\(\frac{cos7a+cos3x-2cos5a}{sin6x-sin4a}=2m\Leftrightarrow\frac{2cos5a.cos2a-2cos5a}{2cos5a.sina}=2m\)
\(\Leftrightarrow\frac{2cos5a\left(cos2a-1\right)}{2cos5a.sina}=2m\Leftrightarrow\frac{cos2a-1}{sina}=2m\)
\(\Leftrightarrow\frac{-2sin^2a}{sina}=2m\Leftrightarrow sina=-m\)
\(\Rightarrow cos2a=1-2sin^2a=1-2m^2\)
rút gọn:
a, A=\(\frac{sina+sin2a+sin3a}{cosa+cos2a+cos3a}\)
b, B=\(\frac{sin^2a+sin^2a.tan^2a}{cos^2a+cos^2a.cot^2a}\)
Chứng minh
\(\frac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)=2cosx
\(\frac{cos4a\cdot tan2a-sin4a}{cos4a\cdot cot2a+sin4a}\)=-tan22a
Giúp mình vs! Mình đang cần gấp :((
\(\frac{2\cos^22a+\sqrt{3}\sin4a-1}{2\sin^22a+\sqrt{3}\sin4a-1}\)
Ai rút gọn giúp mình vs ạ
\(\frac{1-2sin2a+cos2a}{1+2sin2a+cos2a}=tan\left(\frac{\pi}{4}-a\right)\)
M\(=\frac{\left(sina-sin2a\right)}{sina+sin2a}\) biết cos2a=\(\frac{1}{8}\) và π<a<\(\frac{3\pi}{2}\)
cm: \(\frac{\left(1-sin2x.sin3x-cos2x.cos3x\right)}{sinx\left(1-tan^2\left(\frac{x}{2}\right)\right)}=\frac{1}{2}tanx\)
cho tam giác ABC thỏa mãn điều kịên cos2a+cos2b+cos2c=-1 là tam giác gì
Chứng minh rằng:
1 + 4cosa + 6cos2a + 4cos3a + cos4a = \(16\cos2a.\cos^4\frac{a}{2}\)
Cho \(\pi< \alpha< \dfrac{3\pi}{2}\) và sin a = \(\dfrac{-5}{13}\) . Tính cosa , sin2a , cos2a , và sin\(\dfrac{a}{2}\)
Tổng: S = \(\frac{1}{sina}+\frac{1}{sin2a}+\frac{1}{sin4a}+...+\frac{1}{sin2^{2018}a}\) là:
A. tan\(\frac{a}{2}\) - tan\(2^{2018}a\)
B. cot\(\frac{a}{2}\) - cot22018a
C. tan\(\frac{a}{2}\) - tan2018a
D. cot\(\frac{a}{2}\) - cot2018a
