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\(\left(sina-cosa\right)^2=2\Leftrightarrow sin^2a+cos^2a-2sina.cosa=2\)
\(\Leftrightarrow1-sin2a=2\Rightarrow sin2a=-1\)
\(\left(sina+cosa\right)^2=2\Leftrightarrow sin^2a+cos^2a+2sina.cosa=2\)
\(\Leftrightarrow1+sin2a=2\Rightarrow sin2a=1\)
\(\frac{3\pi}{2}< a< 2\pi\Rightarrow cosa>0\Rightarrow cosa=\sqrt{1-sin^2a}=\frac{1}{2}\)
\(\Rightarrow cos\left(a+\frac{\pi}{3}\right)=cosa.cos\frac{\pi}{3}-sina.sin\frac{\pi}{3}\)
\(=\frac{1}{2}.\frac{1}{2}-\left(-\frac{\sqrt{3}}{2}\right).\left(\frac{\sqrt{3}}{2}\right)=...\)
a/ \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos\alpha=\pm\frac{\sqrt{5}}{3}\)
\(\alpha\in\) góc phần tư thứ 3=> \(\cos\alpha=-\frac{\sqrt{5}}{3}\)
\(\tan\left(\alpha+\pi\right)=\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{2}{\sqrt{5}}\)
b/ Nếu bạn học công thức này rồi thì áp dụng vô
\(\sin\left(\alpha+\frac{3\pi}{2}\right)=\sin\alpha.\cos\frac{3\pi}{2}+\cos\alpha.\sin\frac{3\pi}{2}\)
\(\cos\frac{3\pi}{2}=\cos\left(\pi+\frac{\pi}{2}\right)=-\cos\frac{\pi}{2}=0\)
\(\sin\frac{3\pi}{2}=-\sin\frac{\pi}{2}=-1\)
\(\Rightarrow\sin\left(\alpha+\frac{3\pi}{2}\right)=\frac{\sqrt{5}}{3}\)
a)\(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\sin^2\alpha=1-\cos^2\alpha\)
\(\Rightarrow1-2^2=-3\) \(\Rightarrow\cos=-\sqrt{3}\left(0< \alpha< \dfrac{\pi}{2}\right)\)
b) \(\tan\alpha\times\cot\alpha=1\Rightarrow\tan\alpha=\dfrac{1}{\cot\alpha}\Rightarrow\tan=\dfrac{1}{4}\)
a)Do \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
\(cos\alpha=2sin\alpha\)(1)
Nếu \(sin\alpha=0\Rightarrow cos\alpha\) (vô lý).
Vì vậy \(sin\alpha\ne0\) . Từ (1) \(\Rightarrow\dfrac{cos\alpha}{sin\alpha}=2\)\(\Leftrightarrow cot\alpha=2\).
Suy ra: \(tan\alpha=\dfrac{1}{2}\).
\(sin\alpha=\sqrt{\dfrac{1}{1+cot^2\alpha}}=\dfrac{1}{\sqrt{3}}\).
\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{\dfrac{2}{3}}\).
a) Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha< 0;cot\alpha>0;tan\alpha>0\).
Vì vậy: \(sin\alpha=-\sqrt{1-cos^2\alpha}=\dfrac{-\sqrt{15}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\sqrt{15}}{4}:\dfrac{-1}{4}=\sqrt{15}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{\sqrt{15}}\).
b) Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(cos\alpha< 0;tan\alpha< 0;cot\alpha< 0\).
\(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{5}}{3}\);
\(tan\alpha=\dfrac{2}{3}:\dfrac{-\sqrt{5}}{3}=\dfrac{-2}{\sqrt{5}}\); \(cot\alpha=1:tan\alpha=\dfrac{-\sqrt{5}}{2}\).
a) Tồn tại do \(\left|sin\alpha\right|\le1\).
b) Tồn tại do \(\left|cos\alpha\right|\le1\).
c) Tồn tại do\(\left|sin\alpha\right|\le1\).
d) Không tồn tại do \(\left|cos\alpha\right|\ge1\).
e) Không tồn tại do \(\left|sin\alpha\right|\ge1\).
g) Không tồn tại do \(\left|cos\alpha\right|\ge1\).
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
cotα = \(\frac{1}{3}\) \(\Leftrightarrow\frac{cos\alpha}{\sin\alpha}=\frac{1}{3}\Leftrightarrow\sin\alpha=3\cos\alpha\)
cotα =\(\frac{1}{\tan\alpha}=\frac{1}{3}\Rightarrow\tan\alpha=3\)
T = \(\frac{2016}{\sin^2\alpha-\sin\alpha\cos\alpha-\cos^2\alpha}=\frac{2016}{9\cos^2\alpha-3\cos^2\alpha-\cos^2\alpha}\) \(=\frac{2016}{5\cos^2\alpha}=\frac{2016}{5}\times\frac{1}{\cos^2\alpha}=\frac{2016}{5}\times\left(1+\tan^2\alpha\right)\) \(=\frac{2016}{5}\left(1+9\right)=4032\)
\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)
a/ \(\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{3-1}{3+1}\)
b/ \(\frac{2sina+3cosa}{3sina-5cosa}=\frac{3tana+3}{3tana-5}=\frac{3.3+3}{3.3-5}\)
c/ \(\frac{1+2cos^2a}{1-cos^2a-cos^2a}=\frac{1+2cos^2a}{1-2cos^2a}=\frac{1+2.\frac{1}{10}}{1-2.\frac{1}{10}}\)
d/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)