Question

证明：sin^2α+cos^2α=1

利用上述规律解决下列问题：
①求sin^2 1°+sin^2 2°+sin^2 3°+...+sin^2 89°
②若sinα+cosα=3/2，求sinα乘cosα的值
谢谢啦
证明：sin^2α+cos^2α=1 怎么证明，初三刚学到正切，正弦和余弦

Answer 1

sin²1°+sin²2°+sin²3°……sin²89° 共89项，中间项 (89°+1°)/2=45°,除掉sin²45°，有88/2=44对
=sin²(90°-89°)+sin²(90°-88°)+sin²(90°-87°)+.......sin²(90°-46°)+sin²(45°)+sin²46+........+sin²89°
=cos²89°+cos²88°+cos²87°+.....+cos²46°+sin²45°+sin²46°+........+sin²89°
=cos²89°+sin²89°+cos²88°+sin²88°+cos²87°+sin²87°+.......+cos²46°+sin²46°+sin²45°
=1+1+1+.......+1+sin²45°
=44+(√2/2)²
=44+1/2
=44又1/2
第二问平方1+2sinα*cosα=9/4 即sinα*cosα=5/8

Answer 2

sin^2α+cos^2α=1
在圆x^2+y^2=r^2中
sina=y/r,cosa=x/r
易证sin^2α+cos^2α=1
sin^2 1°+sin^2 2°+sin^2 3°+...+sin^2 89°
＝sin^2 1°+sin^2 2°+sin^2 3°+.sin^2 45°＋cos^2 44°..+cos^2 1°
＝44＋1/2
sinα+cosα=3/2
两边平方得
(sinα+cosα)^2=(3/2)^2
1+2sinacosa=9/4
sinacosa=5/8

Answer 3

①sin^2 89°=cos^2 1°
sin^2 88°=cos^2 2°
所以原代数式即是(sin^2 1°+cos^2 1°)+(sin^2 2°+cos^2 2°)+...+(sin^2 44°+cos^2 44°)+sin^2 45°=44.5

②sin^2 α+cos^2 α+2 sinα乘cosα=(sinα+cosα)^2=9/4
sinα乘cosα=(9/4-1)/2=5/8

Answer 4

（1）sin²44º+sin²46º=1
sin²1º+sin²2º+sin²3º+····+sin²88º+sin²89º
=44*1+sin²45º。
又
sin²45º+sin²45º=1
2sin²45º=1
sin²45º=1/2=0.5。
所以sin²1º+sin²2º+sin²3º+····+sin²88º+sin²89º
=44.5。
第二问就算了吧，没推出来

Answer 5

1
sin^2α/（1 cotα) cos^2α/（1 tanα)
=sin^2α·sinα/（sinα cosα) cos^2α·cosα/（cosα sinα)
=(sin^3α cos^3α)/（sinα cosα)
=(sinα cosα)(sin^2α-sinα·cosα cos^2α)/（sinα cosα)
=sin^2α-sinα·cosα cos^2α
=1-sinαcosα

2
tanαsina/(tanα-sina)
=tanαsina/(sinα/cosα-sina)
=(sinα/cosα)/(1/cosα-1)
=sinα/(1-cosα)
=1/tan(α/2);
(tanα sinα)/tanαsinα
=((sinα/cosα) sinα)/tanαsinα
=((1/cosα) 1)/tanα
=((1/cosα) 1)·cosα/tanα·cosα
=(1 cosα)/sinα
=1/tan(α/2);
∴左=右;
tanαsina/(tanα-sina)=(tanα sinα)/tanαsinα

3
(1-sin^4α-cos^4α)/(1-sina^6α-cos^6α)
=[1-(sin^2α cos^2α)^2 2sin^2α·cos^2α]/[1-(sin^2α cos^2α)(sin^4α-sin^2α·cos^2α cos^4α)]
=[1-1 2sin^2α·cos^2α]/[1-1*(sin^4α 2sin^2α·cos^2α cos^4α-3sin^2α·cos^2α)]
=2sin^2α·cos^2α/[1-(sin^2α cos^2α)^2 3sin^2α·cos^2α]
=2sin^2α·cos^2α/[1-1 3sin^2α·cos^2α]
=2/3
看看怎么样？