Cung và góc liên kết

cos(\(\dfrac{3a}{2}\))*cos(\(\dfrac{a}{2}\))=\(\dfrac{1}{2}\left(cos\left(\dfrac{3a}{2}+\dfrac{a}{2}\right)+cos\left(\dfrac{3a}{2}-\dfrac{a}{2}\right)\right)\)=\(\dfrac{1}{2}\left(cos\left(2a\right)+cos\left(a\right)\right)\)=\(\dfrac{1}{2}\left(2cos^2a-1+cosa\right)\)=\(\dfrac{1}{2}\left(2\cdot\left(\dfrac{3}{4}\right)^2-1+\dfrac{3}{4}\right)=\dfrac{7}{16}\)

\(\sin^4x.\sin^2x+\cos^4x.\cos^2x-\left(\sin^4x+\cos^4x+\dfrac{1}{2}\sin^4x+\dfrac{1}{2}\cos^4x-\dfrac{3}{2}\right)-1=-\sin^4x.\left(1-\sin^2x\right)-cos^4x.\left(1-\cos^2x\right)-\dfrac{1}{2}\left(\sin^4x+\cos^4x\right)+\dfrac{1}{2}=-\left(\sin^4x.\cos^2x+\cos^4x.\sin^2x\right)-\dfrac{1}{2}\left(\left(\sin^2x+\cos^2x\right)^2-2\sin^2x.\cos^2x\right)+\dfrac{1}{2}=-\left(\sin^2x.\cos^2x.\left(\sin^2x+\cos^2x\right)\right)-\dfrac{1}{2}.\left(1-2\sin^2x.\cos^2x\right)+\dfrac{1}{2}=-\sin^2x.\cos^2x+\sin^2x.\cos^2x-\dfrac{1}{2}+\dfrac{1}{2}=0\)

Giúp mình với 

mik ko nhìn rõ bài nào cả =)

Mờ ;-;

Có : 0 < \(\alpha< \dfrac{\pi}{2}\Rightarrow cos\alpha>0\)  \(\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{4}\right)^2}=\dfrac{\sqrt{7}}{4}\)

\(\Rightarrow sin2\alpha=2sin\alpha.cos\alpha=2.\dfrac{3}{4}.\dfrac{\sqrt{7}}{4}=\dfrac{3\sqrt{7}}{8}\) 

\(cos2\alpha=1-2sin^2\alpha=1-2.\left(\dfrac{3}{4}\right)^2=-\dfrac{1}{8}\)

Ta có : \(A=\dfrac{cot2\alpha}{tan2\alpha+sin2\alpha}=\dfrac{1}{sin2\alpha+\dfrac{sin^22\alpha}{cos2\alpha}}=\dfrac{1}{\dfrac{3\sqrt{7}}{8}+\dfrac{\dfrac{63}{64}}{-\dfrac{1}{8}}}=\dfrac{8}{-63+3\sqrt{7}}\)

1:

pi/2<x<pi

=>cosx<0

=>\(cosx=-\sqrt{1-\left(\dfrac{5}{12}\right)^2}=-\dfrac{\sqrt{119}}{12}\)

\(tanx=\dfrac{5}{12}:\dfrac{-\sqrt{119}}{2}=\dfrac{-5}{\sqrt{119}}\)

\(cotx=1:\dfrac{-5}{\sqrt{119}}=\dfrac{-\sqrt{119}}{5}\)

2:

3/2pi<x<2pi

=>sin x<0

=>\(sinx=-\sqrt{1-\left(\dfrac{\sqrt{2}}{2}\right)^2}=-\dfrac{\sqrt{2}}{2}\)

tan x=sin x/cosx=1

=>cot x=1