\(\frac{sinx}{1+cosx}+\frac{1+cosx}{sinx}=\frac{sin^2x+\left(1+cosx\right)^2}{sinx\left(1+cosx\right)}\)
\(=\frac{sin^2x+cos^2x+1+2cosx}{sinx\left(1+cosx\right)}=\frac{2+2cosx}{sinx\left(1+cosx\right)}\)
\(=\frac{2\left(1+cosx\right)}{sinx\left(1+cosx\right)}=\frac{2}{sinx}\)
Chứng minh|
a) \(\frac{1+sin2x}{sinx+cosx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=sinx\)
b) \(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
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!
Nếu sinx+cosx=1/2 thì sinx, cosx Bằng
Rút gọn biểu thức
\(T=tanx+\frac{cosx}{1+sinx}\)
Chứng minh :
a) 2(1-sinx)(1+cosx) = (1-sinx+cosx)2
b) 1-cos2x/sin2x = tanx
c) 1+cotx+cot2x+cot3x = cosx+sinx/sin3x
Chứng minh đẳng thức sau: \(\frac{sinx+sin3x+sin5x}{cosx+cos3x+cos5x}=tan3x\)
Rút gọn biểu thức A = \(\frac{sinx+sin2x+sin3x}{cosx+cos2x+cos3x}\)
1. sin^8(x) - cos^8(x) - 4sin^6(x) + 6sin^4(x) - 4sin^2(x) = 1
2. sin6x+sin4x+sin2x/1+cos2x+cos4x = 2sin2x
3. 1+sin2x /cosx+sinx - 1-tan^2(x/2)/1+tan^2(x/2) = sinx
4. cos4x + 4cos2x + 3 = 8cos^4(x)
5. 1+cosx+cos2x+cos3x/ 2cos^2(x)+cosx-1 = 2cosx
1. sin^8(x) - cos^8(x) - 4sin^6(x) + 6sin^4(x) - 4sin^2(x) = 1
2. sin6x+sin4x+sin2x/1+cos2x+cos4x = 2sin2x
3. 1+sin2x /cosx+sinx - 1-tan^2(x/2)/1+tan^2(x/2) = sinx
4. cos4x + 4cos2x + 3 = 8cos^4(x)
5. 1+cosx+cos2x+cos3x/ 2cos^2(x)+cosx-1 = 2cosx
