Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 1 : ta có : \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\left(0,6\right)^2=\dfrac{16}{25}\)
\(\Rightarrow cosa=\pm\dfrac{4}{5}\)
\(\Rightarrow tanx=\dfrac{sinx}{cosx}=\pm\dfrac{3}{4}\) \(\Rightarrow cotx=\dfrac{1}{tanx}=\pm\dfrac{4}{3}\)
bài 2)
ý 1 : a) ta có : \(\dfrac{1}{cos^2a}=\dfrac{sin^2a+cos^2a}{cos^2a}=tan^2a+1\left(đpcm\right)\)
b) ta có : \(\dfrac{1}{sin^2a}=\dfrac{sin^2a+cos^2a}{sin^2a}=1+cot^2a\left(đpcm\right)\)
c) \(cos^4a-sin^4a=\left(sin^2a+cos^2a\right)\left(cos^2a-sin^2a\right)\)
\(=cos^2a-sin^2a=2cos^2a-cos^2a-sin^2a=2cos^2a-1\left(đpcm\right)\)
ý 2 :
ta có : \(tana=2\Rightarrow cota=\dfrac{1}{2}\)
ta có : \(tan^2a+1=\dfrac{1}{cos^2a}\Leftrightarrow cos^2a=\dfrac{1}{tan^2a+1}=\dfrac{1}{5}\)
\(\Rightarrow cosa=\pm\dfrac{1}{\sqrt{5}}\Rightarrow sin^2a=1-cos^2a=\dfrac{4}{5}\) \(\Rightarrow sina=\pm\dfrac{2}{\sqrt{5}}\)
vậy ............................................................................
bài 3 bạn tự luyện tập như bài 2 cho quen nha :)
a.\(1-\sin^2\alpha=\cos^2\alpha\)
b.\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)
c.\(\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=1-\cos^2\alpha=\sin^2\alpha\)
d.\(1+\sin^2\alpha+\cos^2\alpha=1+1=2\)
e.\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha=\tan^2\alpha\left(1-\sin^2\alpha\right)=\tan^2\alpha.\cos^2\alpha=\sin^2\alpha\)
g.\(\cos^2\alpha+\cos^2\alpha.\tan^2\alpha=\cos^2\alpha\left(1+\tan^2\alpha\right)=\cos^2\alpha.\frac{1}{\cos^2\alpha}=1\)
\(=\frac{sin^2x}{cos^2x}\left(cos^2x+sin^2x-1+cos^2x\right)+cos^2x\)
\(=\frac{sin^2x}{cos^2x}\left(1-1+cos^2x\right)+cos^2x\)
\(=\frac{sin^2x.cos^2x}{cos^2x}+cos^2x=sin^2x+cos^2x=1\)
\(\cos^220^o+\cos^240^o+\cos^250^o+\cos^270^o\)
\(=\cos^220^o+\cos^240^o+\sin^250^o+\sin^220^o\)
\(=\left(\cos^220^o+\sin^220^o\right)+\left(\cos^240^o+\sin^240^o\right)\)
\(=1+1\)
\(=2\)
a)
\(A=sin^2\left(10\right)+sin^2\left(20\right)+...+sin^2\left(70\right)+sin^2\left(80\right)\\ A=sin^2\left(10\right)+sin^2\left(20\right)+...+sin^2\left(40\right)+cos^2\left(40\right)+...+cos^2\left(20\right)+cos^2\left(10\right)\\ A=\left(sin^2\left(10\right)+cos^2\left(10\right)\right)+\left(sin^2\left(20\right)+cos^2\left(20\right)\right)+....+\left(sin^2\left(40\right)+cos\left(40\right)\right)\\ A=1+1+1+1+1=4\)câu b tương tự
\(A=sin^210^o+sin^220^o+sin^230^o+sin^240^o+sin^250^o+sin^260^o+sin^270^o+sin^280^o\)
\(A=cos^280^o+cos^270^o+cos^260^o+cos^250^o+sin^250^o+sin^260^o+sin^270^o+sin^280^o\)
\(A=\left(sin^250^o+cos^250^o\right)+\left(sin^260^o+cos^260^o\right)+\left(sin^270^o+cos^270^o\right)+\left(sin^280^o+cos^280^o\right)\)
\(A=1+1+1+1\)
\(A=4\)