已知tanα=3,求(1)(5sinα-2cosα)/4sinα的值,(2)sin²α+2sinαcosα的值
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已知tanα=3,求(1)(5sinα-2cosα)/4sinα的值,(2)sin²α+2sinαcosα的值
已知tanα=3,求(1)(5sinα-2cosα)/4sinα的值,(2)sin²α+2sinαcosα的值
已知tanα=3,求(1)(5sinα-2cosα)/4sinα的值,(2)sin²α+2sinαcosα的值
(1).(5sinα-2cosα)/4sinα 同时除以cosα得
(5tanα-2)/4tanα=13/12
(2)sin²α+2sinαcosα=(sin²α+2sinαcosα)/1=(sin²α+2sinαcosα)/(sin²α+cos²α) 同时除以cos²α得
(tan²α+2tanα)/(tan²α+1)=(9+6)/(9+1)=15/10=3/2
第二题要注意整式除以1,再把1转换成sin²α+cos²α再解答
(5sinα-2cosα)/(4sinα)
=(5tanα-2)/(4tanα)
=(5×3-2)/(4×3)
=13/12
tanα=sinα/cosα=3
sinα=3cosα
sin²α+cos²α=1
(3cosα)²+cos²α=1
10cos²α=1
c...
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(5sinα-2cosα)/(4sinα)
=(5tanα-2)/(4tanα)
=(5×3-2)/(4×3)
=13/12
tanα=sinα/cosα=3
sinα=3cosα
sin²α+cos²α=1
(3cosα)²+cos²α=1
10cos²α=1
cos²α=1/10
sin²α+2sinαcosα
=(3cosα)²+2(3cosα)cosα
=15cos²α
=15/10
=3/2
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(1)原式=(5tanα-2)/4tanα ..........(分数上下同除cosα)
=(5*3-2)/4*3...................(代入tanα=3)
=13/12
(2)原式=(sin²α+2sinαcosα)/(sin²α+cos²α).............
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(1)原式=(5tanα-2)/4tanα ..........(分数上下同除cosα)
=(5*3-2)/4*3...................(代入tanα=3)
=13/12
(2)原式=(sin²α+2sinαcosα)/(sin²α+cos²α)............(原式整体除以1,即sin²α+cos²α)
=(tan²α+2tanα)/(tan²α+1).........................(分数分数上下同除cos²α)
=(3*3+2*3)/(3*3+1)...................................(代入tanα=3)
=1.5
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