题目
sin^2 40+cos^2 70+sin40*cos70的值是?
提问时间:2021-04-08
答案
sin^2 40+cos^2 70+sin40*cos70
=sin^2 40+cos^2 (30+40)+sin40*cos(30+40)
=sin^2 40+[cos30cos40-sin30sin40]^2+sin40*[cos30cos40-sin30sin40]
=sin^2 40+[√3/2cos40-1/2sin40]^2+sin40*[√3/2cos40-1/2sin40]
=sin^2 40+[√3/2cos40-1/2sin40]^2+√3/2sin40cos40-1/2sin^2 40
=sin^2 40+3/4cos^2 40+1/4sin^2 40-√3/2sin40cos40+√3/2sin40cos40-1/2sin^2 40
=sin^2 40+1/4sin^2 40-1/2sin^2 40+3/4cos^2 40
=3/4sin^2 40+3/4cos^2 40
=3/4(sin^2 40+cos^2 40)
=3/4
=sin^2 40+cos^2 (30+40)+sin40*cos(30+40)
=sin^2 40+[cos30cos40-sin30sin40]^2+sin40*[cos30cos40-sin30sin40]
=sin^2 40+[√3/2cos40-1/2sin40]^2+sin40*[√3/2cos40-1/2sin40]
=sin^2 40+[√3/2cos40-1/2sin40]^2+√3/2sin40cos40-1/2sin^2 40
=sin^2 40+3/4cos^2 40+1/4sin^2 40-√3/2sin40cos40+√3/2sin40cos40-1/2sin^2 40
=sin^2 40+1/4sin^2 40-1/2sin^2 40+3/4cos^2 40
=3/4sin^2 40+3/4cos^2 40
=3/4(sin^2 40+cos^2 40)
=3/4
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