Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=4\left(1-sin^2x\right)-6sin^2a=4-10sin^2a=4-10.\left(\frac{1}{5}\right)^2=...\)
\(tana+cota=3\Leftrightarrow\frac{sina}{cosa}+\frac{cosa}{sina}=3\Leftrightarrow\frac{sin^2a+cos^2a}{sina.cosa}=3\)
\(\Leftrightarrow\frac{1}{sina.cosa}=3\Leftrightarrow sina.cosa=\frac{1}{3}\)
\(C=cot^2a-cos^2a.cot^2a=cot^2a\left(1-cos^2a\right)=cot^2a.sin^2a\)
\(=\frac{cos^2a}{sin^2a}.sin^2a=cos^2a=1-sin^2a=1-\left(\frac{3}{4}\right)^2=...\)
Bài 1:
Ta có: \(\tan\alpha+\cot\alpha=3\)
\(\Leftrightarrow\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}=3\)
\(\Leftrightarrow\dfrac{\sin^2\alpha}{\cos\alpha.\sin\alpha}+\dfrac{\cos^2\alpha}{\sin\alpha.\cos\alpha}=3\)
\(\Leftrightarrow\dfrac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha.\cos\alpha}=3\)
\(\Leftrightarrow\dfrac{1}{\sin\alpha.\cos\alpha}=3\)
\(\Leftrightarrow\dfrac{1}{\sin\alpha.\cos\alpha}=\dfrac{3\left(\sin\alpha.\cos\alpha\right)}{\sin\alpha.\cos\alpha}\)
\(\Leftrightarrow1=3\left(\sin\alpha.\cos\alpha\right)\)
\(\Leftrightarrow\sin\alpha.\cos\alpha=\dfrac{1}{3}\)
Vậy \(\tan\alpha+\cot\alpha=3\) thì \(\sin\alpha.\cos\alpha=\dfrac{1}{3}\)
A B C H Chứng minh
\(AB^2=AC^2+BC^2-2AC.BC.\cos C\)
Kẻ \(AH\perp BC\)
Ta có: \(VP=\)\(AC^2+BC^2-2AC.BC.\cos C\)
\(=AC^2+BC^2-2AC.BC.\dfrac{CH}{AC}\)
\(=AC^2+BC^2-2BC.CH\)
\(=AH^2+HC^2+BH^2+HC^2+2BH.CH-2BH.CH-2CH^2\)
\(=AH^2+BH^2\)
\(=AB^2=VT\)
Vậy đẳng thức được chứng minh.
\(\sin\alpha=\frac{5}{13}\) => \(\sin^2\alpha=\frac{25}{169}\)
Mà \(\cos^2\alpha+\sin^2\alpha=1\) nên \(\cos^2\alpha=\frac{144}{169}\) => \(\cos\alpha=\frac{12}{13}\)
Ta có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\) => \(\tan\alpha=\frac{5}{13}:\frac{12}{13}=\frac{5}{12}\)
Lại có \(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}\) => \(\cot\alpha=\frac{12}{13}:\frac{5}{13}=\frac{12}{5}\)
Chúc bạn làm bài tốt
Ta có \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\cos^2\alpha=1-\sin^2\alpha\)
\(\Rightarrow\cos^2\alpha=1-\frac{25}{169}=\frac{144}{169}\Rightarrow\cos\alpha=\frac{12}{13}\)
Ta lại có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{5}{13}.\frac{13}{12}=\frac{5}{12}\Rightarrow\cot\alpha=\frac{12}{5}\)
Bài 2:
a: \(\sin a=\sqrt{1-\left(\dfrac{4}{5}\right)^2}=\dfrac{3}{5}\)
\(P=4\cdot\sin^2a-6\cdot\cos^2a\)
\(=4\cdot\dfrac{9}{25}-6\cdot\dfrac{16}{25}\)
\(=\dfrac{36-64}{25}=\dfrac{-28}{25}\)
b: \(A=\sin^6a+\cos^6a+3\cdot\sin^2a\cdot\cos^2a\)
\(=\left(\sin^2a+\cos^2a\right)^3-3\sin^2a\cdot\cos^2a\cdot\left(\sin^2a+\cos^2a\right)+3\cdot\sin^2a\cdot\cos^2a\)
\(=1-3\sin^2a\cdot\cos^2a+3\sin^2a\cdot\cos^2a\)
=1
Bài 2:
\(\cos a=\sqrt{1-\left(\dfrac{7}{25}\right)^2}=\dfrac{24}{25}\)
\(\tan a=\dfrac{7}{25}:\dfrac{24}{25}=\dfrac{7}{24}\)
\(\cot a=\dfrac{24}{7}\)
1.\(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\cos^2\alpha=1-\sin^2\alpha=1-\left(\frac{3}{5}\right)^2=1-\frac{9}{25}=\frac{16}{25}\)
\(\Rightarrow\cos\alpha=\frac{4}{5}\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}\)
\(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\)
2.\(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\sin^2\alpha=1-\cos^2\alpha=1-\left(0,8\right)^2=1-0,64=0,36\)
\(\Rightarrow\sin\alpha=0,6\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{0,6}{0,8}=\frac{3}{4}\)
\(\tan\alpha.\cot\alpha=1\Rightarrow\cot\alpha=\frac{1}{\tan\alpha}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)
Câu1. Ta có\(\sin^2\alpha+\cos^2\alpha=1\Leftrightarrow\sin^2\alpha=1-\cos^2\alpha=1-\left(\frac{1}{4}\right)^2\)
\(=\frac{15}{16}\Rightarrow\sin\alpha=\frac{\sqrt{15}}{4}\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\sqrt{15}}{4}:\frac{1}{4}=\sqrt{15}\)\(=4\sin\alpha\)
Câu2.
Ta có: \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}=5\Leftrightarrow\cos\alpha+\sin\alpha=5\cos\alpha-5\sin\alpha\)
\(\Leftrightarrow4\cos\alpha=6\sin\alpha\Leftrightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{2}{3}\)
\(\Rightarrow\tan\alpha=\frac{2}{3}\)