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\(\frac{\pi}{2}< a< \frac{3\pi}{4}\Rightarrow\frac{7\pi}{8}< a+\frac{3\pi}{8}< \frac{9\pi}{8}\Rightarrow\frac{\pi}{2}< a+\frac{3\pi}{8}< \frac{3\pi}{2}\)
\(\Rightarrow cos\left(a+\frac{3\pi}{8}\right)< 0\)
\(\frac{\pi}{2}< a< \frac{3\pi}{4}\Rightarrow-\frac{5\pi}{4}< a-\frac{7\pi}{4}< -\pi\Rightarrow-\frac{3\pi}{2}< a< -\pi\)
\(\Rightarrow tan\left(a-\frac{7\pi}{4}\right)< 0\)
\(A=2cosx-3cosx-sin\left(3\pi+\frac{\pi}{2}-x\right)+tan\left(\pi+\frac{\pi}{2}-x\right)\)
\(A=-cosx+sin\left(\frac{\pi}{2}-x\right)+tan\left(\frac{\pi}{2}-x\right)\)
\(A=-cosx+cosx+cotx=cotx\)
\(B=2cosx+sin\left(4\pi+\pi-x\right)+sin\left(2\pi-\frac{\pi}{2}+x\right)-sinx\)
\(B=2cosx+sin\left(\pi-x\right)+sin\left(-\frac{\pi}{2}+x\right)-sinx\)
\(B=2cosx+sinx-sin\left(\frac{\pi}{2}-x\right)-sinx\)
\(B=2cosx-cosx=cosx\)
\(=cos\left(4\pi+\pi+x\right)+sin\left(4\pi+\frac{\pi}{2}-x\right)-tan\left(\pi+\frac{\pi}{2}+x\right).cot\left(\pi+\frac{\pi}{2}-x\right)\)
\(=cos\left(\pi+x\right)+sin\left(\frac{\pi}{2}-x\right)-tan\left(\frac{\pi}{2}+x\right).cot\left(\frac{\pi}{2}-x\right)\)
\(=-cosx+cosx-\left(-cotx\right).tanx\)
\(=1\)
\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)
\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)
\(A=0\)
\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)
\(B=\frac{1}{4}\)
\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)
\(C=\frac{3}{2}\)
\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)
\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)
\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)
\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)
Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là
\(A=cos\left(6\pi+\pi-x\right)+sin\left(2\pi+\frac{\pi}{2}-x\right)+tan^2\left(\pi+\frac{\pi}{2}-x\right)-\frac{1}{sin^2\left(7\pi+\pi+x\right)}\)
\(=cos\left(\pi-x\right)+sin\left(\frac{\pi}{2}-x\right)+tan^2\left(\frac{\pi}{2}-x\right)-\frac{1}{sin^2\left(\pi+x\right)}\)
\(=-cosx+cosx+cot^2x-\frac{1}{sin^2x}\)
\(=cot^2x-\left(1+cot^2x\right)=-1\)
Câu 1:
\(tan\left(a+\frac{\pi}{4}\right)=1\Rightarrow a+\frac{\pi}{4}=\frac{\pi}{4}+k\pi\Rightarrow a=k\pi\) (\(k\in Z\) )
Do \(\frac{\pi}{2}< a< 2\pi\Rightarrow\frac{\pi}{2}< k\pi< 2\pi\Rightarrow\frac{1}{2}< k< 2\Rightarrow k=1\Rightarrow a=\pi\)
\(\Rightarrow P=cos\left(\pi-\frac{\pi}{6}\right)+sin\pi=-\frac{\sqrt{3}}{2}\)
Câu 2:
\(cot\left(a+\frac{\pi}{3}\right)=-\sqrt{3}=cot\left(-\frac{\pi}{6}\right)\)
\(\Rightarrow a+\frac{\pi}{3}=-\frac{\pi}{6}+k\pi\Rightarrow a=-\frac{\pi}{2}+k\pi\) (\(k\in Z\))
\(\Rightarrow\frac{\pi}{2}< -\frac{\pi}{2}+k\pi< 2\pi\Rightarrow-\pi< k\pi< \frac{5\pi}{2}\)
\(\Rightarrow-1< k< \frac{5}{2}\Rightarrow k=\left\{0;1;2\right\}\Rightarrow a=\left\{-\frac{\pi}{2};\frac{\pi}{2};\frac{3\pi}{2}\right\}\) \(\Rightarrow cosa=0\)
\(\Rightarrow P=sin\left(\pi+\frac{\pi}{6}\right)+0=-sin\frac{\pi}{6}=-\frac{1}{2}\)
Vậy đáp án sai
Bạn thay thử \(a=\frac{3\pi}{2}\) vào biểu thức ban đầu coi có đúng \(cot\left(a+\frac{\pi}{3}\right)=-\sqrt{3}\) ko là biết đáp án đúng hay sai liền mà
\(sina.sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)\)
\(=-\frac{1}{2}sina\left[cos\frac{2\pi}{3}-cos2a\right]=-\frac{1}{2}sina\left(-\frac{1}{2}-cos2a\right)\)
\(=\frac{1}{4}sina+\frac{1}{2}sina.cos2a=\frac{1}{4}sina+\frac{1}{4}sin3a-\frac{1}{4}sina\)
\(=\frac{1}{4}sin3a\)
\(sin\frac{\pi}{9}sin\frac{2\pi}{9}sin\frac{4\pi}{9}=sin\frac{\pi}{9}sin\left(\frac{\pi}{3}-\frac{\pi}{9}\right)sin\left(\frac{\pi}{3}+\frac{\pi}{9}\right)=\frac{1}{4}sin\frac{\pi}{3}=\frac{\sqrt{3}}{8}\)
\(cosa.cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)=\frac{1}{2}cosa\left(cos\frac{2\pi}{3}+cos2a\right)\)
\(=\frac{1}{2}cosa\left(cos2a-\frac{1}{2}\right)=\frac{1}{2}cosa.cos2a-\frac{1}{4}cosa\)
\(=\frac{1}{4}cos3a+\frac{1}{4}cosa-\frac{1}{4}cosa=\frac{1}{4}cos3a\)
\(cos\frac{\pi}{18}cos\frac{5\pi}{18}cos\frac{7\pi}{18}=cos\frac{\pi}{18}.cos\left(\frac{\pi}{3}-\frac{\pi}{18}\right).cos\left(\frac{\pi}{3}+\frac{\pi}{18}\right)=\frac{1}{4}cos\frac{\pi}{6}=\frac{\sqrt{3}}{8}\)
\(P=cos\left(-x\right).tanx-cotx.\left(-cotx\right)\)
\(=cosx.tanx+cot^2x=sinx+cot^2x\)
\(=sinx+\frac{1}{1+sin^2x}=-\frac{1}{3}+\frac{1}{1+\frac{1}{9}}=\frac{17}{30}\)
\(\frac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\cosa< 0\\tana< 0\end{matrix}\right.\)
\(M=cos\left(\frac{\pi}{2}-a\right)tan\left(\pi-a\right)=sina.\left(-tana\right)=-sina.tana=-\frac{sin^2a}{cosa}>0\)