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\(M=cos^2\left(\frac{\pi}{4}+a\right)-cos^2\left(\frac{\pi}{4}-a\right)\)
\(=\frac{1}{2}\left(1+cos\left(\frac{\pi}{2}+2a\right)-1-cos\left(\frac{\pi}{2}-2a\right)\right)\)
\(=\frac{1}{2}cos\left(\frac{\pi}{2}+2a\right)-\frac{1}{2}cos\left(\frac{\pi}{2}-2a\right)\)
\(=-\frac{1}{2}sin2a-\frac{1}{2}sin2a=-sin2a\)
\(6sin^4x-2cos^4x=1\Leftrightarrow6sin^4x-2\left(1-sin^2x\right)^2-1=0\)
\(\Leftrightarrow6sin^4x-2\left(sin^4x-2sin^2x+1\right)-1=0\)
\(\Leftrightarrow4sin^4x+4sin^2x-3=0\)
\(\Leftrightarrow\left(2sin^2x+3\right)\left(2sin^2x-1\right)=0\)
\(\Leftrightarrow2sin^2x=1\Rightarrow sin^2x=\frac{1}{2}\Rightarrow cos^2x=\frac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}sin^4x=\frac{1}{4}\\cos^4x=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow C=\frac{1}{4}+3.\frac{1}{4}=1\)
Bài 1:
\(A=\left(1+sinx\right)\left(1-sinx\right)tan^2x=\left(1-sin^2x\right).\frac{sin^2x}{cos^2x}=cos^2x.\frac{sin^2x}{cos^2x}=cos^2x\)
\(B=cot^2x-sin^2x.cot^2x+1-cot^2x=1-sin^2x.\frac{cos^2x}{sin^2x}=1-cos^2x=sin^2x\)
\(C=tan^2x+2+\frac{1}{tan^2x}-\left(tan^2x-2+\frac{1}{tan^2x}\right)=2+2=4\)
Bài 2:
Đề yêu cầu tính giá trị lượng giác nào bạn? sin?cos?tan?cot?
Không hỏi thì làm sao mà biết cần tính gì
Bài 2:
\(0< a< \frac{\pi}{2}\Rightarrow sina>0\)
\(\Rightarrow sina=\sqrt{1-cos^2a}=\frac{\sqrt{3}}{2}\)
\(tana=\frac{sina}{cosa}=\sqrt{3}\)
\(cota=\frac{1}{tana}=\frac{1}{\sqrt{3}}\)
b/ \(-\frac{\pi}{2}< a< 0\Rightarrow cosa>0\)
\(\Rightarrow cosa=\sqrt{1-sin^2a}=\frac{\sqrt{5}}{3}\)
\(tana=\frac{sina}{cosa}=-\frac{2\sqrt{5}}{5}\)
\(cota=\frac{1}{tana}=-\frac{\sqrt{5}}{2}\)
c/ \(-\frac{\pi}{2}< a< 0\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa>0\end{matrix}\right.\)
\(\Rightarrow cosa=\frac{1}{\sqrt{1+tan^2a}}=...\)
Nhìn lại thì đề ảo quá, ai cho số \(\sqrt{\frac{3}{3}}\) bao giờ, như vầy người ta cho luôn số 1 cho rồi. Chắc bạn ghi nhầm, nhưng bạn có thể tự tính bằng cách thay vào công thức, sau đó tính \(sina=cosa.tana\)
d/ \(\frac{3\pi}{2}< 0< 2\pi\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa>0\end{matrix}\right.\)
\(\Rightarrow sina=-\frac{1}{\sqrt{1+cot^2a}}=-\frac{\sqrt{5}}{5}\)
\(cosa=sina.cota=\frac{2\sqrt{5}}{5}\)
\(tana=\frac{1}{cota}=-\frac{1}{2}\)
Câu 1
Ta có vế trái:
\[ VT=\sin\left(\frac{\pi}{2}-\alpha\right) \cos(\pi-\alpha) \] \[ =\cos\alpha.(-\cos\alpha) \] \[ =-\cos^2\alpha \]Ta có vế phải:
\[ VP=-\frac{1}{1+\tan^2(\pi-\alpha)} \] \[ =-\frac{1}{1+\tan^2\alpha} \] \[ =-\frac{1}{\frac{1}{\cos^2\alpha}} \] \[ =-\cos^2\alpha \]Vậy VT = VP (đpcm).
Câu 2
Ta có vế trái:
\[ VT=\sin^2\left(\frac{\pi}{2}-\alpha\right) \] \[ =\cos^2\alpha \]Ta có vế phải:
\[ VP=\frac{1}{1+\tan^2\alpha} \] \[ =\frac{1}{1+\frac{\sin^2\alpha}{\cos^2\alpha}} \] \[ =\frac{1}{\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}} \] \[ =\frac{1}{\frac{1}{\cos^2\alpha}} \] \[ =\cos^2\alpha \]Vậy VT = VP (đpcm).
