sin^2 2+ sin^2 4+ sin^2 6+...+sin^2 84+ sin^2 86+ sin^2 88=?
ans:22
請列明步驟,thank you!!
數學問題sin cos
2011-12-25 9:20 am
回答 (2)
2011-12-25 8:24 pm
✔ 最佳答案
sin²2°+sin²4°+sin²6°+...+sin²84°+sin²86°+sin²88°=sin²2°+sin²4°+sin²6°+...+sin²44° +sin²46°+...+sin²84°+sin²86°+sin²88°=sin²2°+sin²4°+sin²6°+...+sin²44° +sin²(90°-44°)+...+sin²(90°-6°)+sin²(90°-4°)+sin²(90°-2°)=sin²2°+sin²4°+sin²6°+...+sin²44° +cos²44°+...cos²6°+cos²4°+cos²2°=sin²2°+cos²2°+sin²4°+cos²4°+sin²6°+cos²6°+...+sin²44°+cos²44°=1+1+1+...+1 <-[No. of ‘1’is 44/2=22] =22 *** Remember: (a). sin(90°-x°)=cosx°(b). sin²x°+cos²x°=1
參考: Hope I Can Help You ! ^_^ ( From Me )
2011-12-25 10:45 am
因為sinx=cos(90-x)
而sin^2x+cos^2x=1
所以sin^2 2+ sin^2 4+ sin^2 6+...+sin^2 84+ sin^2 86+ sin^2 88
=cos^2 88+cos^2 86 +cos^2 84 +...+cos^2 46 +sin^2 46+...+sin^2 84+ sin^2 86+ sin^2 88
=22
希望你明我講咩啦 ...我知我表逹得唔好
而sin^2x+cos^2x=1
所以sin^2 2+ sin^2 4+ sin^2 6+...+sin^2 84+ sin^2 86+ sin^2 88
=cos^2 88+cos^2 86 +cos^2 84 +...+cos^2 46 +sin^2 46+...+sin^2 84+ sin^2 86+ sin^2 88
=22
希望你明我講咩啦 ...我知我表逹得唔好
參考: lostgor
收錄日期: 2021-04-13 18:26:11
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