Given sinA = (root 2) / 10 and sinB = (root 10) / 10 where A and B are acute angles.
Find A+2B without using a calculatior
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F4 Trigo 問題
2009-02-03 2:38 am
回答 (2)
2009-02-03 3:17 am
✔ 最佳答案
Since sinA = (root 2) / 10 and sinB = (root 10) / 10So cosA=(root 98)/10 and cosB =(root 90)/10
sin(A+2B)
=sin(A)cos(2B)+sin(2B)cos(A)
=sin(A)(2cos^2 B-1)+2cos(A)sin(B)cos(B)
=(root 2) / 10 * 4/5 + 2[(root 98)/10][(root 10) / 10][(root 90)/10]
=sqrt(2)/2
sin(A+2B)=sqrt(2)/2
=>A+2B=45//
2009-02-03 3:23 am
sinA=√2/10,cosB=√10/10
cosA=√98/10,sinB=√90/10
sin2B=2sinBcosB=60/100=3/5
cos2B=4/5
sin(A+2B)
=sinAcos2B+cosAsin2B
=(√2/10)(4/5)+(√98/10)(3/5)
=25√2/50
=√2/2
So A+2B=45
2009-02-02 19:26:15 補充：
Since A and B are very small, so A+2B should still be less than 90
cosA=√98/10,sinB=√90/10
sin2B=2sinBcosB=60/100=3/5
cos2B=4/5
sin(A+2B)
=sinAcos2B+cosAsin2B
=(√2/10)(4/5)+(√98/10)(3/5)
=25√2/50
=√2/2
So A+2B=45
2009-02-02 19:26:15 補充：
Since A and B are very small, so A+2B should still be less than 90
收錄日期: 2021-04-26 14:04:07
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