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a: \(=\left(\sin^210^0+\sin^280^0\right)+\left(\sin^220^0+\sin^270^0\right)+\left(\sin^230^0+\sin^260^0\right)+\left(\sin^240^0+\sin^250^0\right)\)
=1+1+1+1
=4
b: \(=\left(\cos^25^0+\cos^285^0\right)+\left(\cos^215^0+\cos^275^0\right)+\left(\cos^225^0+\cos^265^0\right)+\left(\cos^235^0+\cos^255^0\right)+\cos^245^0\)
\(=1+1+1+1+\dfrac{1}{2}=4+\dfrac{1}{2}=\dfrac{9}{2}\)
Ta có sin100=cos800(vì 100+800=900)⇒sin2100=cos2800
sin200=cos700(vì 200+700=900)⇒sin2200=cos2700
Ta có công thức sin2a+cos2a=1
\(P=cos^210^0+cos^220^0+cos^270^0+cos^280^0=cos^210^0+cos^220^0+sin^220^0+sin^210^0=\left(cos^210^0+sin^210^0\right)+\left(cos^220^0+sin^220^0\right)=1+1=2\)
Bài 1 :
\(D=cos^220^0+cos^230^0+cos^240^0+cos^250^0+cos^260^0+cos^270^0\)
\(=\left(cos^220^0+cos^270^0\right)+\left(cos^230^0+cos^260^0\right)+\left(cos^240^0+cos^250^0\right)\)
\(=1+1+1=3\)
Bài 2 :
\(E=sin^25^0+sin^225^0+sin^245^0+sin^265^0+sin^285^0\)
\(=\left(sin^25^0+sin^285^0\right)+\left(sin^225^0+sin^265^0\right)+sin^245^0\)
\(=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)
Bài 3 :
\(F=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
\(=1-3sin^2\alpha.cos^2\alpha+3sin^2a.cos^2\alpha\)
\(=1\)
Có
A=\(\left(sin^215^o+sin^275^o\right)+\left(sin^240^o+sin^250^o\right)+\left(sin^260^o+sin^230^o\right)\)
\(=\left(sin^215^o+cos^215^o\right)+...\)
\(=1\cdot3=3\)
Câu c tương tự mà mk nghĩ đề sai dấu - trước cos^245độ
Nói chung nếu: a+b=90 độ
thì: \(sin^2a+sin^2b=1\)
b) thì áp dụng nếu a+b=90 độ:
\(tana=cotb\) và ngược lại
Mà \(tana\cdot cota=1\)
Nói chung là công thức......
\(\cos^21^o+\cos^289^o=\cos^21^o+\cos^2\left(90^o-1^o\right)=\cos^21^o+\sin^21^o=1\)
\(\cos^22^o+\cos^288^o=\cos^22^o+\cos^2\left(90^o-2^o\right)=\cos^22^o+\sin^22^o=1\)
.......
\(\cos^244^o+\cos^246^o=\cos^244^o+\cos^2\left(90^o-44^o\right)=\cos^244^o+\sin^244^o=1\)
\(\cos^245^o=\left(\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}\)
=> \(A=1.44+\frac{1}{2}-\frac{1}{2}=44\)
\(\cos^25^o+\cos^210^o+....+\cos^285^o\\ =\left(\cos^25^o+\cos^285^o\right)+\left(\cos^210^o+\cos^280^o\right)+...+\left(\cos^240^o+\cos^250^o\right)+\cos^245^o\\ \\ =\left(\cos^25^o+\sin^25^o\right)+\left(\cos^210^o+\sin^210^o\right)+...+\left(\cos^240^o+\sin^240^o\right)+\frac{1}{2}\\ =1+1+...+1+\frac{1}{2}=16+\frac{1}{2}=\frac{33}{2}\)