\(a,cos\left(\dfrac{21\pi}{6}\right)=cos\left(3\pi+\dfrac{\pi}{2}\right)=cos\left(\pi+\dfrac{\pi}{2}\right)=-cos\left(\dfrac{\pi}{2}\right)=0\\ b,sin\left(\dfrac{129\pi}{4}\right)=sin\left(32\pi+\dfrac{\pi}{4}\right)=sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\\ c,tan\left(1020^o\right)=tan\left(5\cdot180^o+120^o\right)=tan\left(120^o\right)=-\sqrt{3}\)

a) \(cos638^o=cos\left(-82^o\right)=cos\left(82^o\right)=sin8^o\)

b) \(cot\dfrac{19\pi}{5}=cot\dfrac{4\pi}{5}=-cot\dfrac{\pi}{5}\)

Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)

\(\dfrac{sin^42x+cos^42x}{tan\left(\dfrac{\pi}{4}-x\right)tan\left(\dfrac{\pi}{4}+x\right)}=cos^4x\)

\(\Leftrightarrow\dfrac{sin^42x+cos^42x}{cot\left(\dfrac{\pi}{4}+x\right)tan\left(\dfrac{\pi}{4}+x\right)}=cos^4x\)

\(\Leftrightarrow sin^42x+cos^42x=cos^4x\)

Giờ hạ bậc nữa là xong rồi. Làm nốt

Hình như đề bạn bị lỗi, thấy chỗ nào cũng ghi là \(cos^44x\).

ĐK: \(x\ne\dfrac{3\pi}{4}+k\pi;x\ne\dfrac{\pi}{4}+k\pi\)

\(\dfrac{sin^42x+cos^42x}{tan\left(\dfrac{\pi}{4}-x\right).tan\left(\dfrac{\pi}{4}+x\right)}=cos^44x\)

\(\Leftrightarrow\dfrac{sin^42x+cos^42x}{\dfrac{sin\left(\dfrac{\pi}{4}-x\right)}{cos\left(\dfrac{\pi}{4}-x\right)}.\dfrac{sin\left(\dfrac{\pi}{4}+x\right)}{cos\left(\dfrac{\pi}{4}+x\right)}}=cos^44x\)

\(\Leftrightarrow\dfrac{sin^42x+cos^42x}{\dfrac{cosx-sinx}{cosx+sinx}.\dfrac{cosx+sinx}{cosx-sinx}}=cos^44x\)

\(\Leftrightarrow sin^42x+cos^42x=cos^44x\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^24x=cos^44x\)

\(\Leftrightarrow cos^44x-\dfrac{1}{2}cos^24x-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos^24x=1\\cos^24x=-\dfrac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{2}cos8x=\dfrac{1}{2}\)

\(\Leftrightarrow cos8x=1\)

\(\Leftrightarrow x=\dfrac{k\pi}{4}\)

Đối chiều điều kiện ban đầu ta được \(x=\dfrac{k\pi}{2}\)

\(\dfrac{\pi}{2}< x< \pi\Rightarrow cosx< 0\)

\(\Rightarrow cosx=-\sqrt{1-sin^2x}=-\dfrac{20}{29}\)

\(tanx=\dfrac{sinx}{cosx}=-\dfrac{21}{20}\)

\(cotx=\dfrac{1}{tanx}=-\dfrac{20}{21}\)

\(0< a< \dfrac{\pi}{2}\Rightarrow0< \dfrac{a}{2}< \dfrac{\pi}{4}\Rightarrow sin\dfrac{a}{2}>0\)

\(\Rightarrow sin\dfrac{a}{2}=\sqrt{1-cos^2\dfrac{a}{2}}=\dfrac{3}{5}\)

\(sina=2sin\dfrac{a}{2}cos\dfrac{a}{2}=2.\left(\dfrac{4}{5}\right)\left(\dfrac{3}{5}\right)=\dfrac{24}{25}\)

\(cosa=\pm\sqrt{1-sin^2a}=\pm\dfrac{7}{25}\)

\(tana=\dfrac{sina}{cosa}=\pm\dfrac{24}{7}\)

\(\sin^2x=\sqrt{1-\left(-\dfrac{4}{5}\right)^2}=\dfrac{9}{25}\)

mà \(\sin x>0\)

nên \(\sin x=\dfrac{3}{5}\)

=>\(\tan x=-\dfrac{3}{4}\)

\(\Leftrightarrow\cot x=-\dfrac{4}{3}\)

a) Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha< 0;cot\alpha>0;tan\alpha>0\).
Vì vậy: \(sin\alpha=-\sqrt{1-cos^2\alpha}=\dfrac{-\sqrt{15}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\sqrt{15}}{4}:\dfrac{-1}{4}=\sqrt{15}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{\sqrt{15}}\).

b) Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(cos\alpha< 0;tan\alpha< 0;cot\alpha< 0\).
\(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{5}}{3}\);
\(tan\alpha=\dfrac{2}{3}:\dfrac{-\sqrt{5}}{3}=\dfrac{-2}{\sqrt{5}}\); \(cot\alpha=1:tan\alpha=\dfrac{-\sqrt{5}}{2}\).

c) Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
Có \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{1}{tan^2\alpha+1}\)
Vì vậy: \(cos\alpha=\sqrt{\dfrac{1}{tan^2+1}}=\dfrac{\sqrt{58}}{3}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{3}{7}\).
\(sin\alpha=cos\alpha:cot\alpha=\dfrac{\sqrt{58}}{3}:\dfrac{3}{7}=\dfrac{7\sqrt{58}}{9}\).