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\(-\frac{\pi}{4}\le x\le\frac{\pi}{2}\Rightarrow-\pi\le x-\frac{3\pi}{4}\le-\frac{\pi}{4}\)
\(\Rightarrow-1\le cos\left(x-\frac{3\pi}{4}\right)\le\frac{\sqrt{2}}{2}\)
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(A=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{4}+x\right)=0\)
\(x\in\left[-\pi;\frac{2\pi}{3}\right]\Rightarrow2x\in\left[-2\pi;\frac{4\pi}{3}\right]\Rightarrow sin2x\in\left[-1;1\right]\)
a/ \(\pi< x< \frac{3\pi}{2}\Rightarrow sinx< 0\)
\(\Rightarrow sinx=-\sqrt{1-cos^2x}=-\frac{5}{13}\)
\(sin\left(\frac{\pi}{3}-x\right)=sin\frac{\pi}{3}cosx-cos\frac{\pi}{3}sinx=\frac{\sqrt{3}}{2}.\left(-\frac{12}{13}\right)-\frac{1}{2}.\left(-\frac{5}{13}\right)=\frac{5-12\sqrt{3}}{26}\)
b/ \(\pi< x< \frac{3\pi}{2}\Rightarrow cosx< 0\)
\(\Rightarrow cosx=-\sqrt{1-sin^2x}=-\frac{3}{5}\)
\(cot\left(x-\frac{\pi}{4}\right)=\frac{cos\left(x-\frac{\pi}{4}\right)}{sin\left(x-\frac{\pi}{4}\right)}=\frac{sinx+cosx}{sinx-cosx}=7\)
c/ \(cot\left(\frac{5\pi}{2}-x\right)=cot\left(2\pi+\frac{\pi}{2}-x\right)=tanx=2\)
\(\Rightarrow tan\left(x+\frac{\pi}{4}\right)=\frac{tanx+tan\frac{\pi}{4}}{1-tanx.tan\frac{\pi}{4}}=\frac{2+1}{1-2.1}=-3\)
\(sin\left(x+\frac{\pi}{3}\right)-sin\left(x-\frac{\pi}{3}\right)=sin\left(6\pi+\frac{\pi}{2}\right)\)
\(\Leftrightarrow2cosx.sin\frac{\pi}{3}=sin\left(\frac{\pi}{2}\right)\)
\(\Leftrightarrow2cosx.\frac{\sqrt{3}}{2}=1\)
\(\Rightarrow cosx=\frac{1}{\sqrt{3}}\)
\(2sin\left(\frac{\pi}{2}+x\right)+sin\left(3\pi-x\right)+sin\left(\frac{3\pi}{2}+x\right)+cos\left(\frac{\pi}{2}+x\right)\)
\(=2cosx+sinx-cosx-sinx\)
\(=cosx\)
\(\frac{\pi}{2}< a< \frac{3\pi}{4}\Rightarrow\frac{7\pi}{8}< a+\frac{3\pi}{8}< \frac{9\pi}{8}\Rightarrow\frac{\pi}{2}< a+\frac{3\pi}{8}< \frac{3\pi}{2}\)
\(\Rightarrow cos\left(a+\frac{3\pi}{8}\right)< 0\)
\(\frac{\pi}{2}< a< \frac{3\pi}{4}\Rightarrow-\frac{5\pi}{4}< a-\frac{7\pi}{4}< -\pi\Rightarrow-\frac{3\pi}{2}< a< -\pi\)
\(\Rightarrow tan\left(a-\frac{7\pi}{4}\right)< 0\)
\(\pi< x< \frac{3\pi}{2}\Rightarrow sinx< 0;cosx< 0;tanx>0;cotx>0\)
\(tanx-3cotx=6\Leftrightarrow tanx-\frac{3}{tanx}=6\)
\(\Leftrightarrow tan^2x-6tanx-3=0\Rightarrow\left[{}\begin{matrix}tanx=3+2\sqrt{3}\\tanx=3-2\sqrt{3}< 0\left(l\right)\end{matrix}\right.\)
\(\frac{1}{cos^2x}=1+tan^2x\Rightarrow cos^2x=\frac{1}{1+tan^2x}\Rightarrow cosx=\frac{-1}{\sqrt{1+tan^2x}}\) (do \(cosx< 0\))
\(\Rightarrow cosx=\frac{-1}{\sqrt{22+12\sqrt{3}}}\Rightarrow sinx=-\sqrt{1-cos^2x}=-\sqrt{\frac{15+6\sqrt{3}}{26}}\)
\(cotx=\frac{1}{tanx}=\frac{1}{3+2\sqrt{3}}\)
Số xấu dữ dội, bạn tự thay vào kết quả :(
\(0\in\left[-\frac{\pi}{3};\frac{2\pi}{3}\right]\) mà \(cos0=1\Rightarrow cosx\le1\)
Cụ thể hơn thì cung \(x\in\left[-\frac{\pi}{3};\frac{2\pi}{3}\right]\) được tô đậm như hình, gióng xuống trục cos được đoạn \(\left[-\frac{1}{2};1\right]\):
\(x\in\left[-\frac{\pi}{3};\frac{2\pi}{3}\right]\Rightarrow cosx\in\left[-\frac{1}{2};1\right]\)