sin 2alpha=?
\(90^0< a< 180^0\Rightarrow cosa< 0\)
\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{\sqrt{5}}{3}\)
\(sin2a=2sina.cosa=-\frac{4\sqrt{5}}{9}\)
\(sin\left(a+30^0\right)=sina.cos30^0+cosa.sin30^0=\frac{2}{3}.\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{3}.\frac{1}{2}=\frac{\sqrt{3}}{3}-\frac{\sqrt{5}}{6}\)
Câu 6: Cho góc a thỏa cos alpha = 4/5 và 0 < alpha < pi/2 Giá trị cia * sin 2alpha bằng A. - 12/25 B. 24/25 C. - 24/25 D. 12/25
\(0< a< \dfrac{pi}{2}\)
=>\(sina>0\)
\(sin^2a+cos^2a=1\)
=>\(sin^2a=1-\dfrac{16}{25}=\dfrac{9}{25}\)
=>\(sina=\dfrac{3}{5}\)
\(sin2a=2\cdot sina\cdot cosa=2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5}=\dfrac{24}{25}\)
=>Chọn B
1. Đơn giản biểu thức
a. \(\sin\alpha\cdot\cos\alpha\left(\tan\alpha+\cot\alpha\right)\)
b. \(\left(\sin^2\alpha+\cos^2\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
c. \(\tan^2\alpha-\sin^2\alpha\cdot\tan^2\alpha\)
a) ta có : \(sin\alpha.cos\alpha\left(tan\alpha+cot\alpha\right)=sin\alpha.cos\alpha\left(\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}\right)\)
\(=sin^2\alpha+cos^2\alpha=1\)
b) ta có : \(\left(sin^2\alpha+cos^2\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1^2+1-2sin\alpha.cos=2\left(1-2sin\alpha.cos\alpha\right)\)
c) ta có : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
Hãy đơn giản các biểu thức sau:
a) \(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)
b) \(\tan^2\alpha\left(2.\cos^2\alpha+\sin^2\alpha-1\right)\)
\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)
\(\tan^2\alpha\left(2.\cos^2\alpha+\sin^2\alpha-1\right)=\tan^2\alpha\left(\cos^2\alpha+\left(\sin^2\alpha+\cos^2\alpha\right)-1\right)\)\(=\tan^2\alpha.\cos^2\alpha=\left(\frac{1}{\cos^2\alpha}-1\right)\cos^2\alpha=1-\cos^2\alpha=\sin^2\alpha\)
đơn giản biểu thức:
a, \(\left(\frac{sin\alpha+tan\alpha}{cos\alpha+1}\right)^2+1\)
b, \(tan\alpha\left(\frac{1+cos^2\alpha}{sin\alpha}-sin\alpha\right)\)
c, \(\frac{cot^2\alpha-cos^2\alpha}{cot^2a}+\frac{sin\alpha.cos\alpha}{cot\alpha}\)
\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)
\(=tan^2a+1=\frac{1}{cos^2a}\)
\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)
\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)
\(=1-sin^2a+sin^2a=1\)
Hãy đơn giản các biểu thức:
a) \(\sin\alpha-\sin\alpha.\cos^2\alpha\)
b) \(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha\)
c) \(\cos^2\alpha+\tan^2\alpha.\cos^2\alpha\)
d) \(\tan^2\alpha\left(2\cos^2\alpha+\sin^2\alpha-1\right)\)
alibaba nguyễn bạn có thể giải cụ thể đc ko v ?