Tính sin2a, cos2a, tan2a biết
Rút gọn biểu thức:
B = (1+ tan2a).(1- sin2a) \(-\)(1+ cotg2a).(1- cos2a)
\(\left(1+tan^2a\right)\left(1-sin^2a\right)-\left(1+cot^2a\right)\left(1-cos^2a\right)\)
\(=\left(1+\dfrac{sin^2a}{cos^2a}\right).cos^2a-\left(1+\dfrac{cos^2a}{sin^2a}\right).sin^2a\)
\(=cos^2a+sin^2a-sin^2a-cos^2a=\)\(0\)
Vậy B=0
Cho cosa = 3/4 vào 270°<a<370° . Tính
A sina , tana , cota
B sin2a , cos2a , tan2a
B sin( a+ π\3 )
Câu 1 : chứng minh rằng : \(\frac{sina+sin2a+sin3a}{cosa+cos2a+cos3a}=tan2a\)
Câu 2 : chứng minh : \(cos^2\left(\alpha-\frac{\pi}{4}\right)-sin^2\left(\alpha-\frac{\pi}{4}\right)=sin2\alpha\)
\(\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(cos^2\left(a-\frac{\pi}{4}\right)-sin^2\left(a-\frac{\pi}{4}\right)=cos\left(2a-\frac{\pi}{2}\right)\)
\(=cos\left(\frac{\pi}{2}-2a\right)=sin2a\)
Chứng minh tan2a - sin2a .tan2a=(1-cos a)(1+cos a)
\(\tan^2\alpha-\sin^2\alpha\cdot\tan^2\alpha\)
\(=\tan^2\alpha\cdot\left(1-\cos^2\alpha\right)\)
\(=\tan^2\alpha\cdot\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)\)
\(\tan^2\alpha-\sin^2\alpha\cdot\tan^2\alpha\\ =\tan^2\alpha\left(1-\sin^2\alpha\right)=\tan^2\alpha\cdot\cos^2\alpha\\ =\dfrac{\sin^2\alpha}{\cos^2\alpha}\cdot\cos^2\alpha=\sin^2\alpha\\ =1-\cos^2\alpha=\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)\)
Cm:\(\dfrac{1+cos2a+sin2a}{1+sin2a-cos2a}=tana\)
\(VT=\dfrac{1+\cos^2a-\sin^2a+2\cdot\sin a\cdot\cos a}{1+2\cdot\sin a\cdot\cos a-\cos^2a+\sin^2a}\)
\(=\dfrac{2\cdot\cos^2a+2\cdot\sin a\cdot\cos a}{2\cdot\sin^2a+2\cdot\sin a\cdot\cos a}\)
\(=\dfrac{2\cdot\cos a\left(\cos a+\sin a\right)}{2\cdot\sin a\cdot\left(\sin a+\cos a\right)}\)
\(=\dfrac{\cos a}{\sin a}=\cot a\)
Cho sina - cosa =1/5. Tính sin2a, cos2a
(Sina -cosa)^2 =1:25
<=> sin^2a +cos^2a -2sina.cosa =1:25
Ta có sin^2a+cos^2a = 1
<=> 1-2 sina.cosa =1:25
2sina.cosa =24:25
CT : sin2a= 2sina.cosa=24:25
Có sin^2 .2a + co^2.2a = 1
(24:25)^2 + cos^2.2a =1
Từ đây rút cos 2a = căn 1-(24:25)^2 =... bạn tự làm tiếp nha !
rút gọn
\(\dfrac{sin2a+1}{cos2a}-\dfrac{1-sin2a}{sina-cosb}\)
Tính \(\sin2a;\cos2a;\tan2a\) biết :
a) \(\sin a=-0,6\) và \(\pi< a< \dfrac{3\pi}{2}\)
b) \(\cos a=-\dfrac{5}{13}\) và \(\dfrac{\pi}{2}< a< \pi\)
c) \(\sin a+\cos a=\dfrac{1}{2}\) và \(\dfrac{\pi}{2}< a< \dfrac{3\pi}{4}\)
Cho \(\pi< \alpha< \dfrac{3\pi}{2}\) và sin a = \(\dfrac{-5}{13}\) . Tính cosa , sin2a , cos2a , và sin\(\dfrac{a}{2}\)
Lời giải:
$\sin ^2a+\cos ^2a=1$
$\cos ^2a=1-\sin ^2a=1-(\frac{-5}{13})^2=\frac{144}{169}$
Vì $\pi < a< \frac{3\pi}{2}$ nên $\cos a< 0$
Do đó: $\cos a=-\sqrt{\frac{144}{169}}=\frac{-12}{13}$
$\sin 2a=2\sin a\cos a=2.\frac{-5}{13}.\frac{-12}{13}=\frac{120}{169}$
$\cos 2a=\cos ^2a-\sin ^2a=2\cos ^2a-1=2.\frac{144}{169}-1=\frac{119}{169}$
$\cos a=\cos ^2\frac{a}{2}-\sin ^2\frac{a}{2}$
$=1-2\sin ^2\frac{a}{2}$
$\Leftrightarrow \frac{-12}{13}=1-2\sin ^2\frac{a}{2}$
$\Rightarrow \sin ^2\frac{a}{2}=\frac{25}{26}$
Vì $\pi < a< \frac{3\pi}{2}$ nên $\sin \frac{a}{2}>0$
$\Rightarrow \sin \frac{a}{2}=\frac{5}{\sqrt{26}}$