\(\frac{1-2sin2a+cos2a}{1+2sin2a+cos2a}=\frac{1-4sina.cosa+2cos^2a-1}{1+4sina.cosa+2cos^2a-1}=\frac{2cosa\left(cosa-2sina\right)}{2cosa\left(cosa+2sina\right)}\)
\(=\frac{cosa-2sina}{cosa+2sina}\)
Bạn coi lại đề, muốn ra được biểu thức vế phải thì trước sin2a không được có số 2
M\(=\frac{\left(sina-sin2a\right)}{sina+sin2a}\) biết cos2a=\(\frac{1}{8}\) và π<a<\(\frac{3\pi}{2}\)
cm: \(\frac{\left(1-sin2x.sin3x-cos2x.cos3x\right)}{sinx\left(1-tan^2\left(\frac{x}{2}\right)\right)}=\frac{1}{2}tanx\)
Chứng minh:
\(tan^2\left(x-a\right)+tan^2\left(x+a\right)=\frac{2\left(sin^22a+sin^22x\right)}{\left(cos2x+cos2a\right)^2}\)
chứng minh rằng
1) \(tanx=\frac{1-cos2x}{sin2x}\)
2)\(\frac{sin\left(60^0-x\right).cos\left(30^{0^{ }}-x\right)+cos\left(60^{0^{ }}-x\right).sin\left(30^{0^{ }}-x\right)}{sin4x}=\frac{1}{2sin2x}\)
3) \(4cos\left(60^0+a\right).cos\left(60^0-a\right)+2sin^2a=cos2a\)
chứng minh rằng
a)
\(\frac{1-2\text{s}in^2x}{2cot\left(\frac{\pi}{4}+\alpha\right).c\text{os}^2\left(\frac{\pi}{4}-\alpha\right)}=1\)
b)
\(\frac{\frac{\sqrt{3}}{2}c\text{os}2\text{a}-\frac{1}{2}sin2\text{a}}{1-\frac{1}{2}c\text{os}2\text{a}-\frac{\sqrt{3}}{2}sin2\text{a}}=tan\left(a+\frac{\pi}{4}\right)\)
Rút gọn biểu thức sau:\(A=\left[tan\frac{17\pi}{4}+tan\left(\frac{7\pi}{2}-x\right)\right]^2+\left[cot\frac{17\pi}{4}+cot\left(7\pi\right)-x\right]^2\)
Chứng minh rằng:
1 + 4cosa + 6cos2a + 4cos3a + cos4a = \(16\cos2a.\cos^4\frac{a}{2}\)
Chứng minh rằng: (Pls help me)
a, \(\frac{1}{\sin x}+\cot x=\cot\frac{x}{2}\)
b, \(\frac{1-\cos x}{\sin x}=\tan\frac{x}{2}\)
c,\(\tan\frac{x}{2}\left(\frac{1}{\cos x}+1\right)=\tan x\)
d,\(\frac{\sin2a}{2\cos a\left(1+\cos a\right)}=\tan\frac{a}{2}\)
e,\(\cot x+\tan\frac{x}{2}=\frac{1}{\sin x}\)
f,\(3-4\cos2x+\cos4x=8\sin^4x\)
g,\(\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}\)
h,\(\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\)
i,\(\sin x-\cos x=\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\)
l,\(\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)\)
Chứng minh rằng :
a- Sin2α+sin2α.tan2α=tan2α
b- \(\frac{2sin^2\alpha-1}{Sin^2\alpha-sin\alpha.cos\alpha}=1+cot\alpha\)
c- Cos4α - sin4α=cos2a
Rút gọn các biểu thức sau:
1) \(A=2cosx+3cosx\left(\pi-x\right)-sin\left(\frac{7\pi}{2}-x\right)+tan\left(\frac{3\pi}{2}-x\right)\)
2) \(B=2sin\left(\frac{\pi}{2}+x\right)+sin\left(5\pi-x\right)+sin\left(\frac{3\pi}{2}+x\right)+cos\left(\frac{\pi}{2}+x\right)\)
