\(\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosa}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{2}-1\right)}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{a}{2}}}=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{2}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{4}-1\right)}=\sqrt{cos^2\frac{a}{4}}\)
\(=cos\frac{a}{4}\)
1,giá trị lớn nhất cảu biểu thức là:
a, A= sin2x+ 2cosx+1
c, B= cos2x- 2sinx -3
2, kết quả thu gọn của các biểu thức là:
a, A= \(\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosx}}}\) ( 0<x< \(\frac{\pi}{2}\))
b, B= \(\sqrt{2+\sqrt{2+\sqrt{2+2cosa}}}\) ( 0<x< \(\frac{\pi}{2}\))
Rút gọn biểu thức \(\sqrt{\frac{1}{2}-\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\alpha}}\) \(\left(0\le\alpha\le\pi\right)\)
Rút gọn biểu thức :
\(L=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosx}}}\) (0<x <\(\frac{\pi}{2}\))
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,\(\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)\)
1. Rút gọn A= \(\frac{cos4x-cos2x}{2cos2x+1}\)
2. Biết 2cos7x.sin3x-cos10x = \(\frac{1}{\sqrt{6}}\). Tính cos22x
3. Biết cotx = 4. Tính M=\(\frac{sinx+cosx}{cos^3x}\)
P=\(\frac{\text{(sinx+cosx)^2-1 }}{\sqrt{2}cos\left(x+\frac{\Pi}{4}\right).cotx}-\frac{1}{cosx-sinx}\)
ai giúp em bài này với rút gọn biểu thử ạ
Cho \(0< \alpha< \frac{\pi}{2}\). Rút gọn biểu thức \(\sqrt{\frac{1-sin\alpha}{1+sin\alpha}}-\sqrt{\frac{1+sin\alpha}{1-sin\alpha}}\)
Cho tam giác ABC. Hãy rút gọn:
\(a,A=cos^2\left(540^0+\frac{B}{2}\right)+cos^2\frac{1080^0+A+C}{2}+tan\frac{B}{2}tan\frac{A+C}{2}\)
b,\(B=\frac{sin\left(\frac{B}{2}+720^0\right)}{cos\frac{A+C}{2}}+\frac{cos\left(\frac{B}{2}-900^0\right)}{sin\frac{A+C}{2}}-\frac{cos\left(A+C\right)}{sinB}.tanB\)
\(\frac{2\cos^22a+\sqrt{3}\sin4a-1}{2\sin^22a+\sqrt{3}\sin4a-1}\)
Ai rút gọn giúp mình vs ạ
