Bài 6:
a: \(M=cos^215^0+cos^225^0+cos^235^0+cos^245^0+cos^255^0+cos^265^0+cos^275^0\)
\(=\left(cos^215^0+cos^275^0\right)+\left(cos^225^0+cos^265^0\right)+\left(cos^235^0+cos^255^0\right)+cos^245^0\)
\(=1+1+1+\frac12=\frac72\)
b: \(N=\sin^210^0-\sin^220^0+\sin^230^0-\sin^240^0-\sin^250^0-\sin^270^0+\sin^280^0\)
\(=\left(\sin^210^0+\sin^280^0\right)-\left(\sin^220^0+\sin^270^0\right)-\left(\sin^240^0+\sin^250^0\right)+\sin^230^0\)
=1-1-1+1/4
=-1+1/4
=-3/4
Bài 7:
\(\sin^2a+cos^2a=1\)
=>\(\sin^2a=1-0,4^2=0,84\)
=>\(\sin a=\sqrt{0,84}=\frac{\sqrt{21}}{5}\)
tan a=\(\frac{\sin a}{cosa}=\frac{\sqrt{21}}{5}:\frac25=\frac{\sqrt{21}}{2}\)
cot a=\(\frac{cosa}{\sin a}=\frac25:\frac{\sqrt{21}}{5}=\frac{2}{\sqrt{21}}\)
Bài 8:
Ta có: ΔABC vuông tại A
=>\(sinB=cosC=\frac{5}{13}\)
Ta có: \(\sin^2B+cos^2B=1\)
=>\(cos^2B=1-\left(\frac{5}{13}\right)^2=1-\frac{25}{169}=\frac{144}{169}=\left(\frac{12}{13}\right)^2\)
=>cos B=12/13
tan B=sin B:cosB
\(=\frac{5}{13}:\frac{12}{13}=\frac{5}{12}\)
