1) Cho sin (\(\pi\)+a) = \(\frac{-1}{3}\) và \(\frac{\pi}{2}< a< \pi\). Tính P= tan (\(\frac{7\pi}{2}-a\))
2) Cho góc a thỏa mạc \(\frac{\pi}{2}< a< 2\pi\)và tan (\(\left(a+\frac{\pi}{4}\right)\) =1. Tính P= cos \(\left(a-\frac{\pi}{6}\right)+sina\)
3)Cho góc a thõa mãn \(\frac{\pi}{2}< a< 2\pi\) và cot \(\left(a+\frac{\pi}{3}\right)=-\sqrt{3}\) . Tính giá trị biểu thức P= sin\(\left(a+\frac{\pi}{6}\right)+cosa\)
4) Cho góc a thõa mãn sinacosa=\(\frac{12}{25}\) và sin a + cos a>0. Tính P= \(Sin^3a+cos^3a\)
5) Cho góc a thõa mãn sin a+ cos a =m. Tính P=\(\left|Sina-cosa\right|\)
Xin mọi người giải giúp em nha, nếu có thể chi tiết càng tốt. Em xin cảm ơn
Câu 4:
Đặt \(x=sina+cosa>0\Rightarrow x^2=\left(sina+cosa\right)^2\)
\(\Rightarrow x^2=sin^2a+cos^2a+2sina.cosa=1+2.\frac{12}{25}=\frac{49}{25}\)
\(\Rightarrow x=\sqrt{\frac{49}{25}}=\frac{7}{5}\)
\(\Rightarrow P=\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)\)
\(P=\frac{7}{5}\left(1-\frac{12}{25}\right)=\frac{91}{125}\)
Câu 5:
\(sina+cosa=m\Rightarrow\left(sina+cosa\right)^2=m^2\)
\(\Leftrightarrow sin^2a+cos^2a+2sina.cosa=m^2\)
\(\Leftrightarrow1+2sina.cosa=m^2\)
\(\Rightarrow2sina.cosa=m^2-1\)
\(P=\left|sina-cosa\right|\ge0\)
\(\Leftrightarrow P^2=\left(sina-cosa\right)^2=sin^2a+cos^2a-2sina.cosa\)
\(\Leftrightarrow P^2=1-2sina.cosa=1-\left(m^2-1\right)=2-m^2\)
\(\Rightarrow P=\sqrt{2-m^2}\)
Câu 1:
Do \(\frac{\pi}{2}< a< \pi\Rightarrow cosa< 0\)
\(sin\left(\pi+a\right)=-sina\Rightarrow-sina=-\frac{1}{3}\Rightarrow sina=\frac{1}{3}\)
\(\Rightarrow cosa=-\sqrt{1-sin^2a}=\frac{-2\sqrt{2}}{3}\)
\(P=tan\left(\frac{7\pi}{2}-a\right)=tan\left(3\pi+\frac{\pi}{2}-a\right)=tan\left(\frac{\pi}{2}-a\right)=cota\)
\(\Rightarrow P=\frac{cosa}{sina}=-2\sqrt{2}\)
Câu 2:
\(tan\left(a+\frac{\pi}{4}\right)=\frac{tana+tan\frac{\pi}{4}}{1-tana.tan\frac{\pi}{4}}=\frac{tana+1}{1-tana}\)
\(\Rightarrow\frac{tana+1}{1-tana}=1\Rightarrow tana+1=1-tana\Rightarrow tana=0\)
\(\Rightarrow\frac{sina}{cosa}=0\Rightarrow sina=0\)
Do \(\frac{\pi}{2}< a< 2\pi\Rightarrow-1\le cosa< 1\)
\(cos^2a=1-sin^2a=1-0=1\Rightarrow\left[{}\begin{matrix}cosa=-1\\cosa=1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow P=cos\left(a-\frac{\pi}{6}\right)+sina=cosa.cos\frac{\pi}{6}+sina.sin\frac{\pi}{6}+sina\)
\(P=-1.\frac{\sqrt{3}}{2}+0.\frac{1}{3}+0=-\frac{\sqrt{3}}{2}\)
Câu 3:
\(cot\left(a+\frac{\pi}{3}\right)=-\sqrt{3}\Rightarrow\frac{cos\left(a+\frac{\pi}{3}\right)}{sin\left(a+\frac{\pi}{3}\right)}=-\sqrt{3}\)
\(\Leftrightarrow\frac{cosa.cos\frac{\pi}{3}-sina.sin\frac{\pi}{3}}{sina.cos\frac{\pi}{3}+cosa.sin\frac{\pi}{3}}=-\sqrt{3}\Leftrightarrow\frac{1}{2}cosa-\frac{\sqrt{3}}{2}sina=-\sqrt{3}\left(\frac{1}{2}sina+\frac{\sqrt{3}}{2}cosa\right)\)
\(\Leftrightarrow\frac{1}{2}cosa-\frac{\sqrt{3}}{2}sina+\frac{\sqrt{3}}{2}sina+\frac{3}{2}cosa=0\)
\(\Leftrightarrow2cosa=0\Rightarrow cosa=0\)
Do \(\frac{\pi}{2}< a< 2\pi\Rightarrow-1\le sina< 1\)
\(\Rightarrow sin^2a=1-cos^2a=1\Rightarrow\left[{}\begin{matrix}sina=1\left(l\right)\\sina=-1\end{matrix}\right.\)
\(\Rightarrow P=sin\left(a+\frac{\pi}{6}\right)+cosa=sina.cos\frac{\pi}{6}+cosa.sin\frac{\pi}{6}+cosa\)
\(\Rightarrow P=-\frac{\sqrt{3}}{2}+0+0=-\frac{\sqrt{3}}{2}\)
