evaluate cos(pi/11)+cos(3pi/11)+cos(5pi/11)+cos(7pi/11)+cos(9pi/11) without using calculators.
pi=π
AMATHS(sum and product of trigonometric functions)
2008-02-12 6:43 am
回答 (3)
2008-02-12 7:05 am
✔ 最佳答案
求 cosπ/11+cos3π/11+cos5π/11+cos7π/11+cos9π/11 let y=cosπ/11+cos3π/11+cos5π/11+cos9π/11
consider 2ysinπ/11
2ysinπ/11
=sin2π/11+sin4π/11-sin2π/11+sin6π/11-sin4π/11+sin8π/11-sin6π/11sin10π/11-sin8π/11 【sinAcosB=1/2[sin(A+B)+sin(A-B)]】
=sin10π/11
2ysinπ/11=sin10π/11
2ysinπ/11=sin(π-π/11)
2ysinπ/11=sinπ/11
2y=1
y=1/2
thus
cosπ/11+cos3π/11+cos5π/11+cos9π/11=1/2
2008-02-27 5:47 am
good!
2008-02-12 7:11 am
Apply the trigonometric function formula ：sinAcosB=1/2[sin(A+B)+sin(A-B)]and consider
2* sinπ/11*[cosπ/11 + cos3π/11 + cos5π/11 + cos7π/11 + cos9π/11]
= [sin2π/11] + [sin4π/11-sin2π/11] + [sin6π/11-sin4π/11] + [sin8π/11-sin6π/11] + [sin10π/11-sin8π/11]
= sin10π/11
=sin[π-(π/11)]
= sinπ/11
Therefore,
cosπ/11+cos3π/11+cos5π/11+cos9π/11=1/2
2008-02-11 23:12:59 補充：
更正Therefore,cosπ/11 cos3π/11 cos5π/11 cos7π/11 cos9π/11=1/2
2008-02-20 13:07:49 補充：
To 上面回答者 ：myisland8132答案提及 let y=cosπ/11 cos3π/11 cos5π/11 cos9π/11其實應該係 let y=cosπ/11 cos3π/11 cos5π/11 cos7π/11 cos9π/11 先至0岩
2* sinπ/11*[cosπ/11 + cos3π/11 + cos5π/11 + cos7π/11 + cos9π/11]
= [sin2π/11] + [sin4π/11-sin2π/11] + [sin6π/11-sin4π/11] + [sin8π/11-sin6π/11] + [sin10π/11-sin8π/11]
= sin10π/11
=sin[π-(π/11)]
= sinπ/11
Therefore,
cosπ/11+cos3π/11+cos5π/11+cos9π/11=1/2
2008-02-11 23:12:59 補充：
更正Therefore,cosπ/11 cos3π/11 cos5π/11 cos7π/11 cos9π/11=1/2
2008-02-20 13:07:49 補充：
To 上面回答者 ：myisland8132答案提及 let y=cosπ/11 cos3π/11 cos5π/11 cos9π/11其實應該係 let y=cosπ/11 cos3π/11 cos5π/11 cos7π/11 cos9π/11 先至0岩
收錄日期: 2021-04-13 15:07:34
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https://hk.answers.yahoo.com/question/index?qid=20080211000051KK04348