--.-- \(-\pi>-\frac{3}{2}\pi\) 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 \(\frac{\pi}{2}>\frac{a}{2}>\frac{\pi}{4}\)
\(\Rightarrow\cos\frac{a}{2}=\frac{\sqrt{10}}{10}\)
Bài 4:
$\sin a=\frac{1}{2}$ và $0< a< \pi$ nên $a=\frac{\pi}{6}$ hoặc $a=\frac{5}{6}\pi$
Nếu $a=\frac{\pi}{6}$ thì $\tan (2a-\frac{\pi}{2})+\sin a=\tan (2.\frac{\pi}{6}-\frac{\pi}{2})+\frac{1}{2}=\frac{-\sqrt{3}}{3}+\frac{1}{2}=\frac{3-2\sqrt{3}}{6}$
Nếu $a=\frac{5\pi}{6}$ thì:
\(\tan (2a-\frac{\pi}{2})+\sin a=\tan (2.\frac{5\pi}{6}-\frac{\pi}{2})+\frac{1}{2}=\frac{\sqrt{3}}{3}+\frac{1}{2}=\frac{3+2\sqrt{3}}{6}\)
Bài 3:
\(\tan a=\frac{-4}{7}=\frac{\sin a}{\cos a}\)
\(\Rightarrow \frac{\sin ^2a}{\cos ^2a}=\frac{16}{49}\Rightarrow \frac{1}{\cos ^2a}=\frac{65}{49}\) \(\Rightarrow \cos ^2a=\frac{49}{65}\)
Kết hợp điều kiện của $a$ suy ra $\cos a>0\Rightarrow \cos a=\frac{7}{\sqrt{65}}$
$\Rightarrow \sin a=\frac{-4}{7}\cos a=\frac{-4}{\sqrt{65}}$
Do đó:
\(\cos (2a-\frac{\pi}{2})=\cos 2a.\cos \frac{\pi}{2}+\sin 2a.\sin \frac{\pi}{2}\)
\(=(\cos ^2a-\sin ^2a).0+2\sin a\cos a.1=2\sin a\cos a=2.\frac{-4}{\sqrt{65}}.\frac{7}{\sqrt{65}}=\frac{56}{65}\)
Câu 1:
\(\frac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\cosa< 0\end{matrix}\right.\)
Ta có: \(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}\Rightarrow cosa=\frac{-1}{\sqrt{1+tan^2a}}=-\frac{3}{5}\)
\(\Rightarrow sina=cosa.tana=\frac{4}{5}\)
\(\Rightarrow P=\frac{\frac{16}{25}+\frac{3}{5}}{\frac{4}{5}-\frac{9}{25}}=\frac{31}{11}\)
Câu 2:
\(P=sin^4a-cos^4a=\left(sin^2a+cos^2a\right)\left(sin^2a-cos^2a\right)=sin^2a-cos^2a\)
\(P=1-cos^2a-cos^2a=1-2cos^2a\)
Theo cmt ta có \(cos^2a=\frac{1}{1+tan^2a}\Rightarrow P=1-\frac{2}{1+tan^2a}=\frac{12}{13}\)
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:
\(cot\left(a+\frac{\pi}{3}\right)=-\sqrt{3}=cot\left(-\frac{\pi}{6}\right)\)
\(\Rightarrow a+\frac{\pi}{3}=-\frac{\pi}{6}+k\pi\Rightarrow a=-\frac{\pi}{2}+k\pi\) (\(k\in Z\))
\(\Rightarrow\frac{\pi}{2}< -\frac{\pi}{2}+k\pi< 2\pi\Rightarrow-\pi< k\pi< \frac{5\pi}{2}\)
\(\Rightarrow-1< k< \frac{5}{2}\Rightarrow k=\left\{0;1;2\right\}\Rightarrow a=\left\{-\frac{\pi}{2};\frac{\pi}{2};\frac{3\pi}{2}\right\}\) \(\Rightarrow cosa=0\)
\(\Rightarrow P=sin\left(\pi+\frac{\pi}{6}\right)+0=-sin\frac{\pi}{6}=-\frac{1}{2}\)
Câu 1:
\(tan\left(a+\frac{\pi}{4}\right)=1\Leftrightarrow\frac{sin\left(a+\frac{\pi}{4}\right)}{cos\left(a+\frac{\pi}{4}\right)}=1\Leftrightarrow sin\left(a+\frac{\pi}{4}\right)=cos\left(a+\frac{\pi}{4}\right)\)
\(\Leftrightarrow sina.cos\frac{\pi}{4}+cosa.sin\frac{\pi}{4}=cosa.cos\frac{\pi}{4}-sina.sin\frac{\pi}{4}\)
\(\Leftrightarrow\frac{\sqrt{2}}{2}sina+\frac{\sqrt{2}}{2}cosa=\frac{\sqrt{2}}{2}cosa-\frac{\sqrt{2}}{2}sina\)
\(\Rightarrow\sqrt{2}sina=0\Rightarrow sina=0\)
\(\Rightarrow P=cos\left(\pi-\frac{\pi}{6}\right)+0=-cos\frac{\pi}{6}=-\frac{\sqrt{3}}{2}\)
Câu 2:
\(\frac{cos\left(a+\frac{\pi}{3}\right)}{sin\left(a+\frac{\pi}{3}\right)}=-\sqrt{3}\Leftrightarrow cos\left(a+\frac{\pi}{3}\right)=-\sqrt{3}sin\left(a+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(a+\frac{\pi}{3}\right)+\sqrt{3}sin\left(a+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow cosa.cos\frac{\pi}{3}-sina.sin\frac{\pi}{3}+\sqrt{3}sina.cos\frac{\pi}{3}+\sqrt{3}cosa.sin\frac{\pi}{3}=0\)
\(\Leftrightarrow\frac{1}{2}cosa-\frac{\sqrt{3}}{2}sina+\frac{\sqrt{3}}{2}sina+\frac{3}{2}cosa=0\)
\(\Leftrightarrow2cosa=0\Rightarrow cosa=0\)
\(\Rightarrow P=sin\left(\pi+\frac{\pi}{6}\right)+0=-sin\frac{\pi}{6}=-\frac{1}{2}\)