Sin4a/1+cos4a + cos2a/1+cos2a = tana
Rút gọn biểu thức sau:
A=4sinx*cosx*cos2x*cos4x
B=cos^4x -6cos^x*sin^2x+sim^4x
C=\(\frac{\text{cos2a-cos4a}}{sin4a+sin2a}\)
D=\(\frac{\text{cosa+cos3a+cos5a}}{sina+sin3a+sin5a}\)
E=sin^2(\(\frac{\pi}{8}\)+\(\frac{x}{2}\))-sin^2(\(\frac{\pi}{8}\)-\(\frac{x}{2}\))
F=\(\frac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)
Rút gọn các biểu thức sau :
a)\(\dfrac{1+\sin4a-\cos4a}{1+\cos4a+\sin4a}\)
b) \(\dfrac{1+\cos a}{1-\cos a}\tan^2\dfrac{a}{2}-\cos^2a\)
c) \(\dfrac{\cos2x-\sin4x-\cos6x}{\cos2x+\sin4x-\cos6x}\)
Chứng minh
a) \(2sin\left(\frac{\pi}{4}+a\right)sin\left(\frac{\pi}{4}-a\right)=cos2a\)
b) \(tanx-\frac{1}{tanx}=-\frac{2}{tan2x}\)
Cho tam giác ABC CM : Cos2A + Cos2B + Cos2C \(\ge-\dfrac{3}{2}\)
Câu 1 : chứng minh rằng : \(\frac{sina+sin2a+sin3a}{cosa+cos2a+cos3a}=tan2a\)
Câu 2 : chứng minh : \(cos^2\left(\alpha-\frac{\pi}{4}\right)-sin^2\left(\alpha-\frac{\pi}{4}\right)=sin2\alpha\)
Giá trị biểu thức P= \(\left(sin2a+sin2b\right)^2+\left(cos2a+cos2b\right)^2\) BIẾT a-b=\(\frac{\pi}{6}\) là
CM các đẳng thức LG sau:
1)\(\left(cos^4a+sin^4a\right)-2\left(cos^6a+sin^6a\right)=1\)
2) \(\frac{sin^2a+cos^2a}{1+2sina.cosa}=\frac{tana-1}{tana+1}\)
3) \(sin^4a+cos^4a-sin^6a-cos^6a=sin^2a.cos^2a\)
4) \(\frac{cosa}{1+sina}+tana=\frac{1}{cosa}\)
5) \(\frac{tana}{a-tan^2a}.\frac{cot^2a-1}{cota}=1\)
Chứng minh VT=VP:
a) 2.(sinx+cosx+1)2.(sinx+cosx-1)2=1-cos4x
b) \(\frac{\text{3-4cos2a+cos4a}}{3+\text{4cos2a+cos4a}}\)= tan4a
c) (cos2x-sin2x)2+2(sin3x-sinx).cos-sin2x=cos2x
Cần GẤP ạ! Cảm ơn nhiều ạ!
