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\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
Công thức hạ bậc
\(sin^2a=\frac{1}{2}-\frac{1}{2}cos2a\)
Julian Edward
\(\frac{\pi}{2}< a< \pi\Rightarrow cosa< 0\)
\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{4}{5}\)
\(P=1-\left[1-cos\left(\frac{\pi}{2}-2a\right)\right]+sin2a-cos2a-6cota\)
\(=sin2a+sin2a-cos2a-6cota\)
\(=2sin2a-cos2a-6cota\)
\(=4sina.cosa-\left(cos^2a-sin^2a\right)-\frac{6cosa}{sina}\) (thay số và bấm máy)
--.-- \(-\pi>-\frac{3}{2}\pi\) mà
Chắc nhầm đề rồi, phải là \(-\pi>a>-\frac{3}{2}\pi\)mới đúng chứ
\(-\pi>a>-\frac{3}{2}\pi\Leftrightarrow\pi>a>\frac{1}{2}\pi\)
\(\cos a=-\frac{4}{5}\Rightarrow\sin a=\frac{3}{5}\)
\(\sin2a=2\sin a.\cos a=2.\frac{3}{5}.\frac{-4}{5}=-\frac{24}{25}\)
\(\cos2a=2\cos^2a-1=\frac{7}{25}\)
\(\sin\left(\frac{5\pi}{2}-a\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a=-\frac{4}{5}\)
\(\sin\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{3}{5}-\frac{4}{5}.\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{10}\)
\(\cos\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{-4}{5}-\frac{\sqrt{2}}{2}.\frac{3}{5}=-\frac{7\sqrt{2}}{10}\)
\(\Rightarrow\tan\left(a+\frac{\pi}{4}\right)=\frac{1}{7}\)
\(\cos^2\left(\frac{a}{2}\right)=\frac{1+\cos a}{2}=\frac{1}{10}\Leftrightarrow\left|\cos\frac{a}{2}\right|=\frac{\sqrt{10}}{10}\)
Mà \(\frac{\pi}{2}>\frac{a}{2}>\frac{\pi}{4}\)
\(\Rightarrow\cos\frac{a}{2}=\frac{\sqrt{10}}{10}\)
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
Theo mk là A đúng
ta có : cos2x = \(\frac{1+cos2x}{2}\)
=> cos2(\(\frac{\pi}{4}\)+\(\frac{\alpha}{2}\))= \(\frac{1+cos\left(\frac{\pi}{2}+\alpha\right)}{2}\) = \(\frac{1-sinx}{2}\)
\(\frac{\pi}{2}< a< \frac{3\pi}{2}\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{\sqrt{3}}{2}\)
\(A=cosa.cos\frac{4\pi}{3}+sina.sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}.\left(-\frac{1}{2}\right)+\frac{1}{2}.\left(-\frac{\sqrt{3}}{2}\right)=0\)
\(B=cos\left(2a+2019.2\pi\right)=cos2a=1-2sin^2a=1-2\left(\frac{1}{2}\right)^2=\frac{1}{2}\)
\(\left(sina-cosa\right)^2=2\Leftrightarrow sin^2a+cos^2a-2sina.cosa=2\)
\(\Leftrightarrow1-sin2a=2\Rightarrow sin2a=-1\)
\(\left(sina+cosa\right)^2=2\Leftrightarrow sin^2a+cos^2a+2sina.cosa=2\)
\(\Leftrightarrow1+sin2a=2\Rightarrow sin2a=1\)
\(\frac{3\pi}{2}< a< 2\pi\Rightarrow cosa>0\Rightarrow cosa=\sqrt{1-sin^2a}=\frac{1}{2}\)
\(\Rightarrow cos\left(a+\frac{\pi}{3}\right)=cosa.cos\frac{\pi}{3}-sina.sin\frac{\pi}{3}\)
\(=\frac{1}{2}.\frac{1}{2}-\left(-\frac{\sqrt{3}}{2}\right).\left(\frac{\sqrt{3}}{2}\right)=...\)