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Sửa đề
\(A=cos^212+cos^223+cos^234+cos^245+cos^256+cos^267+\)
\(=\left(cos^212+cos^278\right)+\left(cos^223+cos^267\right)+\left(cos^234+cos^256\right)+cos^245\)
\(=\left(cos^212+sin^212\right)+\left(cos^223+sin^223\right)+\left(cos^234+sin^234\right)+cos^245\)
\(=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)

a) sin 40 - cos 50 =0
b) sin230 + sin240 + sin250 + sin260 = 2
c) cos210 - cos220 + cos230 - cos240 - cos250 - cos270 + cos280 = - sin230
\(a.sin40^o-cos50^o=sin40^o-sin40^o=0\)
\(b.sin^230^o+sin^240^o+sin^250^o+sin^260^o=\left(sin^230^0+sin^260^o\right)+\left(sin^240^0+sin^250^o\right)=\left(sin^230^0+cos^230^o\right)+\left(sin^240+cos^240^o\right)=1+1=2\)
\(c.\left(cos^210^o+cos^280^o\right)-\left(cos^220^o+cos^270^0\right)-\left(cos^240^o-cos^250^o\right)+cos^230^o=\left(cos^210^o+sin^210^o\right)-\left(cos^220^o+sin^220^o\right)-\left(cos^240^o+sin^240^0\right)+cos^230^0=1-1-1+\dfrac{3}{4}=-\dfrac{1}{4}\)

=(\(\left(cos^21+cos^289\right)+\left(cos^22+cos^288\right)+...+cos^245\)
=1+1+..+0.5
mà từ 1 đến 89 có 44 cặp
=>=44.5

\(A=\cos^215^o-\cos^225^o+\cos^235^o-\cos^245^o+\cos^255^o-\cos^265^o+\cos^275^o\)
\(A=\sin^275^o-\sin^265^o+\sin^255^o-\sin^245^o+\cos^255^o-\cos^265^o+\cos^275^o\)
\(A=\left(\sin^275^o+\cos^275^o\right)-\left(\sin^265^o+\cos^265^o\right)+\left(\sin^255^o+\cos^255^o\right)-\sin^245^o\)
\(A=1-1+1-\frac{1}{2}\)
\(A=\frac{1}{2}\)
\(tg67-tg67+cos^216\)\(+sin^216-\frac{cotg37}{cotg37}\)=\(0+1-1=0\)