Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sin\alpha=\sqrt{1-\left(\dfrac{20}{29}\right)^2}=\dfrac{21}{29}\)
\(\tan\alpha=\dfrac{21}{20}\)
\(\cot\alpha=\dfrac{20}{21}\)
\(\sin\alpha=\sqrt{1-\dfrac{400}{29^2}}=\dfrac{21}{29}\)
\(\tan\alpha=\dfrac{21}{20}\)
\(\cot\alpha=\dfrac{20}{21}\)
*Ta có \(\cos^2a+\sin^2a=1\)
\(\Rightarrow sina=\sqrt{1-\cos^2a}=\sqrt{1-\left(\dfrac{20}{29}\right)^2}=\dfrac{21}{29}\)
*Ta có \(\tan a=\dfrac{\sin a}{\cos a}=\dfrac{21}{29}:\dfrac{20}{29}=\dfrac{21}{20}\)
*Ta có \(\cot a.\tan a=1\Rightarrow\cot a=\dfrac{1}{\tan a}=\dfrac{20}{21}\)
\(A=\left(sin^212^o+sin^278^o\right)+\left(sin^21^o+sin^289^o\right)+\left(sin^273^o+sin^217^o\right)\)
\(A=\left(sin^290^o\right)+\left(sin^290^o\right)+\left(sin^290^o\right)\)
\(A=1+1+1=3\)
a: \(A=sin^210^0+sin^280^0+cos^220^0+sin^270^0\)
\(=sin^210^0+cos^210^0+sin^270^0+sin^270^0\)
\(=2\cdot sin^270^0+1\)
b: \(=sin^215^0+sin^275^0+sin^235^0+sin^255^0\)
\(=sin^215^0+cos^215^0+sin^235^0+cos^235^0\)
=1+1
=2
\(A=sin^210^0+sin^280^0+cos^220^0+sin^270^0\)
\(=sin^210^0+cos^210^0+sin^270^0+sin^270^0\)
\(=2sin^270^0+1\)
\(B=sin^215^0+sin^275^0+sin^235^0+sin^255^0\)
\(=sin^215^0+cos^215^0+sin^235^0+cos^235^0\)
=1+1
=